91Ó°ÊÓ

According to ohm’s law the relation between the passing current and the potential drop across the wire can be written as :

V = lR

Here I is the current, V is the potential and R is the resistance of the wire.

Let$P_1=25W$ and $P_2=200W$.

In series, the same current flows through both bulbs. The power consumed by the resistor in the circuit depends on the resistance of the resistor and this resistor could be calculated as:

$$\begin{gathered} R_1=\frac{V^2}{P_1} \\ =\frac{{\left(120V\right)}^2}{25W} \\ =576\mathrm{Î\copyright } \end{gathered}$$

And

$$\begin{gathered} R_2=\frac{V^2}{P_2} \\ =\frac{{\left(120V\right)}^2}{200W} \\ =72\mathrm{Î\copyright } \end{gathered}$$

From the above calculations, the bulb 25 W has a higher resistance, therefore, it has a higher voltage drop. The higher voltage bulb will burn the 25 W bulb. As both bulbs are connected in series across 240 V, we can get the current flows through both using Ohm’s law.

$$\begin{gathered} l=\frac{V}{R_1+R_2} \\ =\frac{240V}{576\mathrm{Î\copyright }+72\mathrm{Î\copyright }} \\ =0.37A \end{gathered}$$

The voltage drop across 25 W will be

$$\begin{gathered} V_1=l{R}_1 \\ =\left(0.37A\right)\left(576\mathrm{Î\copyright }\right) \\ =213V \end{gathered}$$

This voltage drop is enough to burn the bulb.

Thus, 25-W bulb burned out immediately.