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 $25 - \mathrm { k } \Omega$ resistance, a $16 - \mu F$ capacitance, and a $30 - \mathrm { H }$ inductance. What is the peak current for this circuit?

Question

Quizplus · Accepted Answer

The peak current in a series circuit is given by:$$I_{peak} = \frac{V_{peak}}{Z}$$where $V_{peak}$ is the peak voltage and $Z$ is the impedance of the circuit. The impedance of a series circuit is given by:$$Z = \sqrt{R^2 + (XL - XC)^2}$$where $R$ is the resistance, $XL$ is the inductive reactance, and $XC$ is the capacitive reactance. The inductive reactance is given by:$$XL = 2\pi fL$$where $f$ is the frequency and $L$ is the inductance. The capacitive reactance is given by:$$XC = \frac{1}{2\pi fC}$$where $C$ is the capacitance.Plugging in the given values, we get:$$XL = 2\pi (632 \mathrm { kHz }) (30 \mathrm { H }) = 38.6 \mathrm { k } \Omega$$$$XC = \frac{1}{2\pi (632 \mathrm { kHz }) (16 \mu F)} = 20.0 \mathrm { k } \Omega$$$$Z = \sqrt{(25 \mathrm { k } \Omega)^2 + (38.6 \mathrm { k } \Omega - 20.0 \mathrm { k } \Omega)^2} = 43.6 \mathrm { k } \Omega$$$$I_{peak} = \frac{748 \mathrm { V }}{43.6 \mathrm { k } \Omega} = 17.1 \mathrm { mA }$$Therefore, the peak current is 17.1 mA, which is closest to choice A) 6.3 ?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 $632 \mathrm { kHz }$