Current Electricity - 04 Advanced Level - Physics

Amith Pussella
Kirchoff Laws of circuits PHT8011 2024Th 2024-04-04

Kirchoff Law of circuits

(a). First Law. (Current Law)
The algebraic sum of currents in any node in a circuit is zero.
I2 i1
I = 0

If currents flowing into the node is considered as positive, currents flowing I1 i2
out of the node is considered as negative.
I = 0 I3
i3
I1+ I2 + I3 - i1- i2- i3 = 0
I1+ I2 + I3 = i1+ i2+ i3
Algebraic sum of current flowing into the node of circuit is equal to the currents flowing out of the node
of circuit.

(b). Kirchoff 's first law implies the conservation of charge.

Electric current is the amount of charge flow in a unit time. Accordingly, amount of charge flow into the
junction is equal to the amount of charge flowing out of the junction. So charges do not remain in the
junction. Or else there's no creation or destruction of charges from the junction. So kirchoff 's first law
implies the consvervation of charges.

(c). Kirchoff's second law (voltage law)

In any closed circuit, the algebraic sum of currents and resistances to a selected direction is equal to the
algebraic sum of electromotive forces towards the same direction.

IR = E --------------(1)
NOTE
 If current to one direction of the closed circuit is positive, currents to other direction must be taken as
negative. Similarly if e.m.forces are positive for one direction e.m. forces to other direction must be taken
as negative.
 When taking the product of current and resistances during kirchoff's second law, values of external
resistorS as well as the internal resistors must be taken.

(d). Kirchoff's second law implies conservation of energy

E.M. forces of a cell is equal to the energy supplied for a unit charge and IR product denotes the work
done when a unit charge moves from higer potential to lower potential. So second law shows the energy
supplied to a unit charge is equal to the energy emitted by it. This shows the conservation of energy .
IR = E

Multiplying both sides by I,

  I2R = E I --------------(2)
Energy dissipated in resistors per unit time = Energy generated by cells per unit time

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 By multiplying equation (2) by t,
 I2Rt = E I t

Algebraic sum of energies dissipated in resistors = Algebraic sum of energies generated from cell

(e). Applying kirchoff law to a certain circuit

R6
 Consider a certain cell pushes current, mark currents G H
according to the kirchoff's first law for the junctions of 3
(I1 - I2) (I1 - I2)
circuit. Here its not necessary to consider the position I2 R2 R4
D I3
to terminals of cells. C E
(I2 - I3) (I1 - I2 + I3)
 Then kirchoff's second law can be applied by selecting 1 R3 2
R1 R5
relavent closed circular circuit parts. r1 E3 r2
r3
B I1 A F
E1 E2
Applying IR = E for ABCDA circular part (circle (1))
I1r1 + I1R1 + I2R2 + (I2 - I3) R3 + (I2 - I3)r3 = E1 - E2 -----------(1)

Applying IR = E for ADEFA circular part (circle (2))

- (I2 - I3)r3 - (I2 - I3)R3 + I3R4 + (I1 - I2 + I3)R5 + (I1 - I2 + I3)r2 = E3 + E2 -----------(2)

Applying IR = E for CGHEDC circular part (circle (3))

(I1 - I2)R6 - I3R4 - I2R2 = 0 -----------(3)
In questions, values of R1 , R2 , R3 , R4 , R5 , R6 " r1 , r2 , r3 are given so I1 , I2 , I3 can be solving above
equations.

Questions on Kirchoff's law

01. (i) State Kirchoff's first law.
(ii) Explain first law is representing conservation of charge.
(iii)
Writedown the relationship between currents.
(Ans :- I1 + I2 + I4 + I6 = I2 + I5)

(iv) Write down relationship between currents

( Ans :- (a) I1 = I2 + I3 (b) I1 = I2 + I3 + I4)

02. Find unknown current in junctions of (a) and (b) circuits.

(Ans :- (a) I1 = 4 A (b) -1 A)

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03.
Find the current I.
(Ans :- I = 8A)

04. (i) State Kirchoff's second law.

(ii) Explain second law representing conservation of energy.

(iii) Find VB.

(Ans :- VB = 10V)

(iv)
Find current flow in the circuit and potential drop through
R1 and R2 resistors.
(Ans :- I = 1 A V1 = 10 V V2 = 20 V)

05.

Find value of R resistor when 6A current flows through the

circuit. (Ans = 5)

06. Find I2 and I3 current and VA voltage

(I2 = 6 A , I3 = 2 A , VA = 152 V)

07.

Find I1 and I2 currents and potential drops through all resistors

V1, V 2, V 3, V 4, V 5 & V 6.
(Ans :- I1 = 5 A , I2 = 3 A , V1 = 30 V , V2 = 30 V ,
V3 = 60 V , V4 = 6 V , V5 = 9 V , V6 = 15 V)

Find the currents through all the resistors.

08. (Ans :- I1 = 3 A , I2 = 1 A , I1 - I2 = 2 A

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09. Find current flowing through each resistor.

(Ans :- I1 = 2A, I2 = -1A, I1 + I2 = 3A)

10.

Find I1 & I2 currents and the current through 20V cell.

(Ans :- I1 = 2 A , I2 = 5 A , I2 - I1 = 3 A )

11. Find I3 & I4 currents in circuit.

(Ans :- I3 = 8 A , I4 = 18 A)

12.

