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)

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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