Resistor Network Answers

1. (a) pd = 3.6 V (1) 1

Example of answer;
p.d. = 0.24 A × 15 Ω = 3.6 V
(b) Calculation of pd across the resistor (1)
[6.0 – 3.6 = 2.4 V]
Recall V = IR (1)
I1 calculated from their pd / 4Ω (1)
[correct answer is 0.60 A. Common ecf is 6V/4Ω gives 1.5 A] 3
Example of answer:
I1 = 2.4 V / 4.0 Ω = 0.6 A
(c) Calculation of I2 from I1 – 0.24 [0.36 A] (1)
[allow ecf of their I1. common value = 1.26 A]
Substitution V = 3.6 V (1)
R = 10 Ω (1) 3
[7]

2. (a) p.d. across 4 Ω resistor

1.5 (A) × 4 (Ω)
= 6 V (1) 1
(b) Resistance R2
Current through R2 = 0.5 A (1)
6 (V)
R2 =
0.5(A)
R2 = 12 Ω (1) 2
[allow ecf their pd across 4 Ω]
(c) Resistance R1
p.d. across R1 = 12 − 6 − 4
= 2 V (1)
Current through R1 = 2 A (1)
2(V)
R1 = = 1Ω (1)
2(A)
[allow ecf of pd from (a) if less than 12 V]
Alternative method
Parallel combination = 3Ω (1)
Circuit resistance = 12(V)/2 (A) = 6Ω (1)
R1 = 6 – (3 + 2) = 1 Ω (1) 3
[allow ecf of pd from (a) and R from (b)]
[6]

3. Charge

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 Charge is the current × time (1) 1
Potential difference
Work done per unit charge [flowing] (1) 1
Energy
9 V × 20 C (1)
= 180 J (1) 2
[4]

4. The power supplies in the two circuits shown below are identical.

+ +
R1 R2 V RT
– –
I1 2 I

Write down the relationship between I1, I2 and I which must hold if the combined resistance of
the parallel pair, R1, and R2, is to equal RT.
I= I1 + I2
(1 mark)
Hence derive the formula for the equivalent resistance of two resistors connected in parallel.
From Ohm’s law:
I = V/RT I1 = V/R1 I2 = V/R2 (1)
∴ V/RT = V/R1 + V/R2 (1)
and 1/RT = 1/R1 + 1/R2 (1)
(3 marks)
Use your formula to show that the resistance between the terminals of a low-resistance
component is hardly changed when a high-resistance voltmeter is connected in parallel with it.
If Rv >>Rlow then 1/Rv >> 1/Rlow (1)
and RT ≈ Rlow (1)
Allow method based on numerical example
(2 marks)
[Total 6 marks]

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5. Resistance calculations
Evidence of 20 Ω for one arm (1)
1 1 1
= + (1)
R 20 20
R = 10 Ω (1) 3

Comment
This combination used instead of a single 10 Ω 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]

6. Circuit
Ammeters and two resistors in series (1) 1
[1 mark circuit penalty for line through cell or resistor]
Cell e.m.f
E= 150 x 10–6 (A) x 40 x 103 (Ω) total R (1)
Powers of 10 (1) 2
E = 6.0 (V)
New circuit
Voltmeter in parallel with 25 (kΩ) resistor (1) 1
Resistance of voltmeter
6(V)
(Total resistance) =
170 × 10 – 6 (A)
= (35.3 kΩ) (1) 

(Resistance of ll combination) = 35 – 15 kΩ 
= (20 Ω) [e.c.f. their total resistance] (1) 
1 1 1 
= + 
20 25 RV 

1
=
5–4 
RV 100 

RV = 100 kΩ [108 kΩ if RT calculated correctly] (1) 


Alternative route 1: 3

p.d. across 15 kΩ = 2.55 V (1) 

(∴p.d. across ll combination = 3.45 V) 

resistance combination = 20 kΩ 
→ RV = 100 kΩ (1) 


(1)
[7]
Alternative route 2: 3
p.d. across parallel combination = 3.45 V (1) 
I through 25 kΩ = 138 µA 
(1) 
→ RV = 100 kΩ 
(1) 

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7. The circuit shows a battery of negligible internal resistance connected to three resistors.
4Ω 0.75 A

9.0 V R 24 Ω

I2 I1

Calculate current I1.

