Question

The total capacity of the system of capacitors shown in the adjoining figure between the points A and B is

• A. 1μF
• B. 2μF
• C. 3μF
• D. 4μF

Answer 1

The Correct Option is B

To find the total capacitance between points A and B in the given system of capacitors, let's analyze the arrangement step-by-step.

The system consists of capacitors arranged in a combination of series and parallel connections. We'll simplify the connections step-by-step:

1. First, observe that there are three capacitors: 1μF, 1μF, and 2μF arranged in a triangular formation at the center.
2. The two 2μF capacitors on the sides are in parallel with the central configuration.
3. Let's first simplify the triangular part:
   • The top branch has the capacitors of 2μF and 1μF in series. The equivalent capacitance is given by: \(C_s = \frac{1}{\frac{1}{2} + \frac{1}{1}} = \frac{2}{3}\text{μF}\).
   • The bottom branch also has a single capacitor of 2μF.
   • The capacitor across the diagonal of the triangle is 1μF.
4. Now, analyzing this configuration:
   • The two results from top \((\frac{2}{3}\text{μF})\) and bottom (2μF) branches are in parallel with the diagonal capacitor (1μF). Calculate the parallel combination: \(C_p = \frac{2}{3} + 1 = \frac{5}{3}\text{μF}\)
5. Finally, calculate the total capacitance considering the capacitors on each side (2μF each), which are in series with \(\frac{5}{3}\text{μF}\): \(C_{\text{total}} = \frac{1}{\frac{1}{2} + \frac{1}{\frac{5}{3}} + \frac{1}{2}}\) \(= \frac{1}{\frac{3}{6} + \frac{3}{5} + \frac{3}{6}}\) \(= \frac{1}{\frac{15 + 18 + 15}{30}}\) \(= \frac{30}{48}\) \(= 2\text{μF}\)

Thus, the total capacity of the system of capacitors between the points A and B is 2μF.