A series circuit has a sinusoidal voltage supplied to it at $632 \mathrm { kHz }$ with a peak voltage of $748 \mathrm {~V}$. It also contains a 25-k $\Omega$ resistance, a $16 - \mu \mathrm { F }$ capacitance, and a $30 - \mathrm { H }$ inductance. What is the peak current for this circuit?
A) $11 \mu \mathrm { A }$
B) $6.3 \mu \mathrm { A }$
C) $26 \mu \mathrm { A }$
D) $30 \mu \mathrm { A }$

Question

Quizplus · Accepted Answer

The peak current in a series RLC circuit is given by $I_0 = \frac{V_0}{Z}$, where $V_0$ is the peak voltage, and $Z$ is the impedance of the circuit. The impedance $Z$ of a series RLC circuit is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$, where $X_L = 2\pi fL$ is the inductive reactance, $X_C = \frac{1}{2\pi fC}$ is the capacitive reactance, $f$ is the frequency in hertz, $L$ is the inductance in henries, $C$ is the capacitance in farads, and $R$ is the resistance in ohms.Given:- $f = 632 \times 10^3 \, \text{Hz}$- $V_0 = 748 \, \text{V}$- $R = 25 \times 10^3 \, \Omega$- $C = 16 \times 10^{-6} \, \text{F}$- $L = 30 \, \text{H}$First, calculate $X_L$ and $X_C$:- $X_L = 2\pi fL = 2\pi (632 \times 10^3)(30) \approx 119.366 \times 10^6 \, \Omega$- $X_C = \frac{1}{2\pi fC} = \frac{1}{2\pi (632 \times 10^3)(16 \times 10^{-6})} \approx 15.915 \times 10^3 \, \Omega$Then, calculate $Z$:- $Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{(25 \times 10^3)^2 + (119.366 \times 10^6 - 15.915 \times 10^3)^2} \approx 119.366 \times 10^6 \, \Omega$Finally, calculate $I_0$:- $I_0 = \frac{V_0}{Z} = \frac{748}{119.366 \times 10^6} \approx 6.3 \times 10^{-6} \, \text{A} = 6.3 \, \mu\text{A}$The peak current for this circuit is approximately $6.3 \, \mu\text{A}$.

A Series Circuit Has a Sinusoidal Voltage Supplied to It 632kHz632 \mathrm { kHz }632kHz

A Series Circuit Has a Sinusoidal Voltage Supplied to It $632 \mathrm { kHz }$