Question

Part A

What is the charge Q on the capacitor immediately after the switch is moved to position b?

Express your answer using two significant figures.

Q =		?C

 

Part B

What is the current I through the resistor immediately after the switch is moved to position b?

Express your answer using two significant figures.

I =		mA

 

Part C

What is the charge Q on the capacitor at t=50?s?

Express your answer using two significant figures.

Q =		?C

 

Part D

What is the current I through the resistor at t=50?s?

Express your answer using two significant figures.

I =		mA

 

Part E

What is the charge Q on the capacitor at t=200?s?

Express your answer using two significant figures.

Q =		?C

 

Part F

What is the current I through the resistor at t=200?s?

Express your answer using two significant figures.

I =		mA

Answer 1

Concepts and reason

The concept used to solve this problem is the charging of an RC circuit. Initially, the capacitor's charge can be calculated by multiplying the capacitance of the capacitor and the voltage of the battery. Later, the current flowing in the circuit can be calculated by using Ohm's law. After that, the instantaneous charge on the capacitor can be calculated by determining the instantaneous voltage and then multiplying that with capacitance. Then, the instantaneous current can be calculated using the ratio of the instantaneous voltage to the resistance. Again, the instantaneous charge on the capacitor can be calculated by determining the instantaneous voltage and then multiplying it with capacitance. Finally, the instantaneous current can be calculated using the ratio of the instantaneous voltage and the resistance.

Fundamentals

The expression for the charge on the capacitor is given below:

\(Q=C V\)

Here, \(V\) is the voltage, \(Q\) is the charge, and \(C\) is the capacitance. The expression of the current in the circuit is as follows:

$$ I=\frac{V}{R} $$

Here, \(I\) is the current, \(R\) is the resistance connected in the circuit. The expression of the instantaneous voltage is given below:

\(V^{\prime}=V e^{(-t / R C)}\)

Here, \(V^{\prime}\) is the instantaneous voltage and \(t\) is time. The expression of the instantaneous charge on the capacitor is as follows:

\(Q^{\prime}=C V^{\prime}\)

Here, \(Q^{\prime}\) is the instantaneous charge. The expression of the instantaneous current is given below:

\(I^{\prime}=\frac{V^{\prime}}{R}\)

Here, \(I^{\prime}\) is the instantaneous current.

 

(A)

The expression for the charge on the capacitor is as follows:

\(Q=C V\)

Substitute \(2.0 \mu \mathrm{F}\) for \(C\) and \(9 \mathrm{~V}\) for \(V\)

$$ Q=\left[(2 \mu \mathrm{F})\left(\frac{1 \times 10^{-6} \mathrm{~F}}{1 \mu \mathrm{F}}\right)\right](9 \mathrm{~V}) $$

\(=18 \times 10^{-6} \mathrm{C}\)

\(\approx 18 \mu \mathrm{C}\)

Part A

The charge on the capacitor is \(18 \mu \mathrm{C}\)

The net charge stored on the capacitor is always zero because there are equal and unlike charges on the plates. It is an electrical energy storing device.

 

(B) The expression for the current is as follows:

\(I=\frac{V}{R}\)

Substitute \(50 \Omega\) for \(R, 9.0 \mathrm{~V}\) for \(V\)

\(I=\frac{(9.0 \mathrm{~V})}{(50 \Omega)}\)

\(=0.18 \mathrm{~A}\)

Part \(b\)

The current in the circuit is 0.18 A.

The current in the circuit is determined by taking the ratio of the voltage to resistance. The relationship between current, voltage and resistance is known as Ohm's law.

 

(c) The expression for the instantaneous voltage is as follows:

\(V^{\prime}=V e^{(-t / R C)}\)

Substitute \(50 \Omega\) for \(R, 9.0 \mathrm{~V}\) for \(V, 2.0 \mu \mathrm{F}\) for \(C,\) and \(50 \mu \mathrm{s}\) for \(t\)

\(V^{\prime}=(9.0 \mathrm{~V})\left(e^{\left(\frac{-(50 \mu \mathrm{s})\left(\frac{\mathrm{Ix} 10^{-6} \mathrm{~s}}{1 \mu \mathrm{s}}\right)}{(50 \Omega)\left[(2.0 \mu \mathrm{F})\left(\frac{1 \times 10^{6} \mathrm{~F}}{1, \mathrm{~L}^{\prime}}\right)\right]}\right)}\right)\)

\(=5.459 \mathrm{~V}\)

The voltage between the two points determines the instantaneous voltage at a particular moment in time.