Voltage drop across 4Ω resistor = 3V (1)
(9 V – 3 V)
I2 = (1)
24 Ω
I1 = 0.25 A (1)
(3 marks)
Calculate resistance R
I2 = 0.75 A – 0.25 A = 0.50 A (1)
R = 6 V / 0.50 A = 12 Ω (1)
R = 12 Ω
(2 marks)
[Total 5 marks]

8. Diagram of torch circuit:

The lamp will light
Correct circuit 2
[Circuit showing one cell only is allowed one mark only unless the cell is labelled 4.5 V.
If a resistor is included, allow first mark only unless it is clearly labelled in some way as
an internal resistance.]

0.3 A

3.5 V

3.5 V/3

Voltage across each circuit component and current in lamp:

Either 3.5 V/3 shown across the terminals of one cell or 3.5 V across
all three cells
3.5 V shown to be across the lamp
0.3 A flowing in the lamp [i.e. an isolated 0.3 A near the lamp does not
score] 3
Calculation of internal resistance of one of the cells:
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 3.5V
Lost volts = 4.5 V - 3.5 V or 1.5 V –
3
or total resistance = (4.5 V)/0.3 A) = 15 KΩ
Internal resistance of one cell = [(1.0 V)/(0.3 A)] ÷ 3
or [(0.33 V) (0.3 A)] or lamp resistance = (3.5 V) / (0.3 A)11.7 Ω
= 1.1 Ω or = (3.3Ω)/3 = 1.1 Ω 3

[Some of these latter marks can be read from the diagram if it is so

labelled]
[8]

9. Proof:
V = V1 + V2 V = V 1 + V2 (1)

V = IR V1 = IR1 V2 = IR2 ÷I (1)

Substitute and cancel I 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./e–1s 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:
ρl
Use of R = with A attempted × sectional area (1)
A
Correct use of 16 (1)
Use of V = IR (1)
0.036 V (1)
4
[10]

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10. No, because V is not proportional to I OR not straight line through origin / (1)
only conducts above 0.5 V / resistance changes 1
Use of R = 0.74 / current from graph (1)
= 9.25 Ω [9.0 – 9.5 Ω] [Minimum 2 significant figures] (1) 2
Calculation of Calculation of total Ratio R: ratio V E = ΣIR (1)
p.d. across R resistance[109 – 115]
[8.26]

÷I – diode resistance [9] Correct Correct

substitutions substitutions (1)

103 Ω [100 – 106] (1)

3

[If not vertical line, 0/2]

(1) (1) (1)

(1) (0) (0)
Anything (gap, curve, below axis) (1)(1) 2
0.7 ≠ 0.7 0.7
[Otherwise 0 0 ]

[8]

11. Circuit diagram

Resistor with another variable resistor/potential divider/variable power
pack (1)
Ammeter reading current through resistor (1)
Voltmeter in parallel with resistor (1) 3
Graph labels
Straight line – resistor 
Both labelled (1) 1
Curve – lamp 
Potential difference
At 0.5 A p.d.= 3.5 V / 3.4 V + 7.8 V / idea of adding p.d. [for same current] (1)
= 11.2 V/11.3 V (1) 2
[Accept 11.0 –11.5 V]
Resistance of lamp
3.5 V
[OR their value of p.d. across lamp ÷ 0.5 A] (1)
0.5 A

= 7.0 Ω (1) 2
[e.c.f. their value]
[8]

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12. Show that
[In diagram or text}
• states p.d. same across each resistor (1)
• use of I = I1 + I2 + I3 [symbols or words] (1)
V = V + V + V
•
R R1 R2 R3 (1) 3

[I = V / R stated somewhere gains one mark]

Networks
First network: 2.5(Ω) (1)
Second network: 25 (Ω) (1)
Third network: 10 (Ω) (1) 3
Meter readings
Ammeter: 25 (mA) (1)
Voltmeter V1: 25 × 10 OR 50 × 5 [ignore powers of 10] (1)
= 0.25 V (1)
Voltmeter V2: 50 × 25 [ignore powers of 10] (1)
= 1.25 V (1) 5
nd
[Allow full e.c.f. for their resistance for 2 network OR their V1 answer]
[11]

13. Readings on voltmeter

Use of any resistor ratio OR attempt to find current in either circuit (1)
At 950 kΩ
10 kΩ × 6 V
V= = 0.063 V (1)
960 kΩ

At 1.0 kΩ
 10 kΩ × 6 V 
V =  11 kΩ  = 5.45 V (1) 3
 
Use of circuit as lightmeter
Maximum resistance corresponds to low light intensity/resistance down as
light intensity up (1)
∴ lightmeter or voltmeter reading will increase as light intensity
increases [or reverse] (1) 2
[Can ecf for 2nd mark if resistance/light intensity incorrect and/or p.d.
calculation wrong]
[5]

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