A 41.1 Q resistor, a 64.5 µF capacitor and a 8.27 mH resistanceless coil are in series with an AC source (Vrms = 213 V, ƒ = 89.4 Hz). Find the rms current, in A, in this circuit.
A 41.1 Q resistor, a 64.5 µF capacitor and a 8.27 mH resistanceless coil are in series with an AC source (Vrms = 213 V, ƒ = 89.4 Hz). Find the rms current, in A, in this circuit.
College Physics
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ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
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![**Problem:**
A 41.1 Ω resistor, a 64.5 µF capacitor, and a 8.27 mH resistance-less coil are in series with an AC source (\(V_{\text{rms}} = 213 \, \text{V}, \, f = 89.4 \, \text{Hz}\)). Find the rms current, in A, in this circuit.
**Solution Explanation:**
To solve this problem, we first need to calculate the impedance of the series circuit, which includes a resistor, capacitor, and inductor.
1. **Reactance of the Capacitor (\(X_C\))**:
\[
X_C = \frac{1}{2\pi fC}
\]
Where \(f = 89.4 \, \text{Hz}\) and \(C = 64.5 \times 10^{-6} \, \text{F}\).
2. **Reactance of the Inductor (\(X_L\))**:
\[
X_L = 2\pi fL
\]
Where \(L = 8.27 \times 10^{-3} \, \text{H}\).
3. **Impedance (\(Z\))**:
The total impedance \(Z\) of the circuit is found using:
\[
Z = \sqrt{R^2 + (X_L - X_C)^2}
\]
Where \(R = 41.1 \, \Omega\).
4. **RMS Current (\(I_{\text{rms}}\))**:
The rms current can be calculated using Ohm’s Law for AC circuits:
\[
I_{\text{rms}} = \frac{V_{\text{rms}}}{Z}
\]
By substituting the appropriate values into these equations, you can find the impedance and then the rms current of the circuit.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2d6524ac-a97a-44db-bfac-cff029bb92d1%2Fe9db509f-ea7f-483e-ae11-0aaaf29daea7%2Fbslabd_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem:**
A 41.1 Ω resistor, a 64.5 µF capacitor, and a 8.27 mH resistance-less coil are in series with an AC source (\(V_{\text{rms}} = 213 \, \text{V}, \, f = 89.4 \, \text{Hz}\)). Find the rms current, in A, in this circuit.
**Solution Explanation:**
To solve this problem, we first need to calculate the impedance of the series circuit, which includes a resistor, capacitor, and inductor.
1. **Reactance of the Capacitor (\(X_C\))**:
\[
X_C = \frac{1}{2\pi fC}
\]
Where \(f = 89.4 \, \text{Hz}\) and \(C = 64.5 \times 10^{-6} \, \text{F}\).
2. **Reactance of the Inductor (\(X_L\))**:
\[
X_L = 2\pi fL
\]
Where \(L = 8.27 \times 10^{-3} \, \text{H}\).
3. **Impedance (\(Z\))**:
The total impedance \(Z\) of the circuit is found using:
\[
Z = \sqrt{R^2 + (X_L - X_C)^2}
\]
Where \(R = 41.1 \, \Omega\).
4. **RMS Current (\(I_{\text{rms}}\))**:
The rms current can be calculated using Ohm’s Law for AC circuits:
\[
I_{\text{rms}} = \frac{V_{\text{rms}}}{Z}
\]
By substituting the appropriate values into these equations, you can find the impedance and then the rms current of the circuit.
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