Question

an rlc series circuit has ( l = 100.0 \text{mh} ) and ( c = 1.00 mu\text{f} ). it is connected to a 1000.0 - hz source emf, and the voltage is found to lead the current by ( 75.0^{circ} ). the value of ( r ) is

( 1750 omega ).
( 126 omega ).
( 1810 omega ).
( 12.6 omega ).
( 175 omega ).

Explanation:

Step1: Calculate inductive and capacitive reactance

• Inductive reactance \(X_L = 2\pi fL\)

Given \(f = 1000\space Hz\), \(L=100\times10^{- 3}\space H\)
\(X_L=2\pi\times1000\times100\times10^{-3}=200\pi\space\Omega\approx 628\space\Omega\)

• Capacitive reactance \(X_C=\frac{1}{2\pi fC}\)

Given \(C = 1\times10^{-6}\space F\)
\(X_C=\frac{1}{2\pi\times1000\times1\times10^{-6}}=\frac{10^{3}}{2\pi}\space\Omega\approx159\space\Omega\)

• Net reactance \(X = X_L - X_C\)

\(X=628 - 159=469\space\Omega\)

Step2: Use phase - angle formula

The phase - angle formula for an RLC series circuit is \(\tan\varphi=\frac{X}{R}\)
Given \(\varphi = 75^{\circ}\), \(X = 469\space\Omega\)
Since \(\tan\varphi=\frac{X}{R}\), then \(R=\frac{X}{\tan\varphi}\)
\(\tan75^{\circ}=\tan(45^{\circ}+ 30^{\circ})=\frac{\tan45^{\circ}+\tan30^{\circ}}{1-\tan45^{\circ}\tan30^{\circ}}=\frac{1+\frac{\sqrt{3}}{3}}{1 - 1\times\frac{\sqrt{3}}{3}}=2+\sqrt{3}\approx3.732\)
\(R=\frac{469}{3.732}\approx126\space\Omega\)

Answer:

126 \(\Omega\) (the second option)