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Question

(c)

Power requirement for $1$ laptop, $P_1=90 \,W$

So, power requirement for $10$ laptops,

$P=10 	imes P_1=10 	imes 90$

$=900 W =0.9 \,k

Vedclass · Accepted Answer

$I$ and $III$ only

Vedclass · Answer

(c)

Power requirement for $1$ laptop, $P_1=90 \,W$

So, power requirement for $10$ laptops,

$P=10 	imes P_1=10 	imes 90$

$=900 W =0.9 \,kW$

In 5h, electrical energy used by all laptops,

$E=P 	imes t=0.9 	imes 5$

$=4.5 \,kWh$

Cost of electrical energy used is

$	ext { Cost }=E 	imes 	ext { Unit cost }$

$=4.5 	imes 5$

$=₹ 22.50$

So, statement $III$ is correct.

For laptop charger, input voltage is $220 \,V .$

So, current when all $10$ laptops are connected through an extension,

$I=\frac{P}{V}=\frac{900}{220} \approx 4.1 \,A$

As, line current exceeds current rating of fuse, therefore $3 A$ fuse cannot be used. So, statement $II$ is incorrect.

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