Question:

Calculate the following in the given circuit:
(A) The equivalent capacitance of the circuit.
(B) The charge on \( 3 \, \mu\mathrm{F} \) and \( 2 \, \mu\mathrm{F} \) capacitors.
\includegraphics[]{equivalent capac.PNG}

Show Hint

For capacitors in series, the equivalent capacitance is given by: \[ \frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots \] For capacitors in parallel, the equivalent capacitance is given by: \[ C_{\text{eq}} = C_1 + C_2 + \cdots \]
Updated On: Mar 3, 2025
Hide Solution
collegedunia
Verified By Collegedunia

Solution and Explanation

(A) Equivalent Capacitance of the Circuit: The given circuit has capacitors arranged in a combination of series and parallel connections. Let us simplify step by step: 1. The \( 6 \, \mu\mathrm{F} \) and \( 12 \, \mu\mathrm{F} \) capacitors are in series: \[ \frac{1}{C_1} = \frac{1}{6} + \frac{1}{12} \quad \Rightarrow \quad C_1 = \frac{12}{3} = 4 \, \mu\mathrm{F}. \] 2. The \( 6 \, \mu\mathrm{F} \) and \( 3 \, \mu\mathrm{F} \) capacitors are in series: \[ \frac{1}{C_2} = \frac{1}{6} + \frac{1}{3} \quad \Rightarrow \quad C_2 = 2 \, \mu\mathrm{F}. \] 3. The \( C_1 = 4 \, \mu\mathrm{F} \), \( 2 \, \mu\mathrm{F} \), and \( C_2 = 2 \, \mu\mathrm{F} \) are in parallel: \[ C_{\text{eq}} = C_1 + C_2 + 2 \quad \Rightarrow \quad C_{\text{eq}} = 4 + 2 + 2 = 8 \, \mu\mathrm{F}. \] Thus, the equivalent capacitance of the circuit is: \[ \boxed{8 \, \mu\mathrm{F}}. \] (B) Charge on \( 3 \, \mu\mathrm{F} \) and \( 2 \, \mu\mathrm{F} \) capacitors: The total charge in the circuit is given by: \[ Q_{\text{total}} = C_{\text{eq}} \cdot V = 8 \, \mu\mathrm{F} \cdot 10 \, \mathrm{V} = 80 \, \mu\mathrm{C}. \] For the \( 3 \, \mu\mathrm{F} \) capacitor, the voltage across \( C_2 = 2 \, \mu\mathrm{F} \) is the same as the voltage across the series combination of \( 6 \, \mu\mathrm{F} \) and \( 3 \, \mu\mathrm{F} \). Using charge conservation: \[ Q = C \cdot V = 3 \, \mu\mathrm{F} \cdot 10 \, \mathrm{V} = 30 \, \mu\mathrm{C}. \] Similarly, for the \( 2 \, \mu\mathrm{F} \) capacitor: \[ Q = C \cdot V = 2 \, \mu\mathrm{F} \cdot 10 \, \mathrm{V} = 20 \, \mu\mathrm{C}. \] Thus, the charges are: \[ \boxed{30 \, \mu\mathrm{C} \text{ on the } 3 \, \mu\mathrm{F} \text{ capacitor, and } 20 \, \mu\mathrm{C} \text{ on the } 2 \, \mu\mathrm{F} \text{ capacitor.}} \]
Was this answer helpful?
0
0