The batteries in
circuit 1 and circuit 2 are identical.
(a) Put ticks in the table below to describe the
connections of the two resistors P and 
| Series | Parallel | |
| Circuit 1 | ||
| Circuit 2 |
are
used as small electrical heaters. State two advantages of connecting them as
shown in circuit 2.
(c) In
circuit 1, the ammeter reads 1.2A when the switch is closed. Calculate the
reading of the voltmeter in this circuit.
(d)
The two switches in circuit 2 are closed. Calculate the combined resistance of
the two resistors in this circuit.
(e)
When the switches are closed in circuit 2, ammeter 1 reads 5A and ammeter 2
reads 2A.
Calculate
(i) the current in resistor P,
(ii) the power supplied to resistor Q,
(iii)
the energy transformed in resistor Q in 300s.
Answer:
(b) (i) If one electrical heater stops working due to some
defect then other keeps working normally.
(ii) Each
electrical appliance (heater) gets the same voltage (220 V) as that of the
power supply line.
(c)
Effective resistance, R=4+6=10
Series
Parallel
Circuit 1
Circuit 2
Current
, =1.2A
Voltmeter
reading = 12 V
(d)
Combined resistance,
(e) Current
=5A
2=2A
(i)
The current in resistor 
= 5A - 2A
=3A
(ii)
The power supplied to resistor 
= 
=12x2
=24W
(iii) The energy transformed in resistor in 300 s = E = P x t
=24 x 300
s= 7200 J
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