Find I1 & I2 currents and potential drop through

each resistor
(Ans :- I1 = -6 A , I2 = 4 A , V1 = 30 V ,
V2 = 50 V, V3 = 60 V , V4 = 80 V)

13. State Kirchoff's laws.

(a). Find magnitudes and direction of current through each part in ABCD circuit.
(b). Can potential of points A,B,C,& D be found ? What are them ? E =4 V
r=2 C
(c). Point D is earthed. What's the current flows to earth from that D
point ?
(d). What's the potential of point D ?
^^^^
^^^^

3 4
(e). What are the potentials at A,B,C & D.
^Ans :- (a). 0.2A anticlockwise (b). Cannot be found. (c). 0 E =2 V
(d). 0 V (e). VD = 0 , VB = -2.8 V, VC = -3.0V,VA = - 0.6V) r=1
A B

14. X
^^^^
2
^^^^
2
^^^^

(a). Calculate currents through the cells. 3

A C
(b). What are the potential differences between terminals of cells ?
E =4 V
(c). What's the potential difference between XY? E = 12 V B E =2 V r=2
(d). Which cell supplies electrical energy to the circuit ? r=2 r=1
Y
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(e). At what rate does it provide electrical energy ?
(f). How much energy is released as heat ?
(g). What happens to the remaining energy ?
^Ans :- (a).1.5 A current through A. (b). 9 V, through a " 3V through B , 5V through C " 1 A Athrough B
(c). 6 V (d).Cell A (e). 18 W , 0.5 A current through C
(f). 14 W (g). Remaining energy is used to exceed the e.m. forces of cells with law e.m.f. )

15.
State Kirchoff's laws.

A 2 B 2 C
^^^^ ^^^^
24 V 16 V Find current through 8V cell given in figure and state it's direction.
8V Calculate power dissipation through  resistor. If point E of above
circuit is earthed. What're the potentials of A & B ?
^^^^
^^^^

2 ^1986 A/L&

1
F E ^^^^
8 D

16.
Find I1, I2, I3 currents and potential difference between
e & b of circuit.
^Ans :- I1 = 2 A , I2 = -8 A , I3 = 6 A , Veb = -13 V&

17.
If E = 13V and R = 10, find I1, I2, I3 currents and
readings of ideal A & V.
^Ans :- 8.4 A , 27 V&

^^^^
R
18. State Kirchoff's laws. 3
^^^^
^^^^

The voltmeter in the given circuit has an infinite resis- R1 R2

tance while the cell and the ammeter A have a
6V
neglegibly small internal resistances. When the key S
V
is open, the V reading is 1V and the A reading is


0.1A. When the key S is closed, V reading is 0. Find S A

^^^^ 

the values of R1 and,R2 and R3. ^1988 A/L&

^^^^

5 20 
^Ans :- R2 = 10  , R1 = 2.5  , R3 = 30 &
S R

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19. ^^^^ l A
State Kirchoff's laws. R1 1V
''Kirchoff's first law describes about the
conservation of charges while kirchoff's second

^^^^
3V R2
law describes about the energy conservation''.
Explain this statement. The internal resistances of both
the cells shown in the diagram are negligibly small. l B
When a voltmeter with infinite internal resistance is
connected between AB, it reads 3V and when an
ammeter with a negligibly small internal resistance is
fixed between the terminals AB it reads a current of
2.25A. Determine the values of R1 and R2.
^1990 A/L& ^Ans :- R1 = 2  , R2 = 4 &
20.

Find VAC, VBC, & VAB of this circuit.

^Ans :- VAC = 7.5 V , VBC = 4 V , VAB = 3.5 V )

21.
Find V1 & E2 of the circuit.
^Ans :- V1 = 4 V , E2 = 14 V )

22.
Find readings of A & V of this circuit.
^Ans :- 0.5 A , 2.5 V)

23.
(a). Find the current I of circuit.
(b). Find the potential difference between 8 resistor.
(c). Find potential difference between 10
^Ans :- (a). 12 A (b). 96 V (c). 60 V )

Ideal ammeter of the given circuit shows 2A reading.

24.
(a). Assume XY as a resistor and find
it's value.
(b). If XY is assumed as a cell with
resistance 2 it's electromotive
force.

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 (c). What's the potential difference between Y and X
under conditions of (b).
^Ans :- (a). 5(b). 6 V (c). - 10 V )

25. Find I1, I2 & I3 currents of circuit.

^Ans :- I1 = 10/7 A , I2 = 12/5A, I3 = - 3.8 A )

26.

(a). Find I1,I2 & I3 currents when S switch is opened.

(b). Find I1, I2 & I3 currents when S is closed.
^Ans :- (a). I1 = I2 = 0.2 A )

A
27. (i) Find the current flowing through 100V cell of the circuit.  1
10 ^ ^^^0 
^Ans :- 7.5 A) 50 V ^^^ ^ 50 V

^^^^
(ii) What's the power dissipated from 10 resistor between 10 
A & B points. ^Ans :- 0 )
1 
^^^0  10 ^
^ ^^^
20 
28. B
10 
30  40  ^^^^ 100 V

50  Find the currents through each resistor.

60  10  ^Ans :- 1.65 A through 60 , 0.15 A through 30 
20 V 0.51 A through 50 0.66 A through 40 
2.16 A through 10  1.5 A through 20 )

100 V 50 V

29. In the circuit R = 5 & E = 20V. Find the readings of

ammeter and voltmeter. (Assume that they're ideal.)

^Ans :- I1 = 3.9 A , Vab = 4.3 V )

30.
(i). Find the current through this circuit and find potential of point P
^Ans :- I = 0.25 A , Vp = 0.25 V )

(ii). Draw a graph to show the variation of electric potential

from the point E throughout the circuit.

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31. Three electric equipments A,B & C rated as 6W,6V,2W,0.5A & 27W, 9V are needed to be connected in
parallel with a cell of e.m.f. 10V and internal resistance 0.5
(i). What's the total current that the cell must provide to work them when there're connected as above ?
(ii). Show that the current needed to work all the equipments as rated cannot be given by only one cell.
(iii). What's the minimum number of cells that must be connected in parallel as a solution for the above problem ?
(iv). If suitable resistors are provided for you, show how you connect them for the cell combination stated in (iii).

