Class 11 - Unit # 09 : Capacitors - Solved Numericals

Physics Numerical Sheet
Capacitors

1. The capacitance of an air-filled parallel-plate capacitor is 1.3 pF. If the separation of the plates is doubled and wax is inserted between them, the new capacitance is 2.6 pF. Find the dielectric constant of the wax.

Data:

• Initial capacitance in air (\(C_{\text{air}}\)) = \(1.3 \text{ pF} = 1.3 \times 10^{-12} \text{ F}\)
• New separation distance (\(d'\)) = \(2d\)
• New capacitance with wax (\(C_{\text{wax}}\)) = \(2.6 \text{ pF} = 2.6 \times 10^{-12} \text{ F}\)
• Dielectric constant of wax (\(\varepsilon_r\)) = ?

Solution:

When air is present between the plates:

$$C_{\text{air}} = \frac{\varepsilon_0 A}{d} = 1.3 \text{ pF}$$

When wax is inserted and the plate separation distance is doubled:

$$C_{\text{wax}} = \frac{\varepsilon_r \varepsilon_0 A}{d'} = \frac{\varepsilon_r \varepsilon_0 A}{2d}$$

Substituting the base air capacitance equation into the formula:

$$C_{\text{wax}} = \frac{\varepsilon_r}{2} \left( \frac{\varepsilon_0 A}{d} \right) = \frac{\varepsilon_r}{2} \cdot C_{\text{air}}$$"

Plugging in the numerical parameters to solve for the target variable:

$$2.6 = \frac{\varepsilon_r}{2} \cdot 1.3 \implies \varepsilon_r = \frac{2.6 \times 2}{1.3} = 4$$

RESULT: The dielectric constant of the wax is 4.

2. Three capacitors have capacitances 10, 50, and 25 \(\mu\text{F}\) respectively. Calculate (i) the charge on each when connected in parallel to a 250 V supply, (ii) the total capacitance in series, and (iii) the potential difference across each when connected in series.

Data:

• Capacitances: \(C_1 = 10 \text{ }\mu\text{F}\), \(C_2 = 50 \text{ }\mu\text{F}\), \(C_3 = 25 \text{ }\mu\text{F}\)
• Voltage supply (\(V\)) = \(250 \text{ V}\)

Solution:

(i) When connected in parallel: The voltage drop stays uniform across every individual component branch (\(V = 250 \text{ V}\)).

$$Q_1 = C_1 \cdot V = (10 \times 10^{-6} \text{ F}) \times 250 \text{ V} = 2500 \text{ }\mu\text{C} = 2.5 \times 10^{-3} \text{ C}$$ $$Q_2 = C_2 \cdot V = (50 \times 10^{-6} \text{ F}) \times 250 \text{ V} = 12500 \text{ }\mu\text{C} = 1.25 \times 10^{-2} \text{ C}$$ $$Q_3 = C_3 \cdot V = (25 \times 10^{-6} \text{ F}) \times 250 \text{ V} = 62500 \text{ }\mu\text{C} = 6.25 \times 10^{-2} \text{ C}$$

(ii) Total capacitance in series (\(C_s\)):

$$\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} = \frac{1}{10} + \frac{1}{50} + \frac{1}{25}$$ $$\frac{1}{C_s} = \frac{5 + 1 + 2}{50} = \frac{8}{50} \implies C_s = \frac{50}{8} = 6.25 \text{ }\mu\text{F}$$

(iii) Potential difference across each in series: Total charge (\(Q_s\)) remains constant across all capacitors in series configuration.

