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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 *= *R*_{A}+ *R*_{B} = 48 Ω*V* = *IR*

220 = 48 *I**I* = 4.58 A

(c) When coils, A and B are used in parallel:

1/*R *= 1/*R*_{A}+ 1/*R*_{B}

Here, *R*_{A }= 24 Ω and *R*_{B} = 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

Concept: System of Resistors - Resistances in Parallel

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