Question

Find out the equivalent capacitance and total energy stored in the given combination of capacitors:

Hint:

For capacitors in series, use \( \frac{1}{C_{\text{eq}}} = \sum \frac{1}{C_i} \). For capacitors in parallel, simply add them: \( C_{\text{eq}} = \sum C_i \).

Answer 1

Capacitor Combination: The given combination of capacitors is shown below:
\[ \text{3 μF} \parallel \text{(4 μF in series with 5 μF)} \] 1. Step 1: Calculate the Equivalent Capacitance of the 4 μF and 5 μF Capacitors in Series:}
For capacitors in series, the equivalent capacitance \( C_{\text{eq, series}} \) is given by: \[ \frac{1}{C_{\text{eq, series}}} = \frac{1}{C_1} + \frac{1}{C_2} \] Substituting the values: \[ \frac{1}{C_{\text{eq, series}}} = \frac{1}{4} + \frac{1}{5} = \frac{5 + 4}{20} = \frac{9}{20} \] Thus: \[ C_{\text{eq, series}} = \frac{20}{9} \approx 2.22 \, \mu\text{F} \] 2. Step 2: Find the Equivalent Capacitance of the Entire Combination:}
Now, the equivalent capacitance of the 3 μF capacitor in parallel with the series combination (2.22 μF) is: \[ C_{\text{eq, total}} = C_1 + C_{\text{eq, series}} = 3 + 2.22 = 5.22 \, \mu\text{F} \] So, the total equivalent capacitance of the combination is 5.22 μF.
3. Step 3: Calculate the Total Energy Stored:}
The energy stored in a capacitor is given by the formula: \[ E = \frac{1}{2} C V^2 \] Substituting the values (\( C = 5.22 \, \mu\text{F} \) and \( V = 2 \, \text{V} \)): \[ E = \frac{1}{2} \times 5.22 \times 10^{-6} \times (2)^2 = \frac{1}{2} \times 5.22 \times 10^{-6} \times 4 = 1.044 \times 10^{-5} \, \text{J} = 10.44 \, \mu\text{J} \] Thus, the total energy stored in the combination is 10.44 μJ.