# Physics 9702 Doubts | Help Page 79

__Question 418: [Current of Electricity]__Electrical device of fixed resistance 20 Ω is connected in series with variable resistor and battery of electromotive force (e.m.f.) 16 V and negligible internal resistance.

What is the resistance of the variable resistor when power dissipated in the electrical device is 4.0 W?

A 16 Ω B 36 Ω C 44 Ω D 60 Ω

**Reference:**

*Past Exam Paper – November 2013 Paper 13 Q34*

__Solution 418:__**Answer: A.**

Power dissipated, P = I

^{2}R
For electrical device,

4.0 = I

^{2}(20)
Current I in circuit = 0.45A

Since this is a series circuit, the
same current flows through the variable resistor.

Ohm’s law: V = IR

p.d. across electrical device = 0.45
(20) = 9.0V

Let the resistance of the variable
resistor = R

From Kirchhoff’s second law, the sum
of p.d. in a loop is equal to the e.m.f. in the circuit.

16 = 9.0 + 0.45R

Resistance R = 16Ω

__Question 419: [Current of Electricity]__Diagram shows electric pump for a garden fountain connected by an 18 m cable to a 230 V mains electrical supply.

Performance of pump is acceptable if potential difference (p.d.) across it is at least 218 V. Current through it is then 0.83 A.

What is maximum resistance per metre of each of the two wires in cable if the pump is to perform acceptably?

A 0.40 Ω m

^{–1}B 0.80 Ω m

^{–1}C 1.3 Ω m

^{–1}D 1.4 Ω m

^{–1}

**Reference:**

*Past Exam Paper – June 2014 Paper 12 Q32*

__Solution 419:__**Answer: A.**

Ohm’s law: V = IR

For the pump to perform acceptably, the
potential difference across it should be at least 218 V. So the potential
difference across the cables is (230 – 218 =) 12 V.

The current through the electric
pump is 0.83A. The same current flows through the cable since it is in series
with the pump.

So, the cable has a total resistance
of

R = (230 – 218) / 0.83 = 14.5Ω

From the diagram, it can be seen
that the CABLE is made up of 2 WIRES. So, the total length of wire in the cable
is 2(18) = 36m. [Notice that the wires are NOT
connected to each other – so, they are not in parallel. Rather, one wire
connects one terminal of the mains to one terminal of the pump and the other
wire connected the other terminal of the mains to the other terminal of the
pump.] This total length of wire corresponds to a total resistance of 14.5Ω
(as calculated above).

Resistance per metre = 14.5 / 36 =
0.4Ωm

^{-1}

__Question 420: [Current of Electricity > Potential divider]__
A 12 V battery is in series with
ammeter, 2 Ω fixed resistor and a 0 – 10 Ω variable resistor. High-resistance voltmeter
is connected across fixed resistor.

Resistance of variable resistor is
changed from zero to its maximum value.

Which graph shows how potential difference
(p.d.) measured by voltmeter varies with the current measured by the ammeter?

**Reference:**

*Past Exam Paper – June 2013 Paper 13 Q36*

__Solution 420:__**Answer: B.**

Note that the graph is that of the
p.d. across the FIXED resistor against the current through it. This current is
the same flowing in the whole circuit since this is a series connection.

When the variable resistor has value
zero, the total resistance in the circuit is 2 Ω. So, there is 12V across the 2
Ω resistor and the current is a maximum (= 12/2 = 6.0A) in this case. On the
graph, this would correspond to point (6.0, 12).

When the variable resistor is at 10Ω
(maximum), the total resistance in the circuit is 12 Ω. So, the current is a
minimum (= 12/12 = 1.0A) and the voltmeter will read 2V. This is obtained from
the potential divider equation.

p.d. across fixed resistor = [2 / (10+2)]
x 12 = 2V

This corresponds to point (1.0, 2).

So, the graph does not start at
point (0, 0) and has a positive gradient since it passes through the points (1.0,
2) and (6.0, 12).

__Question 421: [Current of Electricity]__
A low-voltage supply with e.m.f. of
20 V and internal resistance of 1.5 Ω is used to supply power to a heater of
resistance 6.5 Ω in a fish tank.

What is the power supplied to water
in the fish tank?

A 41 W B 50 W C
53 W D 62 W

**Reference:**

*Past Exam Paper – June 2013 Paper 12 Q33*

__Solution 421:__**Answer: A.**

Power supplied to heater = VI = I

^{2}R = V^{2}/ R
From the potential divider equation,

p.d. across the heater = [6.5 / (6.5+1.5)]
x 20 = 16.25V

Power supplied to heater = VI = V

^{2 }/ R = (16.25)^{2 }/ 6.5 = 40.625W
May I know how do we know which resistance to sub in the equation for question 421?

ReplyDeleteThanks in advance! xx

The power supplied to the water is the power from the heater. Thus, the quantities involved in the equation are for the heater – that is, p.d. across heater, current through heater and resistance of heater.

DeleteOkay I understand now :) Thank you!

Deletefor qs 418, can I use P=VI instead of P= I^2R?

ReplyDeleteyes, but you will need to do more calculations to find the resistance.

Delete