A Level · AS · structured

AS Level foundations · Question 10

AS Level foundations · Original GioPhysics question with a detailed, mark-by-mark answer guide.

Demand: demanding
Marks: 13
Topics: 2
Answer: Complete

Fig. 10.1A single series loop. Along the bottom, a cell with its long plate on the left and a small resistor labelled r sit together inside a dashed rectangle labelled battery, the cell marked e.m.f. 12.0 V. From the dashed box the wire runs to the left, up the left-hand side and along the top through a resistor labelled R that has an arrow drawn across it to show that it is variable, then down the right-hand side through a circle marked A, and back along the bottom into the box. An arrow on the top wire labelled I shows the direction of the conventional current.

structured13 marks

A battery of e.m.f. 12.0 V and internal resistance r is connected in series with a variable resistor R and an ammeter of negligible resistance. When R is set to 4.0 Ω, the ammeter reads 2.0 A.

(a)
Define Define the electromotive force of a source.
2 marks
(b)
Determine Determine the internal resistance r of the battery.
3 marks
(c)
Calculate Calculate the ratio of the power dissipated inside the battery to the total power produced by the battery.
3 marks
(d)
Determine R is now increased to 12.0 Ω. Determine the new terminal potential difference of the battery.
3 marks
(e)
Explain Explain why the terminal potential difference increased when R was increased, even though the e.m.f. of the battery did not change.
2 marks

Your working or explanation

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Answer overviewKey answer: the energy transferred from chemical (or other non-electrical) form to electrical form per unit charge driven through the source E = I(R + r), so 12.0 = 2.0 × (4.0 + r)

(a)

2 marks

Define Define the electromotive force of a source.

How to approach it

Identify what the question is testing, organise the response into distinct mark-earning points, and make every conclusion traceable to a physical principle or to the evidence supplied.

1
the energy transferred from chemical (or other non-electrical) form to electrical form
2
per unit charge driven through the source

(b)

3 marks

Determine Determine the internal resistance r of the battery.

How to approach it

List the given quantities with units, identify the required quantity, write the governing relationship before substituting, and keep extra digits until the final line so rounding does not distort the result.

1
E = I(R + r), so 12.0 = 2.0 × (4.0 + r)
2
4.0 + r = 6.0
3
r = 2.0 Ω

(c)

3 marks

Calculate Calculate the ratio of the power dissipated inside the battery to the total power produced by the battery.

How to approach it

1
power in r = I²r = 2.0² × 2.0 = 8.0 W
2
total power = EI = 12.0 × 2.0 = 24 W
3
ratio = 8.0 / 24 = 0.33 (33%)

(d)

3 marks

Determine R is now increased to 12.0 Ω. Determine the new terminal potential difference of the battery.

How to approach it

1
I = 12.0 / (12.0 + 2.0) = 0.857 A
2
V = IR = 0.857 × 12.0
3
V = 10.3 V

(e)

2 marks

Explain Explain why the terminal potential difference increased when R was increased, even though the e.m.f. of the battery did not change.

How to approach it

State the outcome first, then link cause to effect with the relevant physical principle. Each link in the reasoning should be explicit enough to earn its own marking point.

1
increasing R decreases the current in the circuit
2
so the lost volts Ir across the internal resistance are smaller, and V = E − Ir is closer to E

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