# Physics 9702 Doubts | Help Page 176

__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)
= 1.8×10

^{5}C
(ii)

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

EITHER Energy = EQ OR Energy = EIt

Energy = 14 × (1.8×10

^{5}) OR Energy = 14 × 12.5 × (4×3600)
Energy = 2.52×10

^{6}J
(iii)

{Power, P = I

^{2}R or Power = VI}
Energy dissipated = I

^{2}Rt OR VIt__and__V = IR
Energy dissipated = (12.5

^{2}) × (0.16) × (4×3600) = 3.6×10^{5}J**(c)**

{Energy supplied = 2.52×10

^{6}J. Wasted energy (in internal resistance) = 3.6×10^{5}J.
Useful energy = Energy
supplied – Wasted Energy

Efficiency = Useful energy
/ Total energy supplied}

Percentage efficiency = [(2.52×10

^{6}) – (3.6×10^{5})] / (2.52×10^{6}) = 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 Q / pressure due to the liquid on the base of P ?

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

can u explain q1 a(11) more. I cant understand why u take 42*.16

ReplyDeletewe are using the formula V = IR

Deletewhere I is the current and R the resistance

can you please solve question number 17 from 2011 M/J paper 12.

ReplyDeletethanks in advance!

Go to

Deletehttp://physics-ref.blogspot.com/2014/08/9702-november-2011-paper-12-worked.html

a link is available for Q17