32. State Kirchoff's laws. Internal resistances of the cells given in the P S
circuit are neglegible. 2V
^^^^
3
(i). Calculate the potential of point B relative to potential A.
(ii). If a voltmeter of internal resistance 100 is connected through A B 4V 6
AB, calculate the voltmeter reading obtained. l ^^^^ O
l
(iii). Is it correct if voltmeter is connected between AB to measure
potential difference between A & B as in above (ii) ? Explain your 6V 9
answer. ^Ans :- (i).1 V (ii).0.92 V (iii). Yes& Q ^^^^ R

33. Electromotive force of x battery in the diagram is 20V and V

internal resistance is 2 and the resistance connected with
y battery is 4. E = 20 V
R x r=2
When A & B are connected to a battery as given in the figure,
^^^^
voltmeter reading is 16V. When another 4 is connected to
AB, voltmeter reading drops to 8V. Find e.m.f. of y battery and 4 y r=1
value of R resistance. Find the power of the batteries when ^^^^
4connected to A & B. Assume that the current through volt- l
l
meter is neglegible.(Ans:-15V,18 ,19W) A B
34. 2005 A/L

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35. 2010 A/L (5) (A)

36. 2012 A/L (9) (A)

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37. 2013 A/L (9) (A)

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38. 2015 A/L (9) (A)

39. 2018 A/L (9) (A)

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 Question on capacitors
2 F 4 F When key K of this circuit is closed, what's the charge of each
01. 2 capacitor.
^^^^
8 F 6 F ^Ans :- 3 c , 12 c , 6 c, 9 c)
3
^^^^
E =4 V
r=1
 
3
k ^^^^
3 F 3 F
4
02. What's the charge of each of the capacitors ? ^^^^
3 F 3 F
^Ans :- 6 c) 3
^^^^
R1 ^^ R 2 E =5 V
^^
^^

r=1
^^

03. G

C2
When K1 is closed and readings of G is zero, show that C2 = ( RR21 ) C1
C1
k
 

04. Find I1,I2 & I3 currents of circuit.

Find the capacity of capacitor.
(Ans :- I1 = 1.25 A , Q = 0.5 C )

05. Find I1,I2,I3,I4,I5 currents & capacitance of the

capacitor in circuit.

^Ans :- I1 = - 0.33 A ,
I2 = - 2A, I3 = - 1.7 A ,
I4 = - 0.33 A , Q = 5 C)

06. 1990 A/L Essay (8)

2
Show that the electrical energy of a capacitor with capacity C is given by 1 Q .
Here Q is the charge stored in the capacitor. 2 C

Battery of the circuit shown above has a neglegible internal resistance.

PHT 8011 13 PHYSICS - AMITH PUSSELLA
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 (i) What is the potential difference between the points a and b when the switch S is kept open ? Which are
of the points a and b is at a higher potential ?
(ii) Find the amount of charges and the energy stored in each capacitor when the switch S is kept open.
(iii) What is the final potential at point b when the switch S is closed ?
(iv) When the switch S is closed, by what amount will the charge and the energy stored in the capacitor
charge ?
(v) After charging the capacitors by keeping the switch S is closed, what will be the final charge in each of
the capacitor ?

07. 2014 A/L Essay Question (9)(A)

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 Question on curent electricity
I (1) I (2) I (3)
0 1 ' Equivalent resistance of the resistor network
connected to circuit shown in figure.
(1). R R
^^^^
(2). 2R
^^^2R

^^^^
^^^^
(3). 3 R 2R R P Q R S T P Q R S T P Q R S T
(4). 4 R
^
2R R
(5). 5 R ^^^^ ^^^^ I (4) I (5)

0 2 ' 0.2A current is maintained from a to b e.m.f of

cell is 2V and internal resistance is zero.
Potential difference between ab,
^1&' zero P Q R S T P Q R S T
^2&' 1 V a l l b
(3). 2 V 06' Battery of the given circuit has a negligible
<

(4). 3 V 0.2 A 10 internal resistance. Power dissipated from 1

(5). 4 V ^^^^ resistor is,
2V
^^^^
4 1
03' A potential difference of 120V is applied between ^^^^
P & Q the potential difference between R & S
^^^^
4
is, 3V
10
P l
^^^^
^^^^ ^^^^

5 ^1&' 1$9 W (2). 4/9 W (3). 1 W

^^^^

10  l R (4). 3 W (5). 9 W
5
07' Two identical batteries of negligible internal
Q l l S
resistance are connected to r external resistor is
^1&' 25 V (2). 50 V (3). 100 V shown in figure (A) & (B). Relationship between
(4). 150 V (5). 200 V the currents through R resistor in (A) & (B)
figures is,
0 4 ' A motor works by obtaining 0.5A from 220V
element and provides 90W output. If all wasted
energy are converted to heat, heat produced dur- i1 i2
<

<

ing 10 minutes is, R R

(1). 20 J (2). 90 J (3). 200 J
^^^^ ^^^^
(A) (B)
(4). 12 000 J (5). 54 000 J
^1&' i1 = 2i2 ^2&' i1 = i2 ^3&' i2 = 2i1
05' ^4&' i1 = 2i2 ^5&' i2 = 2i1
x T
P
08' A heating coil is present in an electrical iron to
Q S heat it. It's observed that on considerable length
I R
of heating coil is burned out. If that region is
^^^^ removed and remaining part of the coil is used to
heat the same iron,
A uniform circular motion wire is connected to a ^1&' It works in normal way.
circuit as shown in the figure. Initially x sliding ^2&' It produces less heat but the lifespan increases.
key is connected to point P. Change in I current ^3&' It works in a short time and burn out again.
when sliding key is moved along the PQRST semi ^4&' A small voltage is generated.
circle is, ^5&' It absorbs a small current.

