In a circuit consisting of a capacitance and a generator with alternating emf E g = E g₀ sinωt, V C and I C are the voltage and current. Correct phasor diagram for such circuit is

Solution

Physics sir · 3 steps

For a capacitor, the current must reach its peak first, so draw I_C a quarter-turn ahead of V_C.

1. Current from charge. The capacitor voltage equals the generator emf, V_C = E_(g₀)sin ω t. The charge is q = CV_C, and the current is I_C = dq/dt.

   I_C = ω C E_(g₀)cos ω t = ω C E_(g₀)sin(ω t + π/2)

2. Read the phase. The extra +π/2 means the current leads the voltage by 90°. Phasors rotate anticlockwise, so I_C must sit 90° anticlockwise from V_C.

   Option	Where I_C is drawn	Phase of I_C
   (a)	below the axis, clockwise from V_C	lags V_C
   (b)	along V_C	in phase
   (c)	at a right angle, anticlockwise from V_C	leads by 90°
   (d)	opposite to V_C	180° out of phase

3. So. Only (c) shows I_C perpendicular to V_C and ahead of it in the anticlockwise sense. The answer is (c).

The idea to keep. Capacitor: current leads voltage by 90°. Inductor: current lags by 90°. Resistor: in phase. On a phasor diagram, leading means further round in the anticlockwise direction.

Answer from the NTA final answer key. Solution by PrepWiser.