Electricity Answers

Current, Potential Difference, Resistor Networks, Resistance and

Resistivity
1.

Resistance calculations
Evidence of 20 for one arm (1)
1 1
1
=
+
(1)
R 20 20
R = 10 (1)

Comment
This combination used instead of a single 10 resistor [or same
value as before] (1)
because a smaller current flows through each resistor/reduce heating
in any one resistor/average out errors in individual resistors (1)

2
[5]

2.

Statement 1
Statement is false (1)
Wires in series have same current (1)
Use of I = nAe with n and e constant (1)

[The latter two marks are independent]

Statement 2
Statement is true (1)
Resistors in parallel have same p.d. (1)
Use of Power = V2/R leading to R , power (1)

OR as R , I leading to a lower value of VI 3rd mark consequent

on second
[6]

3.

Charge calculation
Q = 20 000 4.0 104 s [substitution]
Q = 8.0 C/A s

Resistance calculation
R=

l
A

(1.7 10 8 )(50m)
=

(1.0 10 3 m 2 )

R = 8.5 104
Formula

(1)

Correct substitution
Answer
(1)

(1)
3

Potential difference calculation

V = IR
= (20 000 A) (85 105 ) [or their value]
= 17 V [Allow full e.c.f]
(1)
Explanation
For the tree: R or p is larger

(1)

(1)

2
1
[8]

4.

Diagram
Labelled wire and a supply (1)
Ammeter in series and voltmeter in parallel (1)
OR
Labelled wire with no supply (1)
Ohmmeter across wire (1)

Readings
Current and potential difference OR resistance ( consistent with diagram) (1)
Length of wire (1)
Diameter of wire (1)

Use of readings
R = V/I OR = RA/l (1)
Awareness that A is crosssectional area (may be seen above and credited here) (1)
Repetition of calculation OR graphical method (1)

Precaution
Any two from:
Readings of diameter at various places /different orientations
Contact errors
Zeroing instruments
Wire straight when measuring length
Wire not heating up / temperature kept constant (1) (1)

2
[10]

5.

(a)

Io and Jupiter: Time taken for electrons to reach Jupiter

t = s/ = (4.2 108 m)/(2.9 107 m s1) = 14.48 s
Correct substitution in = s/t (ignore powers of ten) (1)
Answer: 14.48 s, 14.5 s [no ue] (1)

(b)

Estimate of number of electrons

Q = ne = It
n = It/e
n = (3.0 106 A) (1s)/(1.6 1019 C)
Use of ne = It (1)
(1.8 2.0) 1025 (1)

(c)

Current direction
From Jupiter (to Io) / to Io / to the moon (1)

1
[5]

6.

Charge
Charge is the current time (1)

Potential difference
Work done per unit charge [flowing] (1)

Energy
9 V 20 C (1)
= 180 J (1)

2
[4]

7.

(a)

p.d. across 4 resistor

1.5 (A) 4 ()
= 6 V (1)

(b)

Resistance R2
Current through R2 = 0.5 A (1)
6 (V)
R2 =
0.5(A)
R2 = 12 (1)

[allow ecf their pd across 4 ]

(c)

Resistance R1
p.d. across R1 = 12 6 4
= 2 V (1)
Current through R1 = 2 A (1)
2(V)
R1 =
= 1 (1)
2(A)
[allow ecf of pd from (a) if less than 12 V]
Alternative method
Parallel combination = 3 (1)
Circuit resistance = 12(V)/2 (A) = 6 (1)
R1 = 6 (3 + 2) = 1 (1)

[allow ecf of pd from (a) and R from (b)]

[6]

8.

Definition of symbols:
n

number of electrons/carriers per unit volume (per m3)

OR
electron (or carrier) density (1)

average (OR drift) velocity (OR speed) (1)

Ratio
ny

Value

Explanation

Same material (1) (1)

Connected in series/Kirchoffs 1st law/conservation of

charge/current is the same (1) (1)

A is halved so double
[Accept qualitative, e.g. A so v , or good
analogy] (1) (1)

nx
ly
lx
vy
vx

6
[Accept e.g. ny = nx.....]
[No e.c.f ]
[NB Mark value first, without looking at explanation. If value correct, mark explanation. If
value wrong, dont mark explanation except: if y/x = or 1:2, see if explanation is correct
physics, and if so give (1). No e.c.f.]
[8]

9.

