Question

Two $75\mathrm{W}(120\mathrm{V})$ lightbulbs are wired in series, then the combination is connected to a $120\mathrm{V}$ supply. How much power is dissipated by each bulb?

Short answer

Power dissipated by each bulb is$18.45\mathrm{W}.$

Step-by-step solution

Given information

We have given,

For one bulb the power dissipation is $75\mathrm{W}$at $120\mathrm{V}.$

We have to find the power dissipated by the each bulb.

Simplify

We know that Power dissipated by one bulb only without connected the other bulb in circuit is given by,

$$\begin{gathered} \mathrm{P}=\frac{{\mathrm{V}}^2}{\mathrm{R}} \\ \mathrm{R}=\frac{{\mathrm{V}}^2}{\mathrm{P}} \\ \mathrm{R}=\frac{{\left(102\right)}^2}{75}=192\Omega \end{gathered}$$

then total resistance in the circuit after connected both the bulb will be equal to the equivalent resistance,

${\mathrm{R}}_{\mathrm{eq}}=2\times 192=384\Omega$

Due to equivalent resistances in series then current flowing in the circuit will be after connecting both bulb,

$$\begin{gathered} {\mathrm{IR}}_{\mathrm{eq}}=\mathrm{V} \\ \mathrm{I}=\frac{120}{384}=0.31\mathrm{A} \end{gathered}$$

Then power dissipated by one bulb in the circuit will be

$\mathrm{P}={\mathrm{I}}^2\mathrm{R}={(0.31)}^{2}192=18.45\mathrm{W}$