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15
09' Equivalent resistance between X and Y points of 14' Equivalent resistance between A & B points is,
the given circuit is, 3
3 ^ ^^^
R
^^^^ ^^^ ^

^^^^
6
X l
R 2R 3
R ^^

^^ 3
^^^^ ^^^^ ^^^^

3
^^

^^
^^^

^^^
^
3
R 2R

^
^^^^ ^^^^ l Y 3 6
R
^^^^ A
^^^^ B
^^^^
^1&' 1  ^2&' 2  ^3&' 3 
(1). 5 R (2). 4R (3). 5R / 2 ^4&' 4  ^5&' 6 
(4). 2 R (5). R
15'

^^^^
100 
10' 20
^^ 
^^ 0

12 V
^^ V
^^
2

6V

^^^^
1 V 20  2V
^^ 00
^^
^^ R
^^

A
Internal resistance of all cells in the circuit are
Internal resistance of the 6V cell in the given neglegible. Voltmeter reading is,
circuit is neglegible and reading of V voltmeter is ^1&' 0 V ^2&' 2 V ^3&' 4 V
zero. Reading of ammeter with neglegible ^4&' 6 V ^5&' 10 V
internal resistance is,
^1&' 0 ^2&' 0'05 A ^3&' 0'1 A 16' 12V battery can provide 1A current for 100 hours.
^4&' 0'6 A If the total energy of the battery can be used to
^5&' Data is insufficient for calculation. lift objects, maximum height that 1200 kg
object can be lifted is,
11' Internal resistance of the cell in the given circuit ^1&' 0.12 m ^2&' 1.2 m ^3&'14.4 m
is neglegible and current through cell is 1.0A. ^4&' 144 m ^5&' 360 m
When additional 2 resistor is added to circuit, 17' When two electric bulbs are connected to 120V
current through cell is 3.0A. Value of R is, power supply seperately, 0.82A and 1.66A
^1&'10  ^2&' 8  R
^^^^ currents flow through them. When those bulbs
^3&' 6  ^ 4&' 4  are connected to 240V in series,
^5&' 2  ^1&' Current through first bulb is 1.66A and through
second bulb is 3.32A.
12' When 1.0A current is placed through PQ which ^2&' Current through 1st bulb is 0.83A an through
is a part of circuit, circuit obtains 5W power. If second bulb is 1.66A.
internal resistance of the cell is neglegible, ^3&' Current through both bulbs is 0.83A.
2 E ^4&' Current through both bulbs is 1.66A.
P l > ^^^^ l Q
^5&' Current through both bulbs is 1.11A.
1A
18' Battery of the 6V , 0.2
^1&' 5 V ^2&' 4 V ^3&' 3 V
^4&' 2 V ^5&' 1 V given circuit has 6V R2
<

2
e.m.f and 0.2 2 A ^^^^
^^^^
13' Three equal resistor of 12 are provided. value internal resistance. R2
of resistance that cannot be obtaining by If current through ^^^^
combing one or more is, cellis 2A the
V
^1&' 36  ^2&' 24  ^3&' 6  voltmeter reading
^4&' 4  ^5&' 2  is,
^1&' 6 V (2). 5.8 V (3). 5.6 V
(4). 5.4 V (5). 2.8 V

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19' If electric conductance and heat conductance are 12 V
compared, electrical quatity that is relevant to the 24' Four equal light bulbs
rate of heat flow is, and R resistor are R
^ 1 & ' Drift velocity of the quantity charge carriers connected to 12V
^ 2 & ' Power dissipation battery as shown in the
^ 3 & ' Electrical conductivity figure. In each bulb
^ 4 & ' Electric current "1.5V, 0.50A" is
^ 5 & ' Potential gradient marked. What is the
value of R to light the
20' A heating coil that can be used by applying a bulbs in normal
constant potential difference is made of a metal brightness ?
wire. At which resistance does the power (1). 4/21  (2). 3/4  (3). 4/3 
generated from coil increases, (4). 21/4  (5). 6 
^ 1 & ' When radius of wire is increased. 6V , 2
^ 2 & ' When length of wire is increased. 25' A battery of 6V e.m.f and
^ 3 & ' When a wire made of a metal with high internal resistance 2 is
resistivity. connected to R resistance as R
^ 4 & ' When wire loops are more closer to each other. shown in the figure. At which
^ 5 & ' When temperature of wire is increased. instance does the rate of heat
generation from R is
21' Which of the following statements shows the maximum ?
increase of resistance of copper with (1). R = 1  (2). R = 2  (3). R = 3 
temperature ? (4). R = 4  (5). R = 
V P
^ 1 & ' Electrons start to move randomly at higher
temperatures. 26' Graph of potential difference Q
^ 2 & ' Ability of positive charges to capture electrons is V and current I for two cop-
high at higher temperatures. per wires P and Q is shown 0
I
^ 3 & ' More positive ions are released against the flow in figure.
of electrons at high temperature.
^ 4 & ' Electrons move in large distance among collisions Except which of the following does the two wires
at higher temperature. are identical ?
^ 5 & ' Positive ions form more barriars for electron ^ 1 & ' Radius of cross section of P is half the radius of
motion at higher temperatures. Q.
^ 2 & ' Radius of cross section P is twice the radius of Q
22' P,Q & R are equal ^ 3 & ' Area of cross section of P is twice the cross
resistors in the Q section of Q.
given circuit. Ratio P ^ 4 & ' Length of P is half the length of Q.
R
between. ^ 5 & ' Length of P is twice the length of Q.
Rate of heat generation of P is, 27' Wire of cross sectional area 1.0mm2 carries 3.2A
Rate of heat generation of Q or R current. Mean drift velocity of an electron of
(1). 0.25 (2). 0.5 (3). 1 charge 1.6 x 10-19 is 2.0mms -1. No. of free
(4). 2 (5). 4 electrons in a unit volume of wire is,
(1). 1.0 x1016m-3 (2).1.0 x1019 m-3
22 -3
23' Main reason for the increase of electrical (3). 1.0 x 10 m (4).1.0 x 1025 m-3
resistance of a metal conductor with the increase (5). 1.0 x 1028 m-3
in temperature is,
28' PQRS is square of
^ 1 & ' Increase in drift velocity of electrons.
side length L cut from Q
L
^ 2 & ' Increase in amplitude of the ions in lattice. R
a uniform metal
^ 3 & ' Increase in cross sectional area of the
resistant plate. Low
conductor. L
resistant connections P
^ 4 & ' Increase in length of the conductor. S
are fixed along PS and
^ 5 & ' Decrease in no. of free electrons.
QR sides. Resistance
between PS and QR
sides.

