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The quantity of a charge that will be transferred by a current flow of \( 20 \) A over \( 1 \) hour \( 30 \)
minutes period is
Options:
minutes period is
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Verified Answer
The correct answer is:
\( 10.8 \times 10^{4} C \)
Given, $I=20 \mathrm{~A} ; \mathrm{t}=1$ hour 30 minutes $=90$ minutes $=90 \times 60$ seconds
Therefore, charge transferred
$Q=I \times t=20 \times 90 \times 60=108 \times 10^{3} \mathrm{C}$
Thus, $Q=10.8 \times 10^{4} \mathrm{C}$
Therefore, charge transferred
$Q=I \times t=20 \times 90 \times 60=108 \times 10^{3} \mathrm{C}$
Thus, $Q=10.8 \times 10^{4} \mathrm{C}$
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