Question

Consider the circuit shown in the following figure. The battery has emf 50.0 V and negligible internal resistance. R₂ = 3.00 Ω, C₁= 3.00 μF, and C₂= 6.00 μF. After the capacitors have attained their final charges, the charge on C₁ is Q₁ = 15.0 μC.

A) What is the final charge on C₂?

B) What is the resistance R₁?

Answer 1

Concepts and reason

The concept required to solve this question is capacitance and Kirchhoff's law. Initially, find the voltage across the first capacitor. Then, find the charge on the second capacitor by using the voltage. Later, find the voltage across resistance 1 by Kirchhoff's law and the current through the circuit containing resistors. Finally, find the resistance of the first resistor by using ohm's law.

Fundamentals

Ohm's law states that the current through a conductor is directly proportional to the voltage across it, where \(\mathrm{R}\) is the constant proportionality. \(V=I R\)

Here, \(V\) is the voltage, \(I\) is current, and \(R\) is the resistance. For example, the capacitance (C) of a capacitor with a charge \(Q\) and voltage \(V\) is given by the following expression:

\(C=\frac{Q}{V}\)

Here, \(C\) is the capacitance, \(V\) is the voltage, and \(Q\) is the charge. Kirchhoff's second law or voltage law states that the algebraic sum of all the voltages around any closed loop in a circuit is equal to zero. This law is a direct consequence of the conservation of energy. Since the path is closed, no charge is lost, and no energy is lost.

(A) The capacitance (C1) of a capacitor with a charge \(Q 1\) and voltage \(V\) is given by the following expression:

\(C_{1}=\frac{Q 1}{V}\)

Rearrange the above expression for V. \(V=\frac{Q_{1}}{C_{1}}\)

Substitute \(15.0 \mu \mathrm{C}\) for \(\mathrm{Q} 1\) and \(3.00 \mu \mathrm{F}\) for \(\mathrm{C} 1\) in the above expression.

$$ \begin{aligned} V &=\frac{15.0 \mu \mathrm{C}}{3.00 \mu \mathrm{F}} \\ &=5.0 \mathrm{~V} \end{aligned} $$

The first and the second capacitors are connected in parallel so that the voltages across them are the same. Rearrange the expression \(C_{2}=\frac{Q_{2}}{V}\) for \(Q 2\)

$$ Q_{2}=C_{2} V $$

Substitute \(6.00 \mu \mathrm{F}\) for \(\mathrm{C} 2\) and \(5.0 \mathrm{~V}\) for \(\mathrm{V}\) in the above expression.

$$ \begin{aligned} Q_{2}=&(6.00 \mu \mathrm{F})(5.0 \mathrm{~V}) \\ &=30.0 \mu \mathrm{C} \end{aligned} $$

The voltage across the parallel components is always the same. For example, capacitor 1 is connected in parallel with capacitor 2 so that the voltage across that capacitor is the same as that of 2 nd capacitor. The charge accumulated on the capacitor depends on the capacitance of the capacitor.

(B) Kirchhoff's second law or voltage law states that the algebraic sum of all the voltages around any closed loop in a circuit is equal to zero. Apply Kirchhoff's law in the resistor circuit. \(-\varepsilon+V_{1}+V_{2}=0\)

Rearrange the above expression for \(V_{1}\). \(V_{1}=\varepsilon-V_{2}\)

The second resistor is in parallel connection with the capacitors so that the voltage across the second resistor is the same as that of the capacitors. Substitute \(50.0 \mathrm{~V}\) for \(\varepsilon\) and \(5.0 \mathrm{~V}\) for \(V_{2}\) in the above expression.

$$ V_{1}=50.0 \mathrm{~V}-5.0 \mathrm{~V} $$

\(=45.0 \mathrm{~V}\)

As the capacitors are fully charged so that no current flows through them. The capacitors branch will behave as open circuits. The current is flowing only in the resistance circuit so that the current flowing through the second resistor is equal to the current flowing through the first resistor. The current flowing through the second resistor is as follows:

\(I_{2}=\frac{V}{R_{2}}\)

Substitute \(5.0 \mathrm{~V}\) for \(\mathrm{V}\) and \(3.0 \Omega\) for \(R_{2}\) in the above expression.

\(I_{2}=\frac{5.0 \mathrm{~V}}{3.0 \Omega}\)

\(=1.67 \mathrm{~A}\)

The resistance of the first resistor can be determined as follows:

\(R_{1}=\frac{V_{1}}{I_{1}}\)

The current through the first resistor is the same as that of the current through the second resistor. Therefore, substitute \(45.0 \mathrm{~V}\) for \(\mathrm{V} 1\) and \(1.67 \mathrm{~A}\) for \(\mathrm{I} 1\) in the above expression.

$$ \begin{array}{l} R_{1}=\frac{45.0 \mathrm{~V}}{1.67 \mathrm{~A}} \\ =26.95 \Omega \end{array} $$

\(=27.0 \Omega\)

The capacitor gets fully charged when the voltage across the capacitor is equal to the power supply voltage. The charging phase of the capacitor ends here, and the current stops flowing into the capacitor so that the circuit breaks at that point.

Part A The final charge on the second capacitor is 30.0μC.

Part B The resistance R1 is 27.0Ω.