Question

an rlc circuit contains a 50.0 mh inductance, a 300 ω resistance and a 200 nf capacitance. it is driven by an emf at 1000 hz with an amplitude of 5.0 v. at time t = 0, the driving emf is zero. what is the power factor of the circuit?
○ 0.20
○ 0.41
○ 0.32
○ 0.64
○ 0.53

Explanation:

Step1: Calculate inductive reactance \(X_L\)

The formula for inductive reactance is \(X_L = 2\pi fL\).
Given \(f = 1000\space Hz\) and \(L=50.0\times10^{- 3}\space H\), then \(X_L=2\pi\times1000\times50\times10^{-3}=100\pi\space\Omega\approx314\space\Omega\)

Step2: Calculate capacitive reactance \(X_C\)

The formula for capacitive reactance is \(X_C=\frac{1}{2\pi fC}\).
Given \(f = 1000\space Hz\) and \(C = 200\times10^{-9}\space F\), then \(X_C=\frac{1}{2\pi\times1000\times200\times10^{-9}}=\frac{10^{6}}{400\pi}\space\Omega\approx796\space\Omega\)

Step3: Calculate impedance \(Z\)

The formula for impedance in an RLC - circuit is \(Z=\sqrt{R^{2}+(X_C - X_L)^{2}}\).
Given \(R = 300\space\Omega\), \(X_L\approx314\space\Omega\), \(X_C\approx796\space\Omega\), then \(Z=\sqrt{300^{2}+(796 - 314)^{2}}=\sqrt{300^{2}+482^{2}}=\sqrt{90000 + 232324}=\sqrt{322324}=568\space\Omega\)

Step4: Calculate power factor

The power factor formula is \(\cos\varphi=\frac{R}{Z}\).
Substitute \(R = 300\space\Omega\) and \(Z = 568\space\Omega\) into the formula, \(\cos\varphi=\frac{300}{568}\approx0.53\)

Answer:

0.53