PHT 8011 17 PHYSICS - AMITH PUSSELLA

17
 (1). Proportional to L-2 10 
(2). Proportional to L-1
(3). Independent of L 10  A 10 
(4). Proportional to L y
(5). Proportional to L2 x 10 

29' A meter of resistance 3 and two resistors of (1). 500 mA (2). 250 mA (3). 100 mA
1 and 2 are provided for you. At which (4). 50 mA (5). 0 mA
combination does the meter shows 0.75V
potential difference when 1A current flows 33' Two wires x and y are made of same material
through it ? and they are in same length. But radius of x is
(1). 3 (2). 3
twice the radius of y. When they are connected
2 1 1 in series and a certain currents is passed, ratio
 
of,
.
3 2 Rate of heat generation in x
(3). is,
(4). Rate of heat generation in y
2  3
1 
(1). 1/8 (2). 1/4 (3).1/2
(5). 3
2 1 (4). 4 (5). 8


1 34' Two wires x and y are made of same material

and they are in same length. But radius of x is
2 twice the radius of y. When they are connected
in parallel and a certain currents is passed, ratio
30' Shape of a conducting sample cut from a thin of,
metal plate is shown in the figure. When I Rate of heat generated in x is,
current flows through this conductor, drift Rate of heat generated in y
velocities of electrons in P,Q and R cross
sections are VP, VQ and VR respectively. which (1). 1/8 (2). 1/4 (3). 1/2
of the gives the correct relationship ? (4). 4 (5). 8
P Q R
35' Resistance of a conductor of 300K temperature
I
is 1. If temperature coefficient of resistance of
the material of wire is 1.25 x 10-3K-1, at what
temperature does the resistance of conductor
d d
become 2 ?
(1). 900 K (2). 1000 K (3). 1100 K
(1). VP = VQ = VR (2). VP > VQ > VR (4). 10000 K (5). 11000 K
(3). VP < VQ < VR (4).VP = VR < VQ
(5). VP = VR > VQ
36' Equivalent resistance between x and y of the
31' When N number of resistors of resistance r are given circuit is approximately,
connected parallely the equivalent resistance is x
R. If they are connected in series, equivalent 2 2 2
resistance is, 
(3). N2R 
2

(1). NR (2). R/N

2

2

2
(4). R/ N (5). R/N
y
32' 10V potential difference is applied between x and
(1). 2.23  (2).3.23  (3). 4.78 
y of the circuit given in figure. What is the
(4).6.67  (5). 
readings of ammeter A with internal resistance
zero ?

PHT 8011 18 PHYSICS - AMITH PUSSELLA

18
 Q
37' If the internal resistance 42' 12V battery is connected to R1 and R2 resistor in
of 6V cell shown in 4  8 series as shown in figure. Potential differences
circuit is neglegilbe, between xz,xy and yz measured by voltmeter are
potential difference P R 12.0V, 1.1V and 2.2V respectively. Reason for
between Q and S is, obtaining these values would be,
8 4 12 V
(1).0 V (2). 1 V
(3). 2 V (4).3 V S
(5). 8 V l R1 l R2 l

x y z
2V
38' A cell of 2V with ^1&' Circuit of battery is broken.
neglegible internal R1 R2 ^ 2 & ' Weak connections of terminals of battery.
resistance is connected in ^^^^ ^^^^ ^3&' Resistance of voltmeter is equal to R1.
series with R 1 and R 2 ^ 4 & ' Resistance of voltmeter is very much greater than
resistor of 20 and 80 V R1 and R2.
as shown in the figure. ^ 5 & ' Resistance of voltmeter is very much less than
If a voltmeter of 80 resistance is connected R1 and R2.
across R2 resistor, reading of it is,
(1). 0.67 V (2).1.33 V (3). 1.67 V 43' 3 accumilators of 2V with a negligible internal
(4). 1.71 V (5). 1.86 V resistance and a rheostat with large resistance is
connected as shown to give a variable potential
39' If 4 resistor connected across terminals of a difference between x and y. When sliding
cell, 2A current passes through the circuit. If 2 contact is moved along the resistor, at which
is connected instead of 4, current is 3A. E.m.f. range does the potential difference varies ?
and internal resistance of cell respectively are, 2V 2V
(1). 15 V, 4   (2).12 V , 2 (3).10 C , 1
(4). 8V , zero (5). 6V , zero

40' P and Q in the given circuit are two resistor wires. X

Ratio between their diameters are 2:1 and ratio Y
between resistances are 1:2. Ratio between 2V

Potential difference across P is, (1). 2 V to 4V (2). 0 V to 4 V

Potential difference across R (3). 0 V to 2 V (4).-2V to + 2V
(5). -2V to + 4V
(1). 1/8 (2).1/4 2V
(3). 1/2 (4). 4
44' Internal resistance of a cell of e.m.f 1.5V is 0.5.
(5). 8
P Q When this cell is connected across 2.5 resistor,
the charge flown through any point of this circuit
for one minutes is,
(1). 5.0 C (2). 22.5 C (3). 25 C
6V (4). 30 C (5). 45 C
41 ' Internal resistance of
^^^^ ^^^^

45' Which value of R gives

the given circuit is 6 l 12  R
x 50mV potential difference
negligible. Potential 6 V d.c.
across 1k resistor ?
difference between x 12  l 6 1 k
and y is, y (Consider the internal resistance
of electric supply is negligible)
(1). 1V (2). 2 V (3). 2.5 V
(1). 1.1 K (2). 1.2 K (3). 11 K
(4). 3.0 V (5). 3.5 V
(4). 12 K (5). 110 K

