Question

Four capacitors are connected to a battery as shown in figure. The ratio of charges on capacitors \(C_2\) and \(C_4\) is

• A. \(\frac{1}{14}\)
• B. \(\frac{3}{22}\)
• C. \(\frac{2}{9}\)
• D. \(\frac{4}{13}\)

Hint:

Find the series equivalent of C1, C2, C3 across V for the charge on C2, and use Q = CV for C4.

Answer 1

The Correct Option is B

Step 1: Read the circuit:
In the figure, C3 (3C), C2 (2C) and C1 (1C) lie one after another in the outer loop, so they are in series. C4 (4C) is on the middle wire joining the two sides, so it is in parallel with that series branch. The battery V sits across both.

Step 2: Series branch:
\[ \frac{1}{C_s} = \frac{1}{C} + \frac{1}{2C} + \frac{1}{3C} = \frac{6+3+2}{6C} = \frac{11}{6C} \Rightarrow C_s = \frac{6C}{11} \]

Step 3: Charge on C2:
Series capacitors carry the same charge, which equals the charge on the equivalent capacitor. The branch is across V, so \(Q_2 = C_sV = \frac{6CV}{11}\).

Step 4: Charge on C4:
C4 is directly across V, so \(Q_4 = 4CV\).

Step 5: Ratio:
\[ \frac{Q_2}{Q_4} = \frac{6CV/11}{4CV} = \frac{6}{44} = \frac{3}{22} \]

Step 6: Check the options:
\(1/14\), \(2/9\) and \(4/13\) do not reduce from \(6/(11\times4)\). A common slip is to give C2 the full V, which would give a ratio of \(1/2\); that is not an option either.

Final Answer:
The ratio \(Q_2 : Q_4 = 3/22\), option (B). \[ \boxed{\frac{3}{22}} \]