MHT CET · Physics · Capacitance
The network of six capacitors is as shown in figure. The equivalent capacitance between \(\mathrm{A}\) and \(\mathrm{B}\) is
- A \(\frac{2 C}{3}\)
- B \(\frac{4 C}{3}\)
- C 2C
- D 3C
Answer & Solution
Correct Answer
(D) 3C
Step-by-step Solution
Detailed explanation
Consider the following diagram with nodes Q having same potential due to the symmetry of the circuit:
Taking the similar nodes together the equivalent diagram can be drawn as:
Therefore,
the eqiuvalent capacitance between \(\mathrm{AQ}=3 C+2 C+C=6 C\)
the equivalent capacitance between \(\mathrm{QB}=3 C+2 C+C=6 C\)
Now, for equivalent capacitance between \(A B\)
\(\begin{aligned}
& \frac{1}{C_{\mathrm{AB}}}=\frac{1}{6 C}+\frac{1}{6 C}=\frac{1}{3 C} \\
& \therefore C_{\mathrm{AB}}=3 C
\end{aligned}\)
Taking the similar nodes together the equivalent diagram can be drawn as:
Therefore,
the eqiuvalent capacitance between \(\mathrm{AQ}=3 C+2 C+C=6 C\)
the equivalent capacitance between \(\mathrm{QB}=3 C+2 C+C=6 C\)
Now, for equivalent capacitance between \(A B\)
\(\begin{aligned}
& \frac{1}{C_{\mathrm{AB}}}=\frac{1}{6 C}+\frac{1}{6 C}=\frac{1}{3 C} \\
& \therefore C_{\mathrm{AB}}=3 C
\end{aligned}\)
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