Figure shows a network of capacitors where the numbers indicate capacitances in microfarad. What must be the…

Figure shows a network of capacitors where the numbers indicate capacitances in microfarad. What must be the value of capacitance \(C\) if the equivalent capacitance between points \(A\) and \(B\) is to be \(1 \mu \mathrm{F} ?\)
  1. \(\frac{31}{23} \mu \mathrm{F}\)
  2. \(\frac{32}{23} \mu \mathrm{F}\)
  3. \(\frac{33}{23} \mu \mathrm{F}\)
  4. \(\frac{34}{23} \mu \mathrm{F}\)

Solution

The series combination of 6 and 12 is equivalent to 4 and the parallel combination of 2 and 2 is also equivalent to 4. Therefore the network can be simplified as shown in Fig. The parallel combination of 4 and 4 is equivalent to 8 and the series combination of 8 and 4 is equivalent to \(8 / 3\). Thus the combination in Fig. reduces to that in Fig. The series combination of 1 and 8 in Fig. yields \(8 / 9\) as shown in Fig.

Now \(8 / 3\) and \(8 / 9\) are in parallel and their equivalent is \(32 / 9\). Therefore, the network finally reduces to that in Fig. Since the total capacitance between \(A\) and \(B\) is to be (i.e. \(1 \mu \mathrm{F}\)), we have \(1=\frac{1}{C}+\frac{9}{32}\)
\(\Rightarrow \quad C=\frac{32}{23} \mu \mathrm{F} .\) Hence the correct choice is (b). ,

Asked in: JEE Mains - Capacitance - Test 2

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