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Question 874: [Current of Electricity]

A car battery has internal resistance of 0.060 Ω. It is re-charged using a battery charger having an e.m.f. of 14 V and an internal resistance of 0.10 Ω, as shown in Fig.1.

(a) At beginning of the re-charging process, current in circuit is 42 A and e.m.f. of the battery is E (measured in volts).

(i) For the circuit of Fig.1, state

1. magnitude of the total resistance

2. total e.m.f. in the circuit. Give answer in terms of E

(ii) Use answers to (i) and data from the question to determine the e.m.f. of the car battery at the beginning of the re-charging process

(b) For majority of the charging time of the car battery, e.m.f. of the car battery is 12 V and the charging current is 12.5 A. Battery is charged at this current for 4.0 hours. Calculate, for this charging time,

(i) charge that passes through the battery

(ii) energy supplied from the battery charger

(iii) total energy dissipated in the internal resistance of battery charger and the car battery

(c) Use answers in (b) to calculate the percentage efficiency of transfer of energy from the battery charger to stored energy in the car battery

Reference: Past Exam Paper – June 2007 Paper 2 Q6

Solution 874:

(a)

(i)

1. Total resistance (= 0.060 + 0.10) = 0.16Ω

2. Total e.m.f. = EITHER (14 – E)     OR (E – 14)

{The difference of the 2 e.m.f. is taken since they are in opposite direction – the flows of current [which is from the positive terminal] from the batteries are in opposite directions – polarities of the battery are opposite. AND either of the 2 answer are accepted since we do not yet know whether E is greater or less than 14V. But logically, current should be flowing from a battery charger to the battery – not the reverse.}

(ii)

{At beginning of re-charging, the current of 42A is from the charger. So, for total emf = (14 – E), the current is taken in the positive direction. For (E – 14), current is taken as a negative value.}

EITHER 14 – E = 42 × 0.16               OR E – 14 = -42 × 0.16

e.m.f. E of the car battery = 7.3V

(b)

(i) Charge, Q = It = 12.5 × (4×60×60) = 1.8×10⁵C

(ii)

{Power = Energy / time. Energy = Power × time}

EITHER Energy = EQ           OR Energy = EIt

Energy = 14 × (1.8×10⁵)         OR Energy = 14 × 12.5 × (4×3600)

Energy = 2.52×10⁶J

(iii)

{Power, P = I²R or Power = VI}

Energy dissipated = I²Rt                    OR VIt and V = IR

Energy dissipated = (12.5²) × (0.16) × (4×3600) = 3.6×10⁵J

(c)

{Energy supplied = 2.52×10⁶J. Wasted energy (in internal resistance) = 3.6×10⁵J.

Useful energy = Energy supplied – Wasted Energy

Efficiency = Useful energy / Total energy supplied}

Percentage efficiency = [(2.52×10⁶) – (3.6×10⁵)] / (2.52×10⁶) = 86%

Question 875: [Measurement > Uncertainty]

An experiment is carried out to measure resistance of a wire.

Current in the wire is (1.0 ± 0.2) A and potential difference across the wire is (8.0 ± 0.4) V.

What is the resistance of the wire and its uncertainty?

A (8.0 ± 0.2) Ω

B (8.0 ± 0.6) Ω

C (8 ± 1) Ω

D (8 ± 2) Ω

Reference: Past Exam Paper – June 2014 Paper 11 Q4

Solution 875:

Answer: D.

Ohm’s law: V = IR

Resistance R = V / I = 8.0 / 1.0 = 8Ω

ΔR / R = ΔV/V + ΔI/I

Uncertainty in resistance, ΔR = R × (ΔV/V + ΔI/I) = 8 × ([0.4/8] + [0.2/1.0]) = ±2Ω

Question 876: [Pressure]

Diagram shows two vessels, P and Q, both with sides inclined at 45°.

Vessel P tapers outwards and vessel Q tapers inwards, as shown.

Both vessels contain a liquid. Depth of the liquid in the vessels is the same. The liquid in vessel P is twice as dense as the liquid in vessel Q.

What is the ratio

pressure due to the liquid on the base of P / pressure due to the liquid on the base of Q ?

A 2 / 1                         B √2 / 1                                   C 1 / √2                                   D 1 / 2

Reference: Past Exam Paper – June 2011 Paper 12 Q20

Solution 876:

Answer: A.

Pressure P = hρg

The pressure P depends only on the depth h, the density ρ of the liquid, and the acceleration of free fall. The volume of the container does not affect the pressure at a point.

Let the density of the liquid in vessel Q = ρ

The depth h of the liquid in the vessels is the same.

For vessel P: Pressure P = h(ρ2)g = 2hρg

For vessel Q: Pressure P = hρg

Ratio = 2hρg / hρg = 2 / 1