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Question
An electric heater which is connected to a 220 V supply line has two resistance coils A and B of 24 Ω resistance each. These coils can be used separately (one at a time), in series or in parallel. Calculate the current drawn when:
(a) only one coil A is used.
(b) coils A and B are used in series.
(c) coils A and B are uses in parallel.
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Solution
(a) When only one coil, A is used:
V = IR
220 = 24 I
I = 9.2 A
(b) When coils, A and B are used in series:
Total resistance R = RA+ RB = 48 Ω
V = IR
220 = 48 I
I = 4.58 A
(c) When coils, A and B are used in parallel:
1/R = 1/RA+ 1/RB
Here, RA = 24 Ω and RB = 24 Ω
1/R = 1/24 + 1/24
or 1/R = 2/24
R = 12 Ω
Total resistance of parallel combination, R = 12 Ω
Now, V = IR
220 = 12 I
I = 18.33 A
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