Electromagnetic Induction and Alternating Current 6 Question 11

11. In the given circuit, the AC source has ω=100rad/s. Considering the inductor and capacitor to be ideal, the correct choice(s) is(are)

(2012)

(a) the current through the circuit, I is 0.3A

(b) the current through the circuit, I is 0.32A

(c) the voltage across 100Ω resistor =102V

(d) the voltage across 50Ω resistor =10V

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Answer:

Correct Answer: 11. (a,c)

Solution:

Circuit 1

XC=1ωC=100ΩZ1=(100)2+(100)2=1002Ωφ1=cos1R1Z1=45

In this circuit, current leads the voltage.

I1=VZ1=201002=152AV100Ω=(100)I1=(100)152V=102V

Circuit 2

XL=ωL=(100)(0.5)=50ΩZ2=(50)2+(50)2=502Ωφ2=cos1R2Z2=45

In this circuit, voltage leads the current.

I2=VZ2=20502=25AV50Ω=(50)I2=5025=102V

Further, I1 and I2 have a mutual phase difference of 90.

I=I12+I22=0.34



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