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Two bulbs are related $(P_{1},V)$ and $(P_{2},V)$ , If they are connected

(i) in series (ii) in parallel across a supply $V$

find the power dissipated in the two combinations in terms of $P_{1}$ and $P_{2}$ .

Answers (1)

S safeer

Electric power is defined as the rate at which electrical energy is consumed in an electrical circuit. Electric power is given as:-

$P=IV=I^{2}R=\frac{V^{2}}{R}$ (as $V=IR$ )

Where V is the potential difference

$I$ is the current and

$R$ is the Resistance

(i) When two bulbs rated $(P_{1},V)$ and $(P_{2},V)$ are connected in series, the current is the same through each bulb.

If $R_{1}$ and $R_{2}$ be the resistances of two bulbs, then effective resistance of the circuit is,

$R=R_{1}+R_{2}$

Using, $P=\frac{V^{2}}{R}$

$\Rightarrow R=\frac{V^{2}}{P}$

we have,

$\frac{V^{2}}{P}=\frac{V_{1}^{2}}{P_{1}}+\frac{V_{2}^{2}}{P_{2}}$

We have given that :- $V_{1}=V_{2}=V$

Therefore -

$\frac{V^{2}}{P}=\frac{V^{2}}{P_{1}}+\frac{V^{2}}{P_{2}}$

$\frac{1}{P}=\frac{1}{P_{1}}+\frac{1}{P_{2}}$

Thus, the effective power when two bulbs are connected in series is

$\frac{1}{P}=\frac{1}{P_{1}}+\frac{1}{P_{2}}$

(ii) When two bulbs rated $(P_{1}V)$ and $(P_{2}V)$ are connected in parallel.

using the relation of parallel connection, we have

$\frac{1}{R}=\frac{P}{V^{2}}$

Thus, $\frac{P}{V^{2}}=\frac{P_{1}}{V_{1}^{2}}+\frac{P_{2}}{V_{2}^{2}}$

given : $V_1=V_2=V$

Thus, $\frac{P}{V^{2}}=\frac{P_{1}}{V^{2}}+\frac{P_{2}}{V^{2}}$

$\Rightarrow \frac{1}{(V^{2})}P=\frac{1}{V^{2}}(P_{1}+P_{2})$

$\Rightarrow P=P_{1}+P_{2}$

Therefore, the effective power or power (dissipated) when two bulbs are connected in parallel is

$P=P_{1}+P_{2}$

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