Question

when electric bulbs of same power,but different marked voltage are connected in series across the power source,their brightness will be

1.proportional to their marked voltage

2.inversely proportional to their marked voltage

3.proportional to the square of their marked voltages

4.inversely proportional to the square of their marked voltage

Answer 1

Here we take, $V_R$ is rated voltage and $P_R$ is rated power, then resistance of bulb,

Let $V_{\left(R\right)1}$ and $V_{\left(R\right)2}$ be the rated voltages of the bulbs and $P_R$ be their rated powers:

$R_1=\frac{{V^2}_{\left(R\right)1}}{P_R}$ ; $R_2=\frac{V_{\left(R\right)2}^2}{P_R}$

Current through circuit, $I=\frac{V}{R_1+R_2}=\frac{V}{\frac{V_{\left(R\right)1}^2}{P_R}}+\frac{V}{\frac{V_{\left(R\right)2}^2}{P_R}}=\frac{V{P}_R}{V_{\left(R\right)1}^2+V_{\left(R\right)2}^2}$

Brightness of bulb is directly proportional to power dissipated across it.

So, brightness of bulb 1 $I^2{R}_1\text{∝}$ ${\left(\frac{V{P}_R}{V_{\left(R\right)1}^2+V_{\left(R\right)2}^2}\right)}^2$ $\times \frac{V_{\left(R\right)1}^2}{P_R}\propto \frac{V^2{P}_R}{1+\frac{V_{\left(R\right)2}^2}{V_{\left(R\right)1}^2}}$

Similarly,

Brightness of bulb 2 $I^2{R}_2\text{∝}$ ${\left(\frac{V{P}_R}{V_{\left(R\right)1}^2+V_{\left(R\right)2}^2}\right)}^2$ $\times \frac{V_{\left(R\right)2}^2}{P_R}\propto \frac{V^2{P}_R}{1+\frac{V_{\left(R\right)2}^2}{V_{\left(R\right)1}^2}}$

Hence, when electric bulbs of same power,but different marked voltage are connected in series across the power source,their brightness will be:

inversely proportional to the square of their marked voltage