In the above circuit, \(C = \sqrt {\frac{3}{2}} \;\mu F,{R_2} = 20\Omega ,\;L = \frac{{\sqrt 3 }}{{10}}H\) and R1 = 10 Ω Current in L – R1 path is I1 and in C - R2 path is I2. The voltage of AC source is given by \(V = 200\sqrt 2 \;sin\left( {100t} \right)\) difference between I1 and I2 is

In the above circuit, \(C = \sqrt {\frac{3}{2}} \;\mu F,{R_2} = 20\Ome

1.	In the above circuit, \(C = \sqrt {\frac{3}{2}} \;\mu F,{R_2} = 20\Omega ,\;L = \frac{{\sqrt 3 }}{{10}}H\) and R1 = 10 Ω Current in L – R1 path is I1 and in C - R2 path is I2. The voltage of AC source is given by \(V = 200\sqrt 2 \;sin\left( {100t} \right)\) difference between I1 and I2 is
A.	30°
B.	60°
C.	150°
D.	90°
Answer» D. 90°

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In the above circuit, \(C = \sqrt {\frac{3}{2}} \;\mu F,{R_2} = 20\Omega ,\;L = \frac{{\sqrt 3 }}{{10}}H\) and R1 = 10 Ω Current in L – R1 path is I1 and in C - R2 path is I2. The voltage of AC source is given by \(V = 200\sqrt 2 \;sin\left( {100t} \right)\) difference between I1 and I2 is
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In the above circuit, \(C = \sqrt {\frac{3}{2}} \;\mu F,{R_2} = 20\Omega ,\;L = \frac{{\sqrt 3 }}{{10}}H\) and R1 = 10 Ω Current in L – R1 path is I1 and in C - R2 path is I2. The voltage of AC source is given by \(V = 200\sqrt 2 \;sin\left( {100t} \right)\) difference between I1 and I2 is

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