How much power is used in moving a $0.03 C$ charge from point A at a potential of $40 V$ to point B at $60 V$ , if this takes 2 seconds?

0.30 W

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0.15 W

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0.6 W

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0.012 W

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Solution

The correct option is A $0.30 W$
Given : $q = 0.03$ C $V_{A} = 40$ V $V_{B} = 60$ V $t = 2$ s
Work done in moving a charge from A to B $W = q (V_{B} - V_{A}) = 0.03 \times (60 - 40) = 0.6$ J
$∴$ Power used $P = \frac{W}{t} = \frac{0.6}{2} = 0.3$ $W$

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How much power is used in moving a 0.03C charge from point A at a potential of 40V to point B at 60V, if this takes 2 seconds?

The correct option is A 0.30WGiven : q=0.03 C VA=40 V VB=60 V t=2 sWork done in moving a charge from A to B W=q(VB−VA)=0.03×(60−40)=0.6 J∴ Power used P=Wt=0.62=0.3 W

How much power is used in moving a $0.03 C$ charge from point A at a potential of $40 V$ to point B at $60 V$ , if this takes 2 seconds?

The correct option is A $0.30 W$
Given : $q = 0.03$ C $V_{A} = 40$ V $V_{B} = 60$ V $t = 2$ s
Work done in moving a charge from A to B $W = q (V_{B} - V_{A}) = 0.03 \times (60 - 40) = 0.6$ J
$∴$ Power used $P = \frac{W}{t} = \frac{0.6}{2} = 0.3$ $W$