Question

With the help of given circuit, find out:
(i) Equivalent capacity of the combination
(ii) Charge on capacitor \(c_1\)
(iii) Total stored energy of the combination.}
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Hint:

When combining capacitors in series and parallel, always calculate the equivalent capacitance step by step. For capacitors in series, use \(\frac{1}{C_{\text{eq}}} = \sum \frac{1}{C}\), and for capacitors in parallel, use \(C_{\text{eq}} = \sum C\).

Answer 1

In the given circuit, we have four capacitors \(c_1 = 10 \, \mu F\), \(c_2 = 20 \, \mu F\), \(c_3 = 20 \, \mu F\), and \(c_4 = 20 \, \mu F\) connected in a combination of series and parallel. The total voltage across the combination is given as \(V = 20 \, \text{V}\). Let's break down the solution into steps:
Step 1: Simplifying the Combination of Capacitors
First, we need to combine capacitors \(c_2\), \(c_3\), and \(c_4\) in parallel. The total capacitance for parallel combination is given by: \[ C_{\text{parallel}} = c_2 + c_3 + c_4 \] Substituting values: \[ C_{\text{parallel}} = 20 \, \mu F + 20 \, \mu F + 20 \, \mu F = 60 \, \mu F \] Step 2: Combining with \(c_1\) in Series
Now, we combine the equivalent capacitance \(C_{\text{parallel}}\) with \(c_1 = 10 \, \mu F\) in series. The equivalent capacitance for capacitors in series is given by: \[ \frac{1}{C_{\text{eq}}} = \frac{1}{c_1} + \frac{1}{C_{\text{parallel}}} \] Substitute the values: \[ \frac{1}{C_{\text{eq}}} = \frac{1}{10} + \frac{1}{60} = \frac{6}{60} + \frac{1}{60} = \frac{7}{60} \] \[ C_{\text{eq}} = \frac{60}{7} \approx 8.57 \, \mu F \] Thus, the equivalent capacitance of the combination is approximately \(8.57 \, \mu F\).
Step 3: Charge on Capacitor \(c_1\)
The charge on the capacitor \(c_1\) is given by: \[ Q_1 = C_{\text{eq}} \times V \] Substitute the values: \[ Q_1 = 8.57 \, \mu F \times 20 \, \text{V} = 171.4 \, \mu C \] So, the charge on capacitor \(c_1\) is \(171.4 \, \mu C\).
Step 4: Total Stored Energy
The total stored energy in the combination of capacitors is given by: \[ U = \frac{1}{2} C_{\text{eq}} V^2 \] Substitute the values: \[ U = \frac{1}{2} \times 8.57 \times 10^{-6} \times (20)^2 = 3.428 \times 10^{-3} \, \text{J} \] Thus, the total stored energy is \(3.428 \, \text{mJ}\).