Sunday, March 8, 2015

Physics 9702 Doubts | Help Page 79

  • 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 = I2R

For electrical device,
4.0 = I2 (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 = I2R = V2 / 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 = V2 / R = (16.25)2 / 6.5 = 40.625W





3 comments:

  1. May I know how do we know which resistance to sub in the equation for question 421?
    Thanks in advance! xx

    ReplyDelete
    Replies
    1. 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.

      Delete
    2. Okay I understand now :) Thank you!

      Delete

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