Question

Two electric bulbs are connected one by one across potential difference V. At that time power consumed in them are $P_1 \, and \, P_2$ respectively. Now, if potential difference V is applied across series combination of these bulbs, what will be total power consumed ?

• A. $\frac {P_1P_2}{P_1+P_2}$
• B. $P_1P_2$
• C. $P_1+P_2$
• D. $\sqrt {P_1P_2}$

Answer 1

Answer: (A)

$\frac {P_1P_2}{P_1+P_2}$

Answer 2

Let $R_1 \, and \, R_2$ be the resistances of the bulbs. Power consumed
\hspace15mm P=\frac {V^2}{R}
So, the power consumed in first bulb is
\hspace15mm P_1=\frac {V^2}{R_1} \hspace15mm ...(i)
and power consumed in second bulb is
\hspace15mm P_2=\frac {V^2}{R_2} \hspace15mm ...(ii)
If these two bulbs are combined in series. So, the total resistance of the combination is
\hspace15mm R=R_1+R_2
Hence, the power consumed in combination is
\hspace15mm P= \frac {V^2}{(R_1+R_2)} \hspace15mm ...(iii)
Now, from Eqs. (i) and (ii)
\hspace15mm \frac {1}{P_1}+ \frac {1}{P_2}= \frac {R_1}{V^2}+ \frac {R_2}{V^2}
\Rightarrow \hspace15mm \frac {1}{P_1}+ \frac {1}{P_2}= \frac {R_1+R_2}{V^2}
\Rightarrow \hspace15mm \frac {1}{P_1}+ \frac {1}{P_2}= \frac {1}{V^2/R_1+R_2}
From E (iii)
\hspace15mm \frac {1}{P_1}+ \frac {1}{P_2}= \frac {1}{P}
\Rightarrow \hspace5mm \frac {P_2+P_1}{P_1P_2}= \frac {1}{P}
\Rightarrow \hspace15mm P=\frac {P_1P_2}{P_1+P_2}