MHT CET · Physics · Capacitance
Three identical capacitors of capacitance ' \(\mathrm{C}\) ' each are connected in series and this connection is connected in parallel with one more such identical capacitor. Then the capacitance of whole combination is
- A \(3 \mathrm{C}\)
- B \(2 \mathrm{C}\)
- C \(\frac{4}{3} \mathrm{C}\)
- D \(\frac{3}{4} C\)
Answer & Solution
Correct Answer
(C) \(\frac{4}{3} \mathrm{C}\)
Step-by-step Solution
Detailed explanation
The equivalent capacitance of three capacitances connected in series is
\(\begin{aligned}
& \frac{1}{\mathrm{C}_{\text {eq }}}=\frac{1}{\mathrm{C}}+\frac{1}{\mathrm{C}}+\frac{1}{\mathrm{C}} \\
& \frac{1}{\mathrm{C}_{\text {eq }}}=\frac{3}{\mathrm{C}} \\
& \mathrm{C}_{\text {eq }}=\frac{\mathrm{C}}{3}
\end{aligned}\)
This capacitance is connected in parallel with another capacitance.
\(\begin{aligned}
& \Rightarrow C_{\text {tatal }}=\frac{C}{3}+C \\
& C_{\text {teal }}=\frac{4 C}{3}
\end{aligned}\)
\(\begin{aligned}
& \frac{1}{\mathrm{C}_{\text {eq }}}=\frac{1}{\mathrm{C}}+\frac{1}{\mathrm{C}}+\frac{1}{\mathrm{C}} \\
& \frac{1}{\mathrm{C}_{\text {eq }}}=\frac{3}{\mathrm{C}} \\
& \mathrm{C}_{\text {eq }}=\frac{\mathrm{C}}{3}
\end{aligned}\)
This capacitance is connected in parallel with another capacitance.
\(\begin{aligned}
& \Rightarrow C_{\text {tatal }}=\frac{C}{3}+C \\
& C_{\text {teal }}=\frac{4 C}{3}
\end{aligned}\)
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