$$Q_s = C_s \cdot V = 6.25 \text{ }\mu\text{F} \times 250 \text{ V} = 1562.5 \text{ }\mu\text{C}$$ $$V_1 = \frac{Q_s}{C_1} = \frac{1562.5 \text{ }\mu\text{C}}{10 \text{ }\mu\text{F}} = 156.25 \text{ V}$$ $$V_2 = \frac{Q_s}{C_2} = \frac{1562.5 \text{ }\mu\text{C}}{50 \text{ }\mu\text{F}} = 31.25 \text{ V}$$ $$V_3 = \frac{Q_s}{C_3} = \frac{1562.5 \text{ }\mu\text{C}}{25 \text{ }\mu\text{F}} = 62.5 \text{ V}$$

RESULT:
(i) Parallel Charges: \(Q_1 = 2500\text{ }\mu\text{C}\), \(Q_2 = 12500\text{ }\mu\text{C}\), \(Q_3 = 62500\text{ }\mu\text{C}\)
(ii) Total Series Capacitance: \(6.25\text{ }\mu\text{F}\)
(iii) Series Voltages: \(V_1 = 156.25\text{ V}\), \(V_2 = 31.25\text{ V}\), \(V_3 = 62.5\text{ V}\)

3. Three capacitors (\(C_1 = 2\text{ }\mu\text{F}\), \(C_2 = 6\text{ }\mu\text{F}\), \(C_3 = 2\text{ }\mu\text{F}\)) are connected through a potential difference of 100 volts such that \(C_2\) and \(C_3\) are in parallel, and this combination is in series with \(C_1\). Find the charges and potential difference across each capacitor.

Data:

• Capacitances: \(C_1 = 2 \text{ }\mu\text{F}\), \(C_2 = 6 \text{ }\mu\text{F}\), \(C_3 = 2 \text{ }\mu\text{F}\)
• Total system voltage (\(V\)) = \(100 \text{ V}\)

Solution:

Equivalent capacitance of parallel block (\(C_2\) and \(C_3\)):

$$C' = C_2 + C_3 = 6 + 2 = 8 \text{ }\mu\text{F}$$

Total equivalent circuit network capacitance (\(C_{eq}\)):

$$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C'} = \frac{1}{2} + \frac{1}{8} = \frac{4 + 1}{8} = \frac{5}{8} \implies C_{eq} = 1.6 \text{ }\mu\text{F}$$

Total operational circuit charge pulled from source:

$$Q_{\text{total}} = C_{eq} \cdot V = 1.6 \text{ }\mu\text{F} \times 100 \text{ V} = 160 \text{ }\mu\text{C}$$

Since \(C_1\) is aligned directly in series within the main line: \(Q_1 = 160 \text{ }\mu\text{C}\).

Potential difference dropped across capacitor \(C_1\):

$$V_1 = \frac{Q_1}{C_1} = \frac{160 \text{ }\mu\text{C}}{2 \text{ }\mu\text{F}} = 80 \text{ V}$$

Remaining residual branch voltage across parallel duo (\(C_2\) and \(C_3\))

$$V_2 = V_3 = V - V_1 = 100 \text{ V} - 80 \text{ V} = 20 \text{ V}$$

Individual internal branch charges:

$$Q_2 = C_2 \cdot V_2 = 6 \text{ }\mu\text{F} \times 20 \text{ V} = 120 \text{ }\mu\text{C}$$ $$Q_3 = C_3 \cdot V_3 = 2 \text{ }\mu\text{F} \times 20 \text{ V} = 40 \text{ }\mu\text{C}$$

Summary Analysis:

Capacitor	Capacitance (\(\mu\)F)	Charge (\(\mu\)C)	Potential Difference (V)
\(C_1\)	2.0	160	80 V
\(C_2\)	6.0	120	20 V
\(C_3\)	2.0	40	20 V

4. A capacitor is charged through a large non-reactive resistance by a battery of constant voltage V. If the capacitor has a capacitance of 10 \(\mu\text{F}\) and the resistance is 1 M\(\Omega\), calculate the time taken for the capacitor to receive 90% of its final charge.

