Q143 / 150 · MHT CET 2026 (19 April Shift 2) · Physics
When a.c. source is connected across a pure capacitor, the correct phase relation between current and emf is shown in figure.
- A(C)
- B(A)
- Correct answerC(B)
- D(D)
Solution
In an AC circuit containing a pure capacitor, the alternating current leads the alternating emf by a phase angle of $\dfrac{\pi}{2}$ or $90^{\circ}$.
Mathematically, if the emf is given by $e_c = E_0 \sin(\omega t)$, the current is given by $I_c = I_0 \sin\left(\omega t + \dfrac{\pi}{2}\right)$.
In a phasor diagram, the phasor representing the current $I_c$ must be rotated counterclockwise by $90^{\circ}$ relative to the phasor representing the emf $e_c$.
Analyzing the given figures:
(A) $I_c$ and $e_c$ are in phase (represents a pure resistor).
(B) $I_c$ leads $e_c$ by $90^{\circ}$ (represents a pure capacitor).
(C) $I_c$ lags $e_c$ by $90^{\circ}$ (represents a pure inductor).
(D) $I_c$ and $e_c$ are out of phase by $180^{\circ}$.
Therefore, figure (B) correctly shows the phase relation for a pure capacitor.
Answer: (B)
Mathematically, if the emf is given by $e_c = E_0 \sin(\omega t)$, the current is given by $I_c = I_0 \sin\left(\omega t + \dfrac{\pi}{2}\right)$.
In a phasor diagram, the phasor representing the current $I_c$ must be rotated counterclockwise by $90^{\circ}$ relative to the phasor representing the emf $e_c$.
Analyzing the given figures:
(A) $I_c$ and $e_c$ are in phase (represents a pure resistor).
(B) $I_c$ leads $e_c$ by $90^{\circ}$ (represents a pure capacitor).
(C) $I_c$ lags $e_c$ by $90^{\circ}$ (represents a pure inductor).
(D) $I_c$ and $e_c$ are out of phase by $180^{\circ}$.
Therefore, figure (B) correctly shows the phase relation for a pure capacitor.
Answer: (B)
Hence, the correct answer is option: 'C'.