Metal wire:
straight line through origin
Semiconductor diode:
line along V axis for negative I
curve up in first quadrant
in gap
p.d. across it (4.5 1.9) V
2 .6 V
= 130
RS =
20 10 3 A

3
[6]

10.

Resistance of strain gauge

l
State R =
(1)
A
Use of formula (1)
x 6 (1)
R = 0.13 [ecf their l] (1)


l 9.9 10 8 m 2.4 10 2 m 6
R =

=
A

1.1 10 7 m 2

129.6 10 3

R = 0.13

Change in resistance
R = 0.13 0.001
R = 1.3 104 () [no e.c.f.]
OR
R = 0.02 0.001
R = 2.0 105
0.1% 0.001 (1)
Correct number for R (1)

Drift velocity
Stretching causes R to increase (1)
Any two from:
Current will decrease
I = nAQ
Drift velocity decreases
nAe constant (1) (1)

3
[9]

[For R decreasing, max 1:

Any one from:
I will increase
I = nAQ
will increase
nAe constant]
11.

Definition of e.m.f. of a cell

Work/energy (conversion) per unit charge

for the whole circuit / refer to total (energy)

OR
Work/energy per unit charge
converted from chemical to electrical (energy)

1
1

OR
E=

W
for whole circuit
Q

All symbols defined

OR
E=

P
for whole circuit
I

All symbols defined

[Terminal p.d. when no current drawn scores 1 mark only]

Circuit diagram

V
R
1
A in series 1

R (can be variable) 1
A and V correct
1
V as shown
Or across R + A
Or across battery

[2nd mark is consequent on R(fixed, variable) or lamp]

Sketch graph

1/I

Graph correctly drawn with axes appropriately labelled and

consistent with circuit drawn
Intercept on R axes
()r

Gradient ()r [Gradient mark consequent

on graph mark]

1
1

[Gradient may be indicated on graph]

[6]

12.

(a)

(i)

Potential difference = work (done)/(unit) charge

OR Potential difference = Power/current (1)

J = kg m 2 s 2 (1)
C = A s or W = J s1 (1)
V = kg m2 A1 s3 (1)
Converts 2 minutes to 120 seconds (1)

(ii)

(b)

Multiplication of VIt or VQ (1)

Energy = 1440 J (1)

Example of answer:
Energy = 6.0 V 2.0 A 120 s
= 1440 J
[7]

13.

Current in heating element

V2
R

p = VI

p=

500 W
I=
230 V

230 2
R=
/ 105.8()
500

I = 2.2 A

I = 2.2 A

1
1

Drift velocity
Drift velocity greater in the thinner wire / toaster filament

Explanation
Quality of written communication

See I = nAQ

I is the same (at all points )

(probably) n (and Q) is the same in both wires

1
[8]

14.

Current:
Conversion, i.e. 0.94 103 m s1
(1)
19
(1)
Use of 1.6 10 C
Answer 3.0 A
1.0 1029 m3 0.20 106 m2 1.6 1019 C 0.94 103 mm s1
Current = 3.0 A [Accept 2.8 A if 0.9 103 used.]

(1)
3

Resistance:

Recall R = l
A

(1)

Substitution:
1.7 10 8 m 4.0 m
R=
0.20 10 -6 m 2
Resistance = 0.34
(1)

(1)

Potential difference:
Potential difference = 3.0 A 0.34
(1)
= 1.0 V (1.02 V)
[Mark for correct substitution of their values or for the answer of 1.0 V]

Explanation:
(Increasing resistivity) increases resistance
Leads to a smaller current
(1)

(1)

Comparison:
Drift velocity decreases (in second wire)
(1)
[Allow V1/V2 = I1/I2]
[Allow e.c.f. answer consistent with their current answer]
[Resistivity up, current down
up, I down / 2 (2nd mark)]

2
1

[10]

15.

Calculation of voltages:
Any use of
Voltage

current x component resistance (1)

Ballast

150 V (1)

Filament

25 V (1)

Voltages on diagram:
3 voltages (150,25,25) marked on diagram near component; ignore units (1)
[Minimum 150 (1 25)]
Vstarter = 30 V (marked on diagram) (1)
Fundamental change necessary:
(Free) charge carriers or free electrons, ionised, particles need to be charged (1) (1)
[NOT T ]

Calculation of power dissipated:

Vballast

230V 110 V (1)

120V/300

0.40 A (1)

230 V 0.40 A [e.c.f for current]

92 W (1)

Power

Faulty component:
Starter is not breaking the circuit/starter still conducting (1)

1
[10]

16.
Word Equation

Quantity Defined

Voltage Current

Resistance

(1)

Voltage Current

Power

(1)

Charge Time

Current

(1)

Work done Charge

Voltage/p.d./e.m.f

(1)
[4]

17.

