Question

A series$RLC$ circuit consists of a$75\Omega$resistor, a $0.12H$ inductor, and a $30mF$ capacitor. It is attached to a$120V/60Hz$ power line. What are (a) the peak current$I$, (b) the phase angle $\varphi$, and (c) the average power loss?

Short answer

a) The peak current is$1.37A$

b) Phase angle is $0.54°$

c) The average power loss is$69.19Watt$

Step-by-step solution

Part(a) Step 1: Given information 

We are given that $75\Omega$resistor , a $0.12H$inductor, and a $30mF$capacitor are there in $RLC$circuit. We are given that frequency of power line source equal to $60Hz$and voltage $120V$.

We need to find peak current.

Part(a) Step 2: Simplify 

Firstly we will find inductive reactance ,$X_L=2\mathrm{\pi fL}=2\mathrm{\pi }\times 60\times 0.12=45.2$$\Omega$

then we will capacitive reactance , $X_C=\frac{1}{2\mathrm{\pi fC}}=\frac{1}{2\mathrm{\pi }\times 60\times 30\times {10}^{-3}}=0.084$$\Omega$

We will calculate impedance by putting the value of $R,X_L,X_C$

$Z=\sqrt{R^2+{\left(X_L-X_C\right)}^2}=87.5$

Peak current = $I=\frac{V_0}{Z}=\frac{120}{87.5}$$=1.37A$

Part(b) Step 1: Given information 

We are given that $75\Omega$resistor ,a $0.12H$inductor, and a $30mF$capacitor are there in $RLC$circuit. We are given that frequency of power line source equal to $60Hz$and voltage $120V$.

We need to find phase angle .

Part(b) Step 2: Simplify 

Firstly we will find inductive reactance ,$X_L=2\mathrm{\pi fL}=2\mathrm{\pi }\times 60\times 0.12=45.2$$\Omega$

then we will capacitive reactance, $X_C=\frac{1}{2\mathrm{\pi fC}}=\frac{1}{2\mathrm{\pi }\times 60\times 30\times {10}^{-3}}=0.084$$\Omega$

Phase angle is given by ,

$\varphi ={\tan }^{-1}\left[\frac{X_L-X_C}{R}\right]={\tan }^{-1}\left[\frac{45.116}{75}\right]={\tan }^{-1}\left(0.601\right)=0.54°$

Part(c) Step 1: Given information 

We are given that $75\Omega$resistor, a $0.12H$inductor, and a $30mF$capacitor are there in $RLC$circuit. We are given that frequency of power line source equal to $60Hz$and voltage $120V$.

We need to find average power loss.

Part(c) Step 2: Simplify

Firstly we need to find , $I_{rms}=\frac{E_{rms}}{Z}$

then we can find ,$E_{rms}=\frac{E_0}{\sqrt{2}}=84.8$$V$

So from here we get, $I_{rms}=\frac{84.8}{87.5}=0.96$$A$

Power factor , $\cos \left(0.54\right)=0.85$

Average power loss $=I_{rms}{E}_{rms}\cos \varphi =84.8\times 0.96\times 0.85=69.19watt$