The wiring of a house has a resistance of 6,O | vedclass

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(B) The rated voltage for both appliances is assumed to be $220\,V$.Resistance of the bulb,$R_{Bulb} = \frac{V^2}{P} = \frac{220^2}{100} = 484\,\Omega

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$23$

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(B) The rated voltage for both appliances is assumed to be $220\,V$.Resistance of the bulb,$R_{Bulb} = \frac{V^2}{P} = \frac{220^2}{100} = 484\,\Omega$.Resistance of the geyser,$R_{Geyser} = \frac{V^2}{P} = \frac{220^2}{1000} = 48.4\,\Omega$.$(i)$ When only the bulb is $ON$,the circuit consists of the wiring resistance $(6\,\Omega)$ in series with the bulb $(484\,\Omega)$.The potential drop across the bulb is $V_{Bulb1} = 220 	imes \frac{484}{484 + 6} = 220 	imes \frac{484}{490} \approx 217.3\,V$.$(ii)$ When the geyser is also switched $ON$,the bulb and geyser are in parallel. Their equivalent resistance is $R_p = \frac{484 	imes 48.4}{484 + 48.4} = \frac{23425.6}{532.4} = 44\,\Omega$.The circuit now consists of the wiring resistance $(6\,\Omega)$ in series with the parallel combination $(44\,\Omega)$.The potential drop across the parallel combination (and thus the bulb) is $V_{Bulb2} = 220 	imes \frac{44}{44 + 6} = 220 	imes \frac{44}{50} = 193.6\,V$.The change in potential drop is $\Delta V = V_{Bulb1} - V_{Bulb2} = 217.3 - 193.6 = 23.7\,V$.This is nearly $23\,V$.

The wiring of a house has a resistance of $6\,\Omega$. $A$ $100\,W$ bulb is glowing. If a geyser of $1000\,W$ is switched on,the change in potential drop across the bulb is nearly .................. $V$.

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