Data:

• Capacitance (\(C\)) = \(10 \text{ }\mu\text{F} = 10 \times 10^{-6} \text{ F}\)
• Resistance (\(R\)) = \(1 \text{ M}\Omega = 1 \times 10^6 \text{ }\Omega\)
• Instantaneous state charge (\(q\)) = \(0.9 Q_0\)

Solution:

The operational charging time model tracking an RC transient state path is expressed as:

$$q = Q_0 \left(1 - e^{-\frac{t}{RC}}\right)$$

Substituting given fractional limit condition values:

$$0.9 Q_0 = Q_0 \left(1 - e^{-\frac{t}{RC}}\right) \implies 0.9 = 1 - e^{-\frac{t}{RC}}$$ $$e^{-\frac{t}{RC}} = 0.1 \implies -\frac{t}{RC} = \ln(0.1) \implies t = 2.3026 \cdot RC$$

Evaluating standard scaling time constant (\(\tau = RC\)):

$$RC = (1 \times 10^6 \text{ }\Omega) \times (10 \times 10^{-6} \text{ F}) = 10 \text{ s}$$ $$t = 2.3026 \times 10 \approx 23.03 \text{ s}$$

RESULT: The time required to charge up to 90% capacity is 23.03 seconds.

5. What capacitance is required to store energy of 10 kWh at a potential difference of 1000 V?

Data:

• Stored Energy (\(E\)) = \(10 \text{ kWh} = 10 \times 1000 \text{ W} \times 3600 \text{ s} = 3.6 \times 10^7 \text{ J}\)
• Potential Difference (\(V\)) = \(1000 \text{ V}\)

Solution:

Using the capacitor electrical energy storage equation:

$$E = \frac{1}{2} C V^2$$ $$3.6 \times 10^7 = \frac{1}{2} \cdot C \cdot (1000)^2 \implies 3.6 \times 10^7 = 0.5 \cdot C \cdot 10^6$$ $$C = \frac{3.6 \times 10^7}{0.5 \times 10^6} = 72 \text{ F}$$

RESULT: The required physical macro capacitance value is 72 F.

6. A 2.0 \(\mu\text{F}\) capacitor and a 4.0 \(\mu\text{F}\) capacitor are connected in parallel across a 300 V potential difference. Calculate the total energy stored in the capacitors.

Data:

• Capacitances: \(C_1 = 2.0 \text{ }\mu\text{F}\), \(C_2 = 4.0 \text{ }\mu\text{F}\)
• Voltage loop (\(V\)) = \(300 \text{ V}\)

Solution:

Equivalent combined capacitance tracking parallel connection components:

$$C_p = C_1 + C_2 = 2.0 + 4.0 = 6.0 \text{ }\mu\text{F} = 6 \times 10^{-6} \text{ F}$$

Total aggregate capacity energy metrics:

$$E = \frac{1}{2} C_p V^2 = \frac{1}{2} \times (6 \times 10^{-6} \text{ F}) \times (300 \text{ V})^2$$ $$E = (3 \times 10^{-6}) \times 90000 = 0.27 \text{ J}$$

RESULT: Total aggregate stored potential energy is 0.27 J.

7. A 12.0-V battery is connected to a capacitor, resulting in 54.0 \(\mu\text{C}\) of charge stored on the capacitor. How much energy is stored in the capacitor?

Data:

• Voltage power profile (\(V\)) = \(12.0 \text{ V}\)
• Stored Charge load (\(Q\)) = \(54.0 \text{ }\mu\text{C} = 54.0 \times 10^{-6} \text{ C}\)

Solution:

Computing potential value matching metrics connecting metrics directly via charge load profiles:

$$E = \frac{1}{2} Q V = \frac{1}{2} \times (54.0 \times 10^{-6} \text{ C}) \times 12.0 \text{ V}$$ $$E = 27.0 \times 10^{-6} \times 12.0 = 3.24 \times 10^{-4} \text{ J} = 324 \text{ }\mu\text{J}$$

RESULT: The accumulated static energy storage value yields 324 \(\mu\text{J}\).