A resistor of resistance 30 Ω is connected in series with a capacitor of capacitance 79.5 µF across a power supply of 50Hz and 100V. Find (a) impedance (b) current (c) phase angle and (c) equation for the instantaneous value of current. Ans (a. 50 Ω, b. 2A, c. 53° lead, d. 2.828 sin (314t + 53°) 



Data Given:

Resistance of resistor = R =  30 Ω

Capacitance of capacitor = C = 79.5 μF = 79.5 x10⁻⁶ F

Frequency of A.C. supply = f = 50 Hz

Voltage (rms) = Vrms = 120 V


To Find:

(a) Impedance = Z = ?

(b) Current = ?

(c) Phase angle = ф = ?

(d) Equation for the instantaneous value of current Iᵢₙₛ = ?


Solution:

(a) Impedance = = ?

The formula for Impedance Z is given by

Z R2+X2C ------------(1)

But XC is unknown So will first find it .

XC  = 1C = 12fC

by putting values

XC  = 123.141650Hz79.510F

XC  = 10.02496 Ω

XC  = 40 Ω 

Now Equation (1)

(30Ω)2+(40Ω)2

Z 900Ω2+1600Ω2

Z 2500Ω2

= 50 Ω -----------------Ans. (1)


(b) Current = ?

Since

Irms VrmsZ

putting values

Irms 100V50Ω

Irms = 2 A -----------------Ans. (2)



(c) Phase angle ф = ?

Phase angle can be calculated as

ф = tan⁻¹(XCR)

ф = tan⁻¹(40Ω30Ω)

ф = tan⁻¹(1.333)

ф = 53.13° ---------------------Ans. (3)


(d) Equation for the instantaneous value of current Iᵢₙₛ = ?

The