KCET · Physics · Capacitance
Five capacitance each of value \(l \mu \mathrm{F}\) are connected as shown in the figure. The equivalent capacitance between \(A\) and \(B\) is

- A \(3 \mu \mathrm{F}\)
- B \(1 \mu \mathrm{F}\)
- C \(2 \mu \mathrm{F}\)
- D \(5 \mu \mathrm{F}\)
Answer & Solution
Correct Answer
(B) \(1 \mu \mathrm{F}\)
Step-by-step Solution
Detailed explanation
Since, the middle branch conducts no current,
therefore the middle capacitor represents open circuit and hence reduced circuit is shown.

Now \(\frac{1}{C_{\text {series }}}=\frac{1}{1}+\frac{1}{1} \)
\( C_{\text {series }}=0.5 \mu \mathrm{F} \)
\( C_{\text {parallel }}=0.5+0.5 \)
or
\( C_{A B}=1 \mu \mathrm{F}\)
therefore the middle capacitor represents open circuit and hence reduced circuit is shown.

Now \(\frac{1}{C_{\text {series }}}=\frac{1}{1}+\frac{1}{1} \)
\( C_{\text {series }}=0.5 \mu \mathrm{F} \)
\( C_{\text {parallel }}=0.5+0.5 \)
or
\( C_{A B}=1 \mu \mathrm{F}\)
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