If a 30, V, 90, W bulb is to be operated on a | vedclass

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(C) The rated power of the bulb is $P = 90\, W$ at a voltage of $V_b = 30\, V$. The current $i$ flowing through the bulb when operating at its rated p

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$30$

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(C) The rated power of the bulb is $P = 90\, W$ at a voltage of $V_b = 30\, V$. The current $i$ flowing through the bulb when operating at its rated power is given by $i = \frac{P}{V_b} = \frac{90\, W}{30\, V} = 3\, A$.Since the bulb is connected in series with a resistor $R$ to a $120\, V$ supply,the voltage drop across the resistor must be $V_R = 120\, V - 30\, V = 90\, V$.Using Ohm's law for the resistor,$V_R = i 	imes R$,we get $90\, V = 3\, A 	imes R$.Therefore,$R = \frac{90\, V}{3\, A} = 30\, \Omega$.

If a $30\, V, 90\, W$ bulb is to be operated on a $120\, V$ line,what resistance in ohms should be connected in series with the bulb?

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