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19
46' Which of the following define the potential
difference between two points of a conductor Answers
which carries current ?
01. 3 02. 5 03. 1 04. 4 05. 3
^ 1 & ' Force needed to move a unit charge between the
two points. 06. 3 07. 1 08. 3 09. 5 10. 3
^ 2 & ' Currents2 x resistance 11. 4 12. 3 13. 5 14. 3 15. 3
^ 3 & ' Ratio of discharge between points to flow
current. 16. 5 17. 5 18. 3 19. 4 20. 1
^ 4 & ' Ratio of rate of discharge of energy between two 21. 5 22. 5 23. 2 24. 4 25. 2
points to charge flow.
^ 5 & ' Ratio of rate of energy discharge between two 26. 5 27. 5 28. 3 29. 3 30. 3

points to flow current. 31. 5 32. 5 33. 2 34. 5 35. 2

36. 2 37. 2 38. 3 39. 2 40. 1

47' When potential difference across metal
conductor is increased, current flow through it 41. 1 42. 5 43. 4 44. 4 45. 3
also increases. Reason is,
46. 5 47. 4 48.1 49. 3 50. 5
^ 1 & ' Decrease of mean time of collision between free
electrons and atoms.
^ 2 & ' Increase in number of charge carries.
^ 3 & ' Increase of speed of free electrons with
temperature.
^ 4 & ' Increase of acceleration of charge carries.
^ 5 & ' Derease of collision of free electrons with atoms
in lattice.
48' Three resistors P,Q and R are connected to a
cell of e.m.f. 60V with neglegible internal
resistance as shown in figure. Current through R
is, 6.0 V

Q
P 2
2 R
2

(1). 1 A (2). 1.5 A (3). 3A

(4). 4.5 A (5). 9 A
Q
2 2
49' In the given circuit, potential
of P relative to R is -2.0V. P R
Potential of Q relative to S is, 4 1
(1). + 1.4 V (2). -1.4 V S
(3). +0.6 V (4). - 0.6 V
(5). zero

50' In the wheatston bridge shown in figure, value

of x to avoid the flow of current through
galvanometer G is,
x
(1). 18 (2). 36 15
72
(3). 48 (4). 54 
G
(5). 72
5 
12

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20
Current electricity - 05
Kirchoff laws on circuits, Principles of galvanometer

(a). Moving Coil Galvanometer

An electro magnetic instrument which gives a deflection proportional to
the current flowing through the coil is called as a galvanometer.

(b). Explanation for the occurance of steady deflection

when a steady current flows through the
galvanometer
 When a direct current is allowed to flow through the coil of
the galvanometer, due to the force couple produced on the
coil, the coil subjects to a deflection. If this force couple is G,
G = BINA -------(1)
B- Magnetic flux density of the field
N- No.of loops in the coil
A- Area of the plane of coil
 Due to the magnetic torque, when the coil rotates, an opposite
torque is created by the hair springs. If that torque is Gl"
G = C -------(2)
C - The torque constant of the sping (rad)
When G = G1 the coil reaches steady state or equilibrium
BINA = C
I = (C 
BNA )
I  
That means when a steady currents flows,a steady deflection occurs. I
(when I is constant  becomes constant)
(c). Features of a moving coil galvanometer
 Two hair springs which are used to keep the aluminium frame with coil in the magnetic field are arranged in
a way that they move in opposite directions to each other. When the coil rotates, one spring unwinds while
the other spring rolls. By doing this, the effect caused by the expansion of spring get cancelled.
 By winding the coil on the aluminium frame, a registance for the rotation of the coil in the magnetic field is
created. From that, after a certain deflection is created by the wire coil, the vibration of it for a small time
around the equilibrium position when it reaches equilibrium position is prevented. That means vibration is
damped quickly.
PHT 8011 21 PHYSICS - AMITH PUSSELLA
21
  A moving coil galvanometer has a linear scale. That means it creates a deflection proportional to the electric
current flowing through the coil. The following two facts are the reason for it.
(1). The first fact is hair springs behave according to Hook's law. Springs create opposite torques which are
proportional to the angle of deflection.
(2). Second fact is that the coil is arranged in a uniform radial magnetic field. Such kind of a field is created by
giving circular poles to the magnetic inside the instrument and by keep a fixed soft-iron cylinder between
poles.

(d). Internal resistance of a galvanometer

The resistance of wire coil etc. of the galvanometer is called the internal resistance of the
galvanometer.'
(e). Explanation for having a deflection proportional to the potential difference between termi-
nals of a galvanometer
 If it is not mentioned specially, galvanometers are considered as instruments which obey the Ohm's law.
V = IR
Therefore V  I -------(1)
Further more I  
  I -------(2)
From (1) and (2), V   
Accordingly it is clear that the deflection of the indicator of ^^^^
the galvanometer is proportional not only to the curent flowing R
> >
through it but also the potential difference between terminals I I
of it. V

(f). Full scale deflection current of a galvanometer

The maximum current that can be sent through to get the maximum deflection without damaging the
galvanometer is called the full scale deflection.
 The potential difference between the terminals of the galvanometer when full scale deflection current flows
through it is the "full scale deflection voltage".

Performance of a galvanometer as an ammeter.

(a). An ammeter is fixed serially with an electrical appliance like a resistor in a circuit to
measure the current flowing it.
Current before fixing the ammeter I= V/R I R

V
Current after fixing the current I1 = r
R+r I R
I1 < I A
(It is considered that the cell does not have a resistance)
V
 Accordingly it is clear that when A is fixed, the current
flowing through the circuit reduces. The reason for this is
due to the internal resistance of A the equivalent resistance
of the circuit increases.
If R >>> r " R + r  R' Then I1  I'

PHT 8011 22 PHYSICS - AMITH PUSSELLA

22
  That means A can be used to measure the current through large resistors relative to the internal resistance
of G accurately.

(b). Ideal ammeter

Ammeters having zero internal resistance are called ideal ammeters. Practically these types of ammeters
can not be produced.
 Practically the internal resistance of an ammeter can not be destroyed completely. But tricks can be used to
reduce the resistance. Resistance can be reduced by fixing a small resistor parallel to A.

(c). Conversion of scale of an ammeter

Let's consider that an ammeter with full scale deflection current I should be R
converted to an ammeter with full scale deflection current 10I. Let R be 10 I I
A
the resistance of the galvanometer.
9I
A shunt resistor is connected parallel to A for the current flowing through the
r
ammeter to be I when a cirrent of 10I flows through the main circuit.
Afte
Since the two resistors are parallel, 5I
r con
versi
9I x r = I x R on
r = R/9 0 10 I

 After it is arranged like this, when a certain reading is 0 I /2

Befo I
showed in the ammeter, the total current flowing in the re co
nver
sion
circuit is ten times of it.

