CBSE Class 12 · 2011 · All India · Current ElectricityShort Answer

Two heating elements of resistances

R_{1}

and

R_{2}

, when operated at a constant supply voltage

V

, consume powers

P_{1}

and

P_{2}

respectively. Deduce expressions for the power of their combination when they are connected in
(i) series and
(ii) parallel across the same supply.

Answer: (i) Series: $P_{series} = \frac{P _{1} P _{2}}{P _{1} + P _{2}}$ . \;
(ii) Parallel: $P_{parallel} = P_{1} + P_{2}$ .

Solution

Here

P_{1} = \frac{V ^{2}}{R _{1}}

and

P_{2} = \frac{V ^{2}}{R _{2}}

. (i) In series

R_{e q} = R_{1} + R_{2}

, so

P_{series} = \frac{V ^{2}}{R _{1} + R _{2}} = \frac{1}{\frac{1}{P _{1}} + \frac{1}{P _{2}}} = \frac{P _{1} P _{2}}{P _{1} + P _{2}}

. (ii) In parallel

\frac{1}{R _{e q}} = \frac{1}{R _{1}} + \frac{1}{R _{2}}

, so

P_{parallel} = \frac{V ^{2}}{R _{1}} + \frac{V ^{2}}{R _{2}} = P_{1} + P_{2}

CBSE Class 12 Board · 2011 · All India · Subjective