A $120 - \mathrm { V }$ rms voltage at $1.00 \mathrm { kHz }$ is applied to a resistor and an inductor in series. If the impedance of this circuit is $110 \Omega$, what is the maximum value of the current?
A) $1.54 \mathrm {~A}$
B) $1.04 \mathrm {~A}$
C) $1.84 \mathrm {~A}$
D) $1.09 \mathrm {~A}$

Question

Quizplus · Accepted Answer

The impedance of a series circuit is given by $Z = \sqrt{R^2 + X_L^2}$, where $R$ is the resistance and $X_L$ is the inductive reactance. In this case, $Z = 110 \Omega$ and $V_{rms} = 120 \mathrm { V}$. The inductive reactance is given by $X_L = 2\pi f L$, where $f$ is the frequency and $L$ is the inductance. In this case, $f = 1.00 \mathrm { kHz}$. Therefore, $X_L = 2\pi (1.00 \mathrm { kHz})(L)$. The maximum value of the current is given by $I_{max} = \frac{V_{rms}}{Z}$. Therefore, $I_{max} = \frac{120 \mathrm { V}}{110 \Omega} = 1.09 \mathrm { A}$.

A 120−V120 - \mathrm { V }120−V Rms Voltage At 1.00kHz1.00 \mathrm { kHz }1.00kHz

A $120 - \mathrm { V }$ Rms Voltage At $1.00 \mathrm { kHz }$