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Chapter 21: Problem 64

For a particular \(R L C\) series circuit, the capacitive reactance is $12.0 \Omega,\( the inductive reactance is \)23.0 \Omega,$ and the maximum voltage across the \(25.0-\Omega\) resistor is \(8.00 \mathrm{V}\) (a) What is the impedance of the circuit? (b) What is the maximum voltage across this circuit?

Short Answer

Expert verified
Answer: The impedance of the circuit is approximately 27.3惟, and the maximum voltage across the circuit is approximately 8.74V.

Step by step solution

01

Find the Impedance

First, we need to find the impedance of the circuit. Use the formula: \(Z = \sqrt{R^2 + (X_L - X_C)^2}\), where \(R = 25.0\Omega\), \(X_L = 23.0\Omega\), and \(X_C = 12.0\Omega\). Calculating the impedance: \(Z = \sqrt{(25.0)^2 + (23.0 - 12.0)^2} = \sqrt{625 + 121} = \sqrt{746} \approx 27.3\Omega\) So, the impedance of the circuit is approximately \(27.3\Omega\).
02

Find the Maximum Current

Now, we need to find the maximum current in the circuit. Use the formula: \(I_{max} = \frac{V_R}{R}\), where \(V_R = 8.00V\) and \(R = 25.0\Omega\). Calculating the maximum current: \(I_{max} = \frac{8.00}{25.0} = 0.32A\) Therefore, the maximum current in the circuit is \(0.32A\).
03

Find the Maximum Voltage across the Circuit

Lastly, we need to find the maximum voltage across the circuit. Use the formula: \(V_{max} = I_{max} \cdot Z\), where \(I_{max} = 0.32A\) and \(Z \approx 27.3\Omega\). Calculating the maximum voltage: \(V_{max} = 0.32 \cdot 27.3 \approx 8.74V\) Hence, the maximum voltage across the circuit is approximately \(8.74V\).

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