Question

If certain number of bulbs rated as ($P_1$ watt, V volt), ($P_2$ watt, V volt)... are connected in series across a potential difference of V volt, then power 'P' consumed by all bulbs is given as

• A. $P=P_1+P_2+P_3$ ........
• B. $\frac{1}{P}=\frac{1}{P_1}+\frac{1}{P_2}+\frac{1}{P_3}$ ........
• C. $P^2=P_1^2+P_2^2+P_3^2$ ........
• D. None of these

Answer 1

Answer: (B)

$\frac{1}{P}=\frac{1}{P_1}+\frac{1}{P_2}+\frac{1}{P_3}$ ........

Power consumed in one bulb rated as $P_{1}W,Vvolt$ having resistance of $R_1\Omega$ is
$P_1=\frac{V^2}{R_1}$ .........(1)
For certain number of bulbs we can write,
$P_2=\frac{V^2}{R_2}$ .........(2)
$P_3=\frac{V^2}{R_3}$ and so on..........(3)
We know in series combination resistances get added, hence

$R_{eq}=R_1+R_2+R_3+...........$

dividing both sides by $V^2$, we get

$\frac{R_{eq}}{V^2}=\frac{R_1}{V^2}+\frac{R_2}{V^2}+\frac{R_3}{V^2}+........$

Referring equations (1),(2) and (3)

$\frac{1}{P_{eq}}=\frac{1}{P_1}+\frac{1}{P_2}+\frac{1}{P_3}+........$