 

The expression for the instantaneous charge is given below:

$$ Q^{\prime}=C V^{\prime} $$

Substitute \(5.459 \mathrm{~V}\) for \(V^{\prime}\) and \(2 \mu \mathrm{F}\) for \(C .\)

$$ \begin{aligned} Q^{\prime} &=(5.459 \mathrm{~V})\left[(2 \mu \mathrm{F})\left(\frac{1 \times 10^{-6} \mathrm{~F}}{1 \mu \mathrm{F}}\right)\right] \\ &=10.918 \times 10^{-6} \mathrm{C} \\ & \approx 11 \mu \mathrm{C} \end{aligned} $$

Part c

The instantaneous charge on the capacitor is \(11 \mu C\).

The instantaneous charge is determined by multiplying the value of the instantaneous voltage and the capacitor's capacitance.

 

(d) The expression for the instantaneous current is as follows:

\(I^{\prime}=\frac{V^{\prime}}{R}\)

Substitute \(5.459 \mathrm{~V}\) for \(V^{\prime}\) and \(50 \Omega\) for \(R\)

\(I^{\prime}=\frac{(5.459 \mathrm{~V})}{(50 \Omega)}\)

\(=0.10918 \mathrm{~A}\)

\(=109.17 \mathrm{~mA}\)

\(\approx 110 \mathrm{~mA}\)

Part d

The instantaneous current in the circuit is \(110 \mathrm{~mA}\).

The instantaneous current is determined by taking the ratio of the instantaneous voltage with the resistance.

 

(e) The expression for the instantaneous voltage is as follows:

$$ V^{\prime}=V e^{(-t / R C)} $$

Substitute \(50 \Omega\) for \(R, 9.0 \mathrm{~V}\) for \(V, 2.0 \mu \mathrm{F}\) for \(C\) and \(200 \mu \mathrm{s}\) for \(t\)

\(V^{\prime}=(9.0 \mathrm{~V})\left(e^{\left(\frac{-(200 \mu \mathrm{s})\left(\frac{1 \times 10^{-6} \mathrm{~s}}{1 \mu \mathrm{s}}\right)}{(50 \Omega)\left[(2.0 \mu \mathrm{F})\left(\frac{1 \times 10^{-6} \mathrm{~F}}{1, \mathrm{w}^{2}}\right)\right]}\right)}\right)\)

\(=1.218 \mathrm{~V}\)

 

The instantaneous voltage is determined by using the relationship between capacitance, battery voltage, resistance, and instantaneous time.

 

The expression for the instantaneous charge is given below:

$$ Q^{\prime}=C V^{\prime} $$

Substitute \(1.218 \mathrm{~V}\) for \(V^{\prime}\) and \(2.0 \mu \mathrm{F}\) for \(C\)

$$ \begin{aligned} Q^{\prime} &=(1.218 \mathrm{~V})\left[(2.0 \mu \mathrm{F})\left(\frac{1 \times 10^{-6} \mathrm{~F}}{1 \mu \mathrm{F}}\right)\right] \\ &=2.436 \times 10^{-6} \mathrm{C} \\ &=2.4 \mu \mathrm{C} \end{aligned} $$

Part e

The instantaneous charge on the capacitor is \(2.4 \mu \mathrm{C}\).

That current flow determines the instantaneous charge onto the capacitor; therefore, the charge stored will be increased.

 

(f) The expression for the instantaneous current is as follows:

\(I^{\prime}=\frac{V^{\prime}}{R}\)

Substitute, \(1.218 \mathrm{~V}\) for \(V^{\prime}\) and \(50 \Omega\) for \(R\)

\(I^{\prime}=\frac{(1.218 \mathrm{~V})}{(50 \Omega)}\)

\(=0.02436 \mathrm{~A}\)

\(\approx 24 \mathrm{~mA}\)

Part \(f\)

The instantaneous current in the circuit is \(24 \mathrm{~mA}\).

The instantaneous current is determined as an alternating quantity of the ac voltage or AC at a particular instant.

 

Part A

The charge on the capacitor is \(18 \mu C\).

Part b

The current in the circuit is \(0.18 \mathrm{~A}\).

Part c

The instantaneous charge on the capacitor is \(11 \mu \mathrm{C}\).

Part d

The instantaneous current in the circuit is \(110 \mathrm{~mA}\).

Part e

The instantaneous charge on the capacitor is \(2.4 \mu \mathrm{C}\).

Part

The instantaneous current in the circuit is \(24 \mathrm{~mA}\).