Performance of a galvanometer as a voltmeter V r

(a). A voltmeter is connected between two points (between R A R B

terminals of a resistor/between terminals of a cell) in I I
parallel way to measure the potential difference between
those two points in a circuit.

When a voltmeter with finite internal resistance (r) is connected in series with resistor R, due to the reduc-
tion of equivalent resistance between points A and B, the accurate potential difference.

If the eqivalent resistance is Rl" 1 1 1

1 = +
R R r

If r >>> R

1 1 1
 0 Therefore 
r R1 R R1  R

 If the internal resistance of the voltmeter is very larger than the resistor which the potential difference is to
be measured, there won't be a significant difference in the equivalent resistance when the voltmeter is
connected. Therefore the reading in the voltmeter is close to the real reading.

 Accordingly it is clear that voltmeter can be used to measure the potential difference between the terminals
of small resistors relative to the internal resistance of the voltmeter.

PHT 8011 23 PHYSICS - AMITH PUSSELLA

23
 The above explanation can be done in the following way too...
 The resistance of the voltmeter is parallel to the resistance between two points in the circuit. If the internal
resistance of the voltmeter is relatively small, due to flowing of large current through it, considerable potential
drop occurs due to the voltmeter.
 If the inetrnal resistance of the voltemeter is very large relative to the resistance between two points, a small
part of current out of the current flowing through the two points flows through the voltmeter. Therefore a
potential drop between two points due to the voltmeter does not occur. Therefore the reading in voltmeter is
very close to the real value.

(b). Ideal voltmeter

A voltmeter with infinite internal resistance is called an ideal voltmeter.

 Since an electric current should flow through the voltemeter for its operation, practically voltmeters having
infinitely large internal resistance (ideal voltmeters) can not be created.

(c). Conversion of scale of a voltmeter

Let's consider that a voltmeter having full scale deflection voltage V should be converted to a voltmeter
having full scale deflection voltage 10V. Let the internal resistance of the voltmeter be R and full scale
deflection current be I.

For this potential is divided by connecting a suitable resistor - inseries in a way that the potential drop across
the voltmeter is V and the total potential difference between terminals is 10V'

R
V = IR 9V = I x r I r
V
9V Ir r
= = 9 A V 9V B
V IR R
r = 9R
After
Thus 9R resistor should be connected in series with the voltmeter. 5V the co
nversi
on
0 10 I
 After it is converted as above, when a certain
reading is showed in the voltmeter the total 0.5 V Befo V
potential difference measured between AB is 0 re the
conv
ten times of it. Thus the scale is converted by e rsion
ten times.

PHT 8011 24 PHYSICS - AMITH PUSSELLA

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 Performance of a galvanometer as an ohm meter

a)  A moving coil galvanometer can be arranged to measure the Ohm meter

resistance and such an instrument is called as an ohm meter.
 When a moving coil galvanometer is arranged to use as an Galvanometer
ohm meter, a variable resistor and a cell are connected in ^^^^
series with the coil of the galvanometer. The resistance that is Rc
to be measured is fixed between A and B. (look at the
diagram) Variable resistor
Rv
E - e.m.f of the cell
Rc - resistance of the coil of the galvanometer
Rv - value of variable resistor E r
r - internal resistance of the cell Cell
R - value of resistance that is to be measured R
l l
A B
E = I (R + Rv + Rc + r)
R = E/I - (Rv+ Rc+ r)
Since I  I= k 

E
R = - (Rv + Rc + r)
k

Therefore R  
The magnitude of the resistance to be measured is inversely proportional to the deflection of the coil. Thus
a linear scale is not obtained.

(b). The way that the scale of a galvanometer is calibrated by 

Let Rc = 1 " r = 1E = 6 V and the full scale deflection current of galvanometer be 1A.

E = I(R + Rv + Rc + r)
6 = I(R + Rv + 1 + 1)
6 = I(R + Rv + 2) -------- (1)
 First the terminals A and B are short circuited,that means the resistor R between A-B is removed and A-B
terminals are connected using a wire and then the terminal of variable resistor RV is adjusted till the full scale
deflection is obtained in the galvanometer.
From (1) 6 = 1(0 + Rv + 2)
Rv = 4 
 Now RV is kept constant and when different resistance values are put between A-B, the relevant deflection
can be read from the galvanometer scale.
 Now if the resistance between AB is R,
6 = I(R + 4 + 2)
6
R = -6
I

PHT 8011 25 PHYSICS - AMITH PUSSELLA

25
Now,

When the deflection becomes full scale deflection I = 1A, R = 0

6
When the deflection becomes 3/4 of full scale deflection = 3/4A , R = - 6 = 2
(3/4)

6
When the deflection becomes 1/2 of full scale deflection = 1/2A , R = - 6 = 6
(1/2)

6
When the deflection becomes 1/4 of full scale deflection = 1/4A , R = - 6 = 18
(1/4)

6
When the deflection becomes 1/8 of full scale deflection = 1/8A , R = - 6 = 42
(1/8)

6
When the deflection becomes 1/16 of full scale deflection = 1/16A , R = - 6 = 90
(1/16)

When the deflection becomes zero, I = o, R 

The scale of the above galvanometer which is in A can be calibrated in . The variation of amphere scale
with  scale is given in the diagram below.

6
18 2 R 
42
90
1
/4 1
/2 3
/4 0
1
/8
1
/ 16
0 1A

I (A)
1A

PHT 8011 26 PHYSICS - AMITH PUSSELLA

26
 Multimeter (AVO meter)
 A galvanometer to measured electric currents in different ranges, potential differences in different ranges
and resistances is considered as a multimeter.
Since this is considered as a combination of ammeter, voltmeter and ohm meter this is called as 'AVO' meter.

A multimeter arranged using a moving coil galvanometer consists of number of shunt resistors with different
values which are connected in parallel to the wire coil, number of multiple resistors with different values
which are connected in series to the coil, a variable resistor and a battery. Multimeter operates as an
ammeter when the shunt resistor S in the diagram is used, multimeter operates as a voltmeter when the
variable resistor R and the battery are used.

