KCET · Physics · Current Electricity
The quantity of a charge that will be transferred by a current flow of \( 20 \) A over \( 1 \) hour \( 30 \)
minutes period is
- A \( 10.8 \times 10^{3} C \)
- B \( 10.8 \times 10^{4} C \)
- C \( 5.4 \times 10^{3} \mathrm{C} \)
- D \( 1.8 \times 10^{4} \mathrm{C} \)
Answer & Solution
Correct Answer
(B) \( 10.8 \times 10^{4} C \)
Step-by-step Solution
Detailed explanation
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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