Question

An ac source producing emf $e=E_0\left[cos\right(100\pi t)+cos(500\pi t\left)\right]$ is connected in series with a capacitor and a resistor. The steady state current in the circuit is found to be $i=I_{1}cos(100\pi t+{\phi }_1)+I_{2}cos(500\pi t+{\phi }_2)$

• A. $I_1>I_2$
• B. $I_1=I_2$
• C. $I_1<I_2$
• D. nothing can be said

Answer 1

Answer: (C)

$I_1<I_2$

We can assume two different sources connected in series with same orientation.

$E_1=E_0\cos 100\pi t$

$E_2=E_0\cos 500\pi t$

such that $e=E_0\cos 100\pi t+E_0\cos 500\pi t$

becomes,$e=E_1+E_2$

Now, solving using principle of superposition we get,

$i_1=100\pi C\cos (100\pi t+{\varphi }_1)$

$i_2=500\pi C\cos (500\pi t+{\varphi }_2)$

Final current will be,

$i=i_1+i_2$

From this we get

$I_1=100\pi C$

and $I_2=500\pi C$

From this we get,

$I_1<I_2$.