Question

Figure shows two capacitors connected in series and joined to a battery. The graph shows the variation in potential as one moves from left to right on the branch containing the capacitors.

• A. $C_1>C_2$
  
• B. $C_1=C_2$
  
• C. $C_1<C_2$
  
• D. The information is not sufficient to decide the relation between $C_1$ and $C_2$.

Answer 1

The correct option is C $C_1<C_2$


As the two capacitors are in series, the charge on both of them will be equal. $Q=CV$ concludes that the capacitor which has a lower potential drop across it , will have more capacitance and vice versa. From the graph it’s evident that $V$ across $C_1$ is more and hence $C_1<C_2$

Hence, option (c) is the correct alternative.

Key concept- In series combination, charge on the capacitors are the same. And don’t forget, $Q=CV$