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A30V-90W lamp is operated on a $120 \mathrm{VDC}$ line. A resistor is connected in series with the lamp in order to glow it properly. The value of resistance is
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The correct answer is:
$30 \Omega$
Resistance of lamp
$\mathrm{R}_{0}=\frac{\mathrm{V}^{2}}{\mathrm{P}}=\frac{(30)^{2}}{90}=10 \Omega$
Current in the lamp
$\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}_{0}}=\frac{30}{10}=3 \mathrm{~A}$
As the lamp is operated on $120 \mathrm{~V} \mathrm{DC}$, then resistance becomes
$\mathrm{R}^{\prime}=\frac{\mathrm{V}^{\prime}}{\mathrm{i}}=\frac{120}{3}=40 \Omega$
For proper glow, a resistance $\mathrm{R}$ is joined in series with the bulb
$\begin{aligned}
& \mathrm{R}^{\prime}=\mathrm{R}+\mathrm{R}_{0} \\
\Rightarrow & \mathrm{R}^{\alpha}=\mathrm{R}^{\prime}-\mathrm{R}_{0}=40-10=30 \Omega
\end{aligned}$
$\mathrm{R}_{0}=\frac{\mathrm{V}^{2}}{\mathrm{P}}=\frac{(30)^{2}}{90}=10 \Omega$
Current in the lamp
$\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}_{0}}=\frac{30}{10}=3 \mathrm{~A}$
As the lamp is operated on $120 \mathrm{~V} \mathrm{DC}$, then resistance becomes
$\mathrm{R}^{\prime}=\frac{\mathrm{V}^{\prime}}{\mathrm{i}}=\frac{120}{3}=40 \Omega$
For proper glow, a resistance $\mathrm{R}$ is joined in series with the bulb
$\begin{aligned}
& \mathrm{R}^{\prime}=\mathrm{R}+\mathrm{R}_{0} \\
\Rightarrow & \mathrm{R}^{\alpha}=\mathrm{R}^{\prime}-\mathrm{R}_{0}=40-10=30 \Omega
\end{aligned}$
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