MHT CET · Physics · Capacitance
Two condensers \(C_1\) and \(C_2\) in a circuit are joined as shown below.
The potential of point A is \(V_1\) and that of B is \(V_2\). The potential of point D will be
- A \(\frac{C_1 V_2+C_2 V_1}{C_1+C_2}\)
- B \(\frac{1}{2}\left(V_1+V_2\right)\)
- C \(\frac{C_2 V_1-C_1 V_2}{C_1+C_2}\)
- D \(\frac{C_2\left(V_1+V_2\right)}{C_1+C_2}\)
Answer & Solution
Correct Answer
(D) \(\frac{C_2\left(V_1+V_2\right)}{C_1+C_2}\)
Step-by-step Solution
Detailed explanation
\(V_2-V_1\) will be divided between \(C_1\) and \(C_2\). Let \(V_D\) be potential at point D.
Let \(\mathrm{Q}\) be the charge on each capacitor. Then the potential drop can be written as:
\(\begin{aligned} & V_2-V_1=V_2-V_D+V_D-V_1 \\ & \Rightarrow V_2-V_1=\frac{Q}{C_2}+\frac{Q}{C_1} \\ & \Rightarrow Q=\frac{C_1 C_2\left(V_2-V_1\right)}{C_1+C_2}\end{aligned}\)
Potential difference across \(C_1\) is
\(V_D=V_1+\frac{C_2\left(V_2-V_1\right)}{C_1+C_2}=\frac{C_2\left(V_1+V_2\right)}{C_1+C_2}\)
Let \(\mathrm{Q}\) be the charge on each capacitor. Then the potential drop can be written as:
\(\begin{aligned} & V_2-V_1=V_2-V_D+V_D-V_1 \\ & \Rightarrow V_2-V_1=\frac{Q}{C_2}+\frac{Q}{C_1} \\ & \Rightarrow Q=\frac{C_1 C_2\left(V_2-V_1\right)}{C_1+C_2}\end{aligned}\)
Potential difference across \(C_1\) is
\(V_D=V_1+\frac{C_2\left(V_2-V_1\right)}{C_1+C_2}=\frac{C_2\left(V_1+V_2\right)}{C_1+C_2}\)
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