A coil of resistance 3 Ω and inductive reactance 4 Ω (at the supply frequency) is connected across a sinusoidal supply of 50 V (rms). A variable…

The coil admittance is $Y_L=\dfrac{1}{R+jX_L}=\dfrac{R-jX_L}{R^2+X_L^2}=\dfrac{3-4j}{25}\ \mathrm{S}$, i.e. conductance $G=\tfrac{3}{25}\ \mathrm{S}$ and inductive susceptance $B_L=\tfrac{4}{25}\ \mathrm{S}$ (lagging). The capacitor in parallel contributes a capacitive susceptance $B_C=1/X_C$ which is leading. For the line current to be in phase with the voltage the net susceptance must vanish: $B_C=B_L=\tfrac{4}{25}\ \mathrm{S}$ (so $X_C=6.25\ \Omega$). The total admittance is then purely conductive, $Y=G=\tfrac{3}{25}\ \mathrm{S}$, and the line current is $I=VG=50\times\dfrac{3}{25}=6\ \mathrm{A}$. For comparison, the coil alone draws $50/5=10\ \mathrm{A}$; the reactive component is supplied by the capacitor. The answer is $6$.