Question 1 of 16beginner🔧 ApplyNumerical3 marks

A light bulb is rated at 100W 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)
Answer

Resistance of the bulb We are given power P=100 WP = 100 \text{ W} and voltage V=220 VV = 220 \text{ V}. Using the formula for power P=V2/RP = V^2 / R, we rearrange it to find the resistance:

R=V2P=(220 V)2100 W=484ΩR = \frac{V^2}{P} = \frac{(220 \text{ V})^2}{100 \text{ W}} = 484 \Omega
(b)
Answer

Peak voltage of the source The given voltage of 220 V is the rms value. The peak voltage (vmv_m) is related to the rms voltage by the relation vm=2Vv_m = \sqrt{2} V.

vm=2×220 V=1.414×220 V=311 Vv_m = \sqrt{2} \times 220 \text{ V} = 1.414 \times 220 \text{ V} = 311 \text{ V}
(c)
Answer

RMS current through the bulb Using the relation P=IVP = IV, we can calculate the rms current:

I=PV=100 W220 V=0.454 AI = \frac{P}{V} = \frac{100 \text{ W}}{220 \text{ V}} = 0.454 \text{ A}

Explanation

The solution is derived directly from the provided context for Example 7.1. Part (a) utilizes the power formula P=V2/RP=V^2/R to solve for resistance. Part (b) uses the standard relationship between peak and rms values for an AC supply, where peak voltage is 2\sqrt{2} times the rms voltage. Part (c) rearranges the power formula P=IVP=IV to solve for current.

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