Demonstration that resistance is 0.085 :

R

l/A (1)

1.7 108 m 20 m / (4.0 106 m2) (1)

Calculation of voltage drop:

V

37 A 0.085 (1)

3.1 V 2 = 6.3 V [Not if Vshower then found] (1)

[Only one conductor, leading to 3.1 V, gets 1st mark]

[Nothing if wires in parallel]
Explanation:
Lower resistance/R = 0.057 /less voltage drop/new V=

2
old V (1)
3

Power dissipated in cable/energy wasted/wire not so hot

OR more p.d/current/power to shower
OR system more efficient (1)

2
[6]

18.

Proof:
V = V1 + V2
V = IR
IR2

V1 = IR1

Substitute and cancel I

V2 =

V = V 1 + V2

(1)

(1)

Sub using R =

(1)
3

Explanation of why it is a good approximation:

Resistance of connecting lead is (very) small

(1)

So I R(very) small = (very) small p.d./e1s do little work so p.d. small/r small

(1)

compared with rest of the circuit so p.d. small

2

Circumstances where approximation might break down:

If current is large OR resistance of rest of circuit is small

(1)

[Not high voltage/long lead/thin lead/high resistivity lead/hot lead]

1
Calculation:
Use of R =

l
A

with A attempted sectional area

(1)

Correct use of 16

(1)

Use of V = IR

(1)

0.036 V

(1)
4
[10]

19.

Number of carriers or electrons per unit volume / per m3 /carrier density/electron density (1)
[Not charge density / concentration]
Drift velocity OR drift speed OR average/mean/net/overall velocity (1)

[Not just velocity; not speed unless drift]

m3 (1)
m2 As m s1 (1)
Multiply and reduce to A (1)

[Base units not needed]

[Mixed units and symbols could get the third mark]
[mA = m1 loses 1 mark]
Metal:

M: n large so there is a current

n: n in metal much larger (1)

Insulator
I: n zero (negligible)/very small so less
current (or zero current)

Current in metal is larger (1)

2

[Ignore anything about v. Allow e.g. electron density for n]

[7]

20.

No, because V is not proportional to I OR not straight line through origin / (1)
only conducts above 0.5 V / resistance changes

Use of R = 0.74 / current from graph (1)

= 9.25 [9.0 9.5 ] [Minimum 2 significant figures] (1)

Calculation of
p.d. across R
[8.26]

Calculation of total
resistance[109 115]

Ratio R: ratio V

E = IR (1)

diode resistance [9]

Correct
substitutions

Correct
substitutions (1)

103 [100 106] (1)

3
[If not vertical line, 0/2]

(1)
(1)

(1)
(0)

(1)
(0)
Anything (gap, curve, below axis) (1)(1)

0.7

0.7

0.7

[Otherwise 0 0 ]
[8]

21.

Use R = l/A OR correct rearrangement OR plot R l gradient = /A (1)

[Symbols or words]
With A = tw (1)

l = RA/ [Rearrangement mark symbols or numbers] (1)

Use of A = tw (1)
[Correct physical quantities substituted but ignoring unit errors, powers of 10]
= 110 m
[111 m] (1)

Reduce width/w of strip OR use thinner/t foil [Not reduce A; not increase T, V, I] (1)
Smaller w/t/A will be less accurate OR have larger error OR larger R
will be more accurate (1)

[Increase w or t, could give e.c.f. to increased accuracy]

[7]

22.

I 2 R / (I I 2 r) /
I 2 r / (I I 2 r)

( Ir ) 2
(1)
R

( Ir ) 2
(1)
R

I OR I 2 R + I 2 r / 2 / (R + r) (1)
I = I 2 R + I 2 r OR (It = I 2 RT + I 2 rt / their (iii) = their (i) + their (ii) (1)
Cancel I (OR I and t) and arrange [only if energy equation is correct] (1)

Maximum current occurs when R = 0 (1)

Imax = /r (1)

OR larger r means smaller I (1 mark)

1 M [Could be underlined OR circled] (1)
It gives the smallest current (1)
[If 100 k this reason: 1 only]

2
[9]