An inductor of reactance XL=4Ω and resistor of resistance R=3Ω are connected in series with a voltage source of emf ε=(20V)[sin(100πrad/s)t]. The current in the circuit at any time t will be?
A
I=(4A)[sin(100πrad/s)t+37o]
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B
I=(4A)[sin(100πrad/s)t−37o]
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C
I=(4A)[sin(100πrad/s)t+53o]
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D
I=(4A)[sin(100πrad/s)t−53o]
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Solution
The correct option is AI=(4A)[sin(100πrad/s)t+37o] The inductor and the resistor are in series. Given, RL=4Ω,R=3Ω
Resultant impedance R=√R2L+R2=√42+32=5Ω
In RL circuit, current due to inductor leads by resistance.
Therefore phase angle, ϕ=tan−134
= 370
Peak voltage, eo=20V
Current, I=e0R=205=4A
Since the current is leading. Therefore value of I is 4A[sin(100π)t+370]