An alternating voltage v(t)=220sin100πtvolt is applied to a purely resistive load of 5Ω. The time taken for the current to rise from half of the peak value to the peak value is
A
2.2ms
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B
5ms
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C
3.3ms
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D
7.2ms
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Solution
The correct option is C3.3ms V(t)=220sin(100πt)volt I(t)=2205sin(100πt)volt =44sin(100πt)volt
Comparing with I=I0sin(θ), I=I02 when θ=300 I=I0 when θ=900 ∴ phase to be covered θ=600=π3
Time taken, t=θω=π3100π=1300sec =3.3ms.