MHT-CET Physics Electrostatics PYQ - Combined Capacitors

11) When three capacitors of equal capacitance are connected in parallel and one capacitor of the same capacitance is connected in series with this combination, the resultant capacitance is \(4.5\,\mu\text{F}\). What is the capacitance of each capacitor?

A. \( 5\,\mu\text{F} \)

B. \( 6\,\mu\text{F} \)

C. \( 7\,\mu\text{F} \)

D. \( 8\,\mu\text{F} \)

Solution:

Let the capacitance of each capacitor be \(C\).

First, three capacitors of capacitance \(C\) connected in parallel have an equivalent capacitance given by: \[ C_{\text{parallel}} = C + C + C = 3C. \]

Next, this combination is connected in series with another capacitor of capacitance \(C\). For two capacitors connected in series, the equivalent capacitance is given by: \[ \frac{1}{C_{\text{total}}} = \frac{1}{C_{\text{parallel}}} + \frac{1}{C} = \frac{1}{3C} + \frac{1}{C}. \]

Simplify the expression: \[ \frac{1}{C_{\text{total}}} = \frac{1}{3C} + \frac{3}{3C} = \frac{4}{3C}. \]

Therefore, the total capacitance is: \[ C_{\text{total}} = \frac{3C}{4}. \]

We are given that: \[ C_{\text{total}} = 4.5\,\mu\text{F}. \] Thus: \[ \frac{3C}{4} = 4.5\,\mu\text{F}. \]

Solving for \(C\): \[ C = \frac{4 \times 4.5\,\mu\text{F}}{3} = \frac{18\,\mu\text{F}}{3} = 6\,\mu\text{F}. \]

Hence, the capacitance of each capacitor is \(6\,\mu\text{F}\), which corresponds to option B.