Four bulbs B_1, B_2, B_3 and B_4 of 100, W ea | vedclass

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(C) The power of each bulb is $P = 100\, W$ and the voltage is $V = 220\, V$.The current flowing through each bulb is given by $I = \frac{P}{V} = \fra

Vedclass · Accepted Answer

$1.35$

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(C) The power of each bulb is $P = 100\, W$ and the voltage is $V = 220\, V$.The current flowing through each bulb is given by $I = \frac{P}{V} = \frac{100}{220} \approx 0.4545\, A$.From the circuit diagram,the ammeter is connected in series with the three bulbs $B_2, B_3$,and $B_4$,while bulb $B_1$ is connected in parallel to this combination.Therefore,the current through the ammeter is the sum of the currents flowing through bulbs $B_2, B_3$,and $B_4$.Total current $I_{total} = I_{B_2} + I_{B_3} + I_{B_4} = 3 	imes 0.4545\, A = 1.3636\, A$.Rounding to two decimal places,the reading is approximately $1.36\, A$. Given the options,$1.35\, A$ is the intended answer based on the approximation $100/220 \approx 0.45\, A$.

Four bulbs $B_1, B_2, B_3$ and $B_4$ of $100\, W$ each are connected to a $220\, V$ main supply as shown in the figure. The reading in an ideal ammeter will be ............... $A$.

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