Question

A coil of resistance 10 ohm is connected in series with inductance 0.1 H and a condenser of capacitance of 150 micro farad across a 200 volt, 50Hz supply. Find i) total impedance ii) current iii) power factor iv) voltage across coil v) voltage across condenser.​

Answer 1

Answer:

from a look forward to hearing back in the

Explanation:

minds, I am sure guys soon for fighting

Answer 2

Answer

I = 10A

V coil = 481.73 V

Solution

Given:

Vs = 100 V rms

L = 0.15 H

f=50 Hz

R=10 Ω

From the question I and Vs are in phase.

As per the given question the Phasor diagram is shown below:

Vs and I are in same phase, VL and VC must be equal.

i.e. VL  =  Vc

⇒ I* X_L = I*X_c

⇒ X_L = X_c

total impedance Z of the circuit is:

⸫ Z = R = 10 Ω

We know that, Vs = I*Z

⸫ I = Vs / Z = 10 A

I = 10 A

VL = I*(XL) = I*(2*π*f*L) = 10* (2*π*50*0.15)

VL = 471.238898 V

VL = 471.238898 j

VR = I*R = 100 V

V coil = VR + VL j = 100 + 471.238898 j

V coil = SQRT [(100)^2 + (471.238898)^2]

V_coil = 481.7323936 V

Also we can find out value of C

We have, X_L = X_c

so, 2*π*f*L = 1/(2*π*f*C)

C= 67.5474 μF

Also phase angle lagging in V_coil is:

Phase angle in V_coil = 78.0191

Because I lags V in inductor or coil.

Circuit:

C = 67.5474 μF

RMS values:

Phase plot of VS and I

The results match with Simulation results.

Z = R+j*(w*L-1/(w*C))

V = Z*I V and I being in phase Z should be equal to R then

I = V/R = 10 Amps

V = I*(R+j*w*L)=10*(10+j*2*PI*50*0.15)=100+j*150*PI=100+j*471,24

V = 481.73 Volts with a lagging PHI=-78 degrees

#SPJ3