Download Solution of Exercise 20 (Electric Circuits)

Solution of Exercise 15 (Electrical power and energy)
1. (a)
E
L
N
(b) Cost of electricity = 0.87 (2.5) (150) = $ 326.25
2. (a) (i) A To Y, B To Z, C to X
(ii) If a fault develops, resulting in the live wire touching the metal body, a large current flows to the
earth since there is no resistance in the conducting path to the earth. If there were no earth wire, the metal
body would become high potential and anyone who happened to touch it could receive a fatal electric shock.
(b) (i) Live
(ii) Current drawn from each kettle = P/V
= 1500 / 200 = 7.5 A
maximum no. of kettles  15 / 7.5 = 2
(c) (i) Pt = mcΔT
1500 (0.8) t = 1 (4200) (100-20)
t = 280 s.
(ii) Cost = 0.4 (1.5) (280 / 3600) = $ 0.0467
3. (a) (i) M is kilowatt-hour meter. It measures the electrical energy used.
(ii) Total current drawn from the mains supply I
= total power / voltage = (600 + 2000 + 1500) / 220 = 18.6 A
(iii) It is because the current drawn from the water heater is very large, if it is connected in the ring
circuit, overloading may occur.
(iv) 1. The current flows from the consumer unit to the sockets via two paths. Each path carries
only half of the current. The chance of overloading the circuit is reduced.
2. Since the current is smaller, so thinner cables can be used.
Multiple Choice
1-5
CCBDA
11-15 C E A A D
6-10 B E E A C
16-17 B A
Explanations to m.c.
1. Resistance R = V2 / P = 2002 / 60
So power = V2/R = 2202 / ( 2002 / 60)
= 60 (220/200)2 > 60.
Hence the power is greater, i.e. more electrical energy is converted to light energy per second.
2. Let V be the battery’s e.m.f., R be the resistance of each bulb.
In case A, total power = V2/R’ = V2 / 3R (since 3R is the equivalent resistance)
In case B, total power = V2 / R’ = V2 /2R (since 2R is the equivalent resistance, the LHS bulb is shorted)
In case C, total power = V2/R + V2/2R = 3V2/2R = 1.5 V2 / R
In case D, total power = V2 / R (since the top two bulbs are shorted)
In case E, current = V / (0.5R + R) = 2V/3R
total power = I2 (R/2 + R) = 2V2 / 3R
So C is the brightest.
3. V2/4 = 12 => V2 = 48
Power dissipated by 5 = V2/5 = 48/5 = 9.6 W
4. Power = V2/R = 2002 / 30 = 1333 W
Operating current = V/R = 200/30 = 6.7 A
So suitable fuse = 10 A
5. If C2 is connected to the earth pin, the circuit can be redrawn as follows:
C2
C1
E
L
N
6. Energy converted = Pt = 2 (1000) (60) J
8. Let the resistance and the voltage of power supply be R and V respectively.
When connected in series, combined resistance = 2R
Power dissipated P = V2 / 2R
When connected in parallel, combined resistance= R/2
Power dissipated = V2 / (R/2) = 4P
9. Statement (1). When the contact point X touches the metal case and the switch is closed, a large current
flows through the metal case to the earth without flowing through the heating element and the fuse will blow
because of excessive current.
Statement (2). Whent the contact point Y touches the metal case and the switch is closed, current flows
through the heating element as usual and then to the earth. The fuse will not blow as the size of current is the
same as usual.
Statement (3). There will be no current if the heating element is broken since the circuit is open.
11. Installing a fuse cannot prevent short circuit. Short circuit would still occur but if it occurs, excessive
current flows in the circuit and the fuse would blow.
13. When connected in parallel, P and Q work at rated voltages, power of Q is larger than that of P, so Q is
brighter than P.
When connected in series, the currents through P and Q are the same, so we compare their resistances.
RP = V2 / PP, RQ = V2 / PQ.
Since Q has a higher power, Q must have a lower resistance. By P = I2R, Q is less brighter.
14. Total resistance = 2002 / 1000 + 2002 / 2000 = 60 
P = V2 / R = 2002 / 60 = 0.667 kW
15. Energy required in both cases are the same. So
(V12 /R) t1 = (V22 /R) t2
2402 (5) = 2002 t2 => t2 = 7.2 min.
16. If the lamp is to work as rated, its p.d. = 6 V,
so p.d. across R = 12 - 6 = 6V.
Current I = P/V = 24 / 6 = 4A.
Hence R = V/I = 6 / 4 = 1.5 
17. Power consumed by motor = power delivered by battery – power dissipated by 10 
= 12 (0.5) – 0.52 (10) = 3.5 W