Question

Three capacitors $(2.0 \mu \mathrm{F}, 3.0 \mu \mathrm{F}, 6.0 \mu \mathrm{F})$ are connected in series to an ac voltage source with amplitude \(12.0 \mathrm{V}\) and frequency \(6.3 \mathrm{kHz}\). (a) What are the peak voltages across each capacitor? (b) What is the peak current that flows in the circuit?

Short answer

Answer: The peak voltages across the capacitors are 6.0 V for the first capacitor (C1), 4.0 V for the second capacitor (C2), and 2.0 V for the third capacitor (C3). The peak current that flows in the circuit is 0.475 A.

Step-by-step solution

Calculate the equivalent capacitance

To determine the equivalent capacitance of the capacitors connected in series, we can use the formula for series capacitance: \( \dfrac{1}{C_eq} = \dfrac{1}{C_1} + \dfrac{1}{C_2} + \dfrac{1}{C_3}\) Where, \(C_1 = 2.0\;\mu F\) (The first capacitor) \(C_2 = 3.0\;\mu F\) (The second capacitor) \(C_3 = 6.0\;\mu F\) (The third capacitor) Solve for \(C_eq\): \(C_eq = \dfrac{1}{(\dfrac{1}{2.0 \times 10^{-6}} + \dfrac{1}{3.0 \times 10^{-6}} + \dfrac{1}{6.0 \times 10^{-6}})}\) \(C_eq = 1.0 \mu F\)

Calculate the angular frequency

Now we need to find the angular frequency (\(\omega\)) of the AC voltage source. The relationship between the angular frequency and frequency is: \(\omega = 2\pi f\) Where, \(f = 6.3\;kHz\) (Frequency of the AC voltage source) \(\omega = 2\pi \times 6.3 \times 10^{3} = 39600\;rad/s\)

Calculate the impedance of the circuit

Next, calculate the impedance (Z) of the circuit using the equivalent capacitance and the angular frequency: \(Z = \dfrac{1}{\omega C_eq}\) \(Z = \dfrac{1}{39600 \times 1.0 \times 10^{-6}}\) \(Z = 25.3\;\Omega\)

Calculate the peak current flowing in the circuit

Now we can determine the peak current (I) in the circuit using Ohm's law: \(I = \dfrac{V}{Z}\) Where, \(V = 12.0\;V\) (Amplitude of the AC voltage source) \(I = \dfrac{12}{25.3} = 0.475\;A\) The peak current that flows in the circuit is \(0.475\;A\).

Calculate the peak voltage across each capacitor

Finally, we can calculate the peak voltage across each capacitor using the peak current and individual capacitor's impedance: \(V_{C_1} = I \times Z_{C_1}\) \(V_{C_2} = I \times Z_{C_2}\) \(V_{C_3} = I \times Z_{C_3}\) Where, \(Z_{C_1} = \dfrac{1}{\omega C_1}\) \(Z_{C_2} = \dfrac{1}{\omega C_2}\) \(Z_{C_3} = \dfrac{1}{\omega C_3}\) Solve for each peak voltage: \(V_{C_1} = 0.475 \times \dfrac{1}{39600 \times 2.0 \times 10^{-6}} = 6.0\;V\) \(V_{C_2} = 0.475 \times \dfrac{1}{39600 \times 3.0 \times 10^{-6}} = 4.0\;V\) \(V_{C_3} = 0.475 \times \dfrac{1}{39600 \times 6.0 \times 10^{-6}} = 2.0\;V\) The peak voltages across the capacitors are \(6.0\;V\), \(4.0\;V\), and \(2.0\;V\) for \(C_1\), \(C_2\), and \(C_3\), respectively.