Using the rotating key in the multimeter the quantity that is expected to be measured from the instrument and
the relevant range of values is selected. When the multimeter is used as an ohm meter, as described above
first the terminals of the instrument are short circuited and the variable resistor is adjusted so that the
indicator comes to the zero mark in the ohm scale. Multimeter is created to measure direct electric currents
and potential differences.

Questions related to galvanometer

01'(i). What is meant by a galvanometer ? Draw a diagram to show the internal structure of a galvanometer and
label it.
(ii). Explain that when a steady current flows through a galvanometer it shows a steady deflection.
(iii). Mention the features of a moving coil galvanometer.
(iv). What is meant by the internal resistance of a galvanometer ?
(v). Explain the occurance of a deflection which is proportional to the potential difference between terminals in
a galvanometer.
(vi). What is meant by full scale deflection current and full scale deflection voltage of a galvanometer ?
(vii). The internal resistance of a galvanometer is 20. The full scale deflection current of it is 0.1A. Label its
scale in ampheres and volts.
(viii). Label the scale in ampheres and volts of G of internal resistance 10 and full scale deflection current 1A.

PHT 8011 27 PHYSICS - AMITH PUSSELLA

27
02'(i). How does an ammeter connect to a circuit to measure the current flowing through a certain part of the
circuit ?
(ii). Explain whether the real current is measured when an ammeter with internal resistance is connected to the
circuit.
(iii). What is meant by an ideal ammeter ?
(iv). (a). Two resistors of 400 and 800 are connected in series to a 6V battery as given in the diagram. What
is the current flowing in the circuit ? 800 
(b). To measure the current flowing through the circuit an ammeter with internal
resistance 10 is connected in series as given in the diagram. What is the
6V 400 
reading of the ammeter ?
Calculate by what percentage the current in the circuit have changed
after connecting the ammeter. 800 

03'(i). Explain how scale conversions of an ammeter is done. A

(ii). A galvanometer of internal resistance r shows full scale deflection for a
current i. To make it an ammeter that can measure 10 i, 50 i, 100 i, 1000 i, 6V
400 
explain how resistors should be connected with it
^Ans(- r , r , r , r
9 49 99 999 resistors in parallel&
(iii). The internal resistance of a galvanometer is 5 and the current that gives maximum deflection is 0.01 A.
In order to increase it up to 10 A, how should you arrange it ?
^Ans (- by connecting 5/999 resistor in parallel&

04' A rheostat of resistance 12 000 (from a to c) is connected to

a direct voltage provider of 240V.
A
(a). If the voltmeter is ideal draw the graph to show the variation
of reading of A and reading of V when the sliding key b is
moved from a to c. If the ammeter is ideal, what will be the
c
reading of A ?
240
12000  b
(b). If the internal resistance of V is 6000, find the readings of
V and A in the following instances.
(i). When point b is at a. a V
(ii). When point b is at a distance which is 1/4th of length of ac.
(iii). When point b is at the exact middle point of ac.
(iv). When point b is at a distance which is 3/4th of length of ac.
(v). When point b is at c.

05' A cell e.m.f 1.4V and internal resistance 2 is connected to a 100 resistor. An ammeter with internal
resistance 4/3 is used to measure the current in the circuit while a voltmeter is used to measure the
potential difference difference across 100 resistor.
(i). Using standard symbols draw the relevant circuit.
(ii). If the reading of ammeter is 0.02A, find the resistance of the voltmeter. What is the potential difference
across the terminals of the cell at that time ?
(iii). If the reading of the voltmeter is 1.10V, how much is the error in the reading ?
^Ans :- (ii). 200  (iii). 0.23 V &

06' As given in the diagram when V is connected across 400 resistor, it shows V
30V reading. When this V is connected across 300 resistor what will be the
reading of it ?
^Ans :- 22.5 V & 300  400 

60 V

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07' 1997 A/L Structured Essay

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PHT 8011 29 PHYSICS - AMITH PUSSELLA

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 ......................................................................................................................................................................

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08' 2002 A/L Structured Essay

PHT 8011 30 PHYSICS - AMITH PUSSELLA

30
(b)

(c)

................................................................................................................................................................................................................

Quarter of the full scale deflection

................................................................................................................................................................................................................

Write the above values of R also in the corresponding boxes in figure 2.

PHT 8011 31 PHYSICS - AMITH PUSSELLA

31
09' Three resistors of resistance R,2R and 3R respectively
are connected with them in series. The potential
difference across 2r is measured using a voltmeter of
resistance 10R. What is the percentage error ?
^Ans :- 11.76 %&

10' 120V simple current supply is connected to a large resistor (X). A voltmeter with resistance 10k which is
connected in parallel to the circuit shows a reading of 4V. What is the value of X ? Give your idea about the
use of a voltmeter other than an ammeter to find the value of a large resistor (X).
(Ans :- X = 290k, the current of the circuit is very low. (4 x 10-4A) By connecting an ammeter to a circuit
with a large resistor, the reading can not be obtained accurately.)

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32
11' A battery with e.m.f 9V and negligble internal resistance is connected to 3k resistor.

To measure the potential difference of a part of the resistor (Between points A and B in the diagram)
(i) a voltmeter with 20k rsistance is used.
(ii) a voltmeter with 1k resistance is used.
(iii) both votmeters are connected across AB. In which of the above instances we can get the,
(a). highest reading ? (b). lowest reading ?
^Ans :- (a) (i) Shows the highest reading (b) (iii) Shows the lowest reading )

12' Find these in relevant to the given circuit.

(i). Potential at A.
(ii). Potential at D.
(iii). reading of the voltmeter connected with 10V battery.
^Ans :- (i). 4 V , (ii). 0.5 V , (iii). 11 V&

13' The direct current voltmeter connected across ab terminals shows

a reading of 5V in the range 5 - V. What is the value of potential
of the battery ? (Ans :- 9V)

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14' 2001 A/L Essay (5)(a)

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