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Question

An ac source producing emf e=E0[cos(100πt)+cos(500πt)] is connected in series with a capacitor and a resistor. The steady state current in the circuit is found to be i=I1cos(100πt+φ1)+I2cos(500πt+φ2)

A
I1>I2
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B
I1=I2
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C
I1<I2
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D
nothing can be said
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Solution

The correct option is D I1<I2
We can assume two different sources connected in series with same orientation.
E1=E0cos100πt
E2=E0cos500πt
such that e=E0cos100πt+E0cos500πt
becomes,e=E1+E2
Now, solving using principle of superposition we get,
i1=100πCcos(100πt+ϕ1)
i2=500πCcos(500πt+ϕ2)
Final current will be,
i=i1+i2
From this we get
I1=100πC
and I2=500πC
From this we get,
I1<I2.

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