$A$ light bulb is rated at $100\;W$ for a $220\;V$ supply. Find
$(a)$ the resistance of the bulb;
$(b)$ the peak voltage of the source; and
$(c)$ the rms current through the bulb.
$(a)$ the resistance of the bulb;
$(b)$ the peak voltage of the source; and
$(c)$ the rms current through the bulb.
(N/A) Given power $P = 100\;W$ and rms voltage $V = 220\;V$.
The resistance of the bulb is calculated as:
$R = \frac{V^2}{P} = \frac{(220\;V)^2}{100\;W} = 484\;\Omega$
$(b)$ The peak voltage $(V_m)$ of the source is given by:
$V_m = \sqrt{2} \times V = 1.414 \times 220\;V \approx 311\;V$
$(c)$ The rms current $(I)$ through the bulb is calculated using $P = I \times V$:
$I = \frac{P}{V} = \frac{100\;W}{220\;V} \approx 0.454\;A$
The resistance of the bulb is calculated as:
$R = \frac{V^2}{P} = \frac{(220\;V)^2}{100\;W} = 484\;\Omega$
$(b)$ The peak voltage $(V_m)$ of the source is given by:
$V_m = \sqrt{2} \times V = 1.414 \times 220\;V \approx 311\;V$
$(c)$ The rms current $(I)$ through the bulb is calculated using $P = I \times V$:
$I = \frac{P}{V} = \frac{100\;W}{220\;V} \approx 0.454\;A$
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