Alternating Current

Step: Find the phase difference between the current and the voltage across the capacitor.
The circuit consists of a resistor \((R)\) in series with a capacitor \((C)\) and another resistor \((R)\) in parallel below it.
This implies the current in the vertical branch (the \(R\) at \(X\)) is zero. Therefore, the voltage at the junction \(C\) must be zero (the same as the ground potential at \(B\)).
The given condition states that the RMS voltage across \(R\) is equal to the RMS voltage across \(C.\)
For an RC circuit, the voltage across \(R\) and \(C\) are in quadrature (\(90^\circ\) phase difference).
The voltage across \(R\): \(V_R = I R,\) in phase with the current. 
The voltage across \(C\): \(V_C = I / (\omega C),\) leading the current by \(90^\circ.\)

The current through the capacitor \(C\) leads the voltage across \(C\) by \(90^\circ\) in a purely capacitive branch. The phase difference between the current through the capacitor and the voltage across it is \(90^\circ.\)
Hence, option (3) is the correct answer.

Hint: For an ideal transformer; \( I_pV_p=I_sV_s
\)

Step: Find the current in the primary circuit. 
The power in the secondary circuit is given by;
\(P=VI\)
\(P_s=I_sV_s=44~\text W\)
The current in the primary circuit is given by;
As there is no power loss we know that;
\( I_pV_p=I_sV_s
\)
\(\Rightarrow I_p=\frac{I_sV_s}{V_p}\)
\(\Rightarrow I_p=\frac{44}{220}=0.2~\text A\)
Hence, option (3) is the correct answer.