CBSE Class 12 · 2019 · Current ElectricityShort Answer

Two bulbs are rated

(P_{1}, V)

and

(P_{2}, V)

. If they are connected
(i) in series and
(ii) in parallel across a supply of voltage

V

, find the power dissipated in the two combinations in terms of

P_{1}

and

P_{2}

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

Solution

Each bulb has resistance

R_{1} = \frac{V ^{2}}{P _{1}}

R_{2} = \frac{V ^{2}}{P _{2}}

. (i) In series

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

, so

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

. (ii) In parallel each bulb runs at its rated power, so

P_{parallel} = P_{1} + P_{2}

CBSE Class 12 Board · 2019 · Subjective