Three capacitors of capacitances \( 2\,\mu\mathrm{F} \), \( 3\,\mu\mathrm{F} \) and \( 6\,\mu\mathrm{F} \) are connected in series. The combination is connected to a 12 V supply. The respective potential differences across them will be
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 6\,\mathrm{V},\ 4\,\mathrm{V} \) and \( 2\,\mathrm{V} \)
Capacitors in series carry the same charge \( Q \), and the supply voltage is shared in inverse proportion to the capacitances, since \( V = Q/C \).
\( \displaystyle \frac{1}{C_{\mathrm{eq}}} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3 + 2 + 1}{6} = 1 \;\Rightarrow\; C_{\mathrm{eq}} = 1\,\mu\mathrm{F} \)
\( \displaystyle Q = C_{\mathrm{eq}}\,V = 1\,\mu\mathrm{F} \times 12\,\mathrm{V} = 12\,\mu\mathrm{C} \)
\( V_1 = Q/C_1 = 12/2 = 6\,\mathrm{V} \)
\( V_2 = Q/C_2 = 12/3 = 4\,\mathrm{V} \)
\( V_3 = Q/C_3 = 12/6 = 2\,\mathrm{V} \)
The three voltages add up to \( 12\,\mathrm{V} \), as they must.