Two parallel plate capacitors of capacitances \( 2\,\mu\mathrm{F} \) and \( 3\,\mu\mathrm{F} \) are joined in series and the combination is connected to a battery of V volts. The values of potential across the two capacitors \( V_1 \) and \( V_2 \) and energy stored in the two capacitors \( U_1 \) and \( U_2 \) respectively are related as _______________________.
Fill in the blank with the correct answer from the options given below.
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \displaystyle\frac{V_1}{V_2} = \frac{U_1}{U_2} = \frac{3}{2} \)
Capacitors in series carry the same charge \( Q \). So the potential \( V = Q/C \) and the energy \( U = Q^{2}/2C \) are both inversely proportional to the capacitance.
\( C_1 = 2\,\mu\mathrm{F} \)
\( C_2 = 3\,\mu\mathrm{F} \)
\( \displaystyle \frac{V_1}{V_2} = \frac{Q/C_1}{Q/C_2} = \frac{C_2}{C_1} = \frac{3}{2} \)
\( \displaystyle \frac{U_1}{U_2} = \frac{Q^{2}/2C_1}{Q^{2}/2C_2} = \frac{C_2}{C_1} = \frac{3}{2} \)