Explanation:
By Joule's law, the power
dissipated through a resistor \(R\), having
a potential difference \(V\) is
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,P = \frac{{{V^2}}}{{{R^\prime }}}\)
When bulbs are connected in series, then
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{R^\prime } = R + R + R = 3R\)
power dissipated,
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,P = \frac{{{V^2}}}{{3R}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 1 \right)\)
When they are connected in parallel, then
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{1}{{{R^{\prime \prime }}}}{\text{ }} = \frac{1}{R} + \frac{1}{R} + \frac{1}{R} = \frac{3}{R}\)
\( \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{R^{\prime \prime }} = \frac{R}{3}\)
Power dissipated,
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{P^\prime } = \frac{{{V^2}}}{{R/3}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 2 \right)\)
From Eqs. (1) and (2), we have
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{P^\prime } = 9P\)
