Combination of Resistances
(a) Series Combination
Rerien =R, + R, + R, . . .
Series connection means current is same in all the resistors.
R Ra R R
A B
WW
Let the equivalent resistance between Aand B is equal to R.. Then by definition,
Reuhalert=T ...(1)
Using Kirchhoff's 2e Rule for the loop shown in figure,
V= IR, +IR, +IR, +...+R, ..(2)
From equation (1) and (2),
Re =R +R, +R..R,
(b) Parallel Combination
parallel combination, potential difference will remain same across each resistor.
R
R2
R3
Rn
Here again, Re = ..(1)
I=+, +i, +...+, ...(2)
From equation (1) and (2)
34.
tis fiven that the resistance of the series combination of two resistances is
$. When they are joined in parallel, the total resistance is become P If S e nP,
then find out the minimum possible value of n is
(A) 4 (B) 3 (C) 2 (D) 1
Sol. For two resistances R, and R,
S=R,+ R, (in series), 1+ (in parallel)
According to S = nP, R, +R,
If nis minimum, R, = R, = Rthenn=4
Given that a hollowcylinder of length Land inner radius R, and outer radius
R,, as shown in figure. The material hasesstivity RO.
() Suppose a potential difference source (battery) ls applled between
the ends of the cylnder body and producesa current which is flowin
parallel to the ais. Find the resistance?
(b) If the potential difference source (battery) is applled
ner and outer surfaces so that current flows radially between the in
resistance measured in outward, find the
this casst
Sol. () we observe that when a potential difference is applied between the
two ends of the given cylinderical body, current flows parallel to the
axis of cylinder.
p=resistivity of
material
In this case,the cross-sectional area is A= (R-R),and the resistance
pL
is given by R= P=
A (R-R)
(b) Consider a differential element which is made up of a thin cylinder of
inner radius r and outer radius r+ dr and length L. Its contribution to
the resistance of the system is given by
Direction of flow of
current
A. +8
pdr
dR = Pdl Direction of
2rrL
current flow
Where A = 2xrL is the area normal to the current
flow direction. The total resistance of the system
becomes
Ydr
36.
Consider a material of resistivity r, in a shape of a truncated cone of altitude
; and radii b anda, for the ri ht and the left ends respectively%, s shown in
the figure.
Assuming that thecurrent is distributed uniformly throu hthe cross-section
of the cone, what is the resistance between the two ends.
y-axis
radius
-axis
Sol. y-axis
radius
b
X -axis
dx
Consider a thin disk of radius 'r' at a distance x from the left end of the
figure
shown.
Equation of line:
y= mx +c
r=
b
..()
Now differentiate this equation :
ldr
(b-a)
Ryea = dR, + dR, + dR, +..dR,
37.
dRa ***** **
Combination of disks
{current is same in all disks so they must be connected in series)
ldr
R=Po (Total resistance)
(ab)
Two resistors with temperature coefficient of reistancea, and a,
have
resistances R, and Ro t 0°C. What is the temperature coefficient of the
compound resistor consistingg of the two resistors connected, (a) in series (b)
in parallaL
Sol. In series : At 0°C Ro Roz Ro=Re + Roz
At t°c, Ro(1+ a,t) +Re(1+ a,t) =R,(1+ at)
R(1+ a,t) +Res(1+ at) =(Ro + ResX1+ at)
Ro +Roa,t +Ren +Rea,t =R +Reg +(Rn +Ry Jat
Rera, +Rosa,
Ro + Ro
38.
In Parallel :
At t °C,
R,(1+at) Ro(i+a.t) Reu(l+ ast)
R +Rez 1 1
RRo(l+ at) R(1+a,t)' Re(1+a,t)
R=oRo
Ro +Roa
-
Since at <<1, so by using the Binomial expansion, we get
l-at)*-at)- -at)"-a,t)
Rep
Rga R +Roz
Battery & E.M.F
Battery
Battery is a device which maintains a potential difference across its two terminals.
Electromotive force (o.m.f)
Definition 1: Electromotive force the capacity of the system battery to make the charge
flow.
Definition 2: It is the work done by the battery for the flow of 1coulomb charge from lower
potential terminal to higher potential terminal inside the battery.
Mechanism of Battery
Condition 1:
When battery is not connected to external load.
Foe ol
lo
O+F
Battery Battery
Battery
F =Non-electrostatic =Electric field At full charge condition
generated due to of battery net force on
force developed by
charge separation charge becomes zero
Battery
Initially, non-Electrostatic force F, developed by battery mechanism, come in picture and
starts pulling positive charge on one side and -ve charge on another side of battery. Due to
this, electric field develop in the battery. At some instant.
At this condition battery is fully charged.
When battery is connected to external load 'R' electrons move from negative plate of battery
through wire and reach at positive plate of battery so charge on the plates decreases and
so does the .
(R>)
Due to this, F, again pull some charge to positive plate and negative plate, again = q and
this process going on. Here work done by F, come from chemical energy of battery.
Important Point :
(1) F does not decrease with time.
(2) due to charge separation.
(3) Charges are not created only separated.
(4) Battery is source of energy not charge.
40.
(5) F, depends on chemical energy.
(6) Inside the battery direction of "I" and Eis different.
(7) Electro motive force is measured in volt.
Representation for Battery
(1) Ideal cell : Cell in which there is no heating effect.
VA Vo Lower potential
Higher potential
difference difference
ldeal cell
(2) Non ideal cell :
Internal resistance () : The potential of battery (real
source in a circuit) is not equal to the emf of the cell.
The reason is that charge moving through the material Vae Ve
of battery (electrolyte) of the cell encounters collision
resistance
causing resistance. We call this the internal
of the source.
Case 1: Battery acting as a source (or battery is discharging)
(V, - V) =t - ir V=g - ir
V, - V, V= called terminal voltage.
Charge flow in time 'dt' dq = idt
Energy supply to load (Power)lead ci
sw = (t- r) (idt) Internal power
Power supplied Power dissipation
Aw = tidt - Prdt to load developed
battery
dw =d-ir
dt
Case 2: Battery acting as a load (or battery charging)
V, - V, =c+ ir
cell
The rate at which chemical energy is stored in the r
Thermal power inside the cell = ir
Electrical power input = [i + Pr
= (c + ir)i
= (V, - VJi
41.
For Electrolyte Cell :
ra d(distance between electrode)
r«- (area of electrode)
r«c(concentration)
1
temperature
In the iven electric circuit calculate the followin parts R
(a) Current flow in the circut. D
(b) Power output delvered to load R
(c) Relation between internal resistance r and R so
that the electrc power output (that means power
glven to R) ls maximum.
(d) Find value of maxdmum power output.
Plot graph between output power and resistance load
From graph we see that for a dven power output there exists two val
ues of external resistance, prove that thee product of these resistance
quals P.
What is the efficiency of the cell when t is used to supply madmum
power.
Sol. (a) In the circuit shown if we assume that potential at A is zero then
potential at B is [ - ir. Now since the connecting wires are of zero
resistance.
.: V, = V=0»V= V, =g - ir
Now current through CD is also I (: it is in series with the cell)
: izk-V(e-ir)-0
R R
Current i=
r+R
Note: After learning the concept of series combination we will able to
calculate the current dire ctly.
(b) Power output P=PR =
42.
(c)
dP 2:'R
For maximum power supply dP =0 ’r+R - 2R =0’r= R
dR
Here for maximum power output, outer resistance
should be equal to internal resistance.
(d) Pmas 4r
P
Pman
R
R, R=r R,
(e) Graph between 'P' and R, where maximum power
output at R=r, i.e., Prax
Relation between power output &resistance load
is given as,
'R
P=
(r+R
sR
() Power output P=
(r+R
P( + 2rR+R)=R R
Above quadratic equation in Rhas two roots R,and R, for given values of
[, P and r such that
R,R, =r (product of roots), r= RR,
Power of battery spent =
(r+r Power (output) =
power output 4r
Sx 100
Efficiency= c e l l , x100 = 50%
total power spent by
43.
Q. In the figure given below, find out the current in the wire BD.
50
C B
20V
10V
20
Sol. V=+20 52 V=0
C 4A B
5A
4A
10V
20V 1A $+10V
D A
292
Let at point D, potential = 0 and write the potential of other points, Then
current in wire AD =2= 5A from A to D
Current in wire CB = 20 = 4A from C to F
Current in wire BD =1A from D to B
Kirchhoff'sLaws
Kirchhoff's current law: This law is based on law of conservation
of charge. It state that "The algebraic sum of the currents meeting
at a point of the circuit is zero" or total currents entering a
junction equals total current leaving the junction.
Incoming current = outgoing current
Calculate the value of current I in the circuit shown in figure.
2A
1A
2A
Sol. Apply KCL at A: 2 + 4=, , = 6A 2A
Apply KCL at B: 1, =1+, 6=1+, 1A
I, = 5A
4A
Apply KCL at C: I, +|=2 5A+|= 2 2A
Answer is I = -3A
i.e., current I is 3A and is leaving the
point C.
Find the potential at point A in the ven part of circuit.
22
+10V 12 20
-50V
-3ov
45.
20V
Sol.
22
15V
+10V 10 20V + 22
A soV
19
-30V
Steps for problem solving:
Step 1: Assume outgoing current in all the branches.
Step 2: let potential of point A is V,.
Step Apply KCL at junction "A:
20V
22
10V 1Q 20v*a 22
-50V
12
-30V
Apply KCL at junction 'A'
Outgoing current = incoming current
1,+1, + I, +I, = 0
V, -20-10 Y-15-20 V, -5 +50 V, +300
12 20 22 12
M-0-0
60+ 35-45-60 -10
M.I9]-3.4530
2 2 1 2
13]=-5 »--volt
46.
Kirchhofs voltage law
The algebraic sum of all the potential drop across each element
along a closed loop is zero.
AV =0
"It is based on the law of conservation of energy".
Traversing
Ri direction
WW R
Assuming
clockwise current
Ez
+ IR, + E, + IR, +E, = 0
For "R' and emf we can write as IR+EMF =0
The Rule for determining AV (Potential drop) across a resistor and
For Resistors : a battery
Lower
b
Traversing direction
AV = V, - V,= -R
Lower''
Higher V
Traversing direction
AV =V, - V, = +R
47.
For Batteries :
Lower"V' HigherV
()
Traversing direction
AV =V, - y,= +¬
Higher"V' Lower "V'
Traversing direction
AV =V, - V,= -E
Summary of sign conventions :
-E
(ii) +¬
(ii) -IR
(iv) +IR
Steps to solve numerical problems using KVL:
Step 1: Assign direction of current in the loop/closed circuit (if not given).
Step 2: Choose the traversing direction.
Step 3: Apply KVL in the loop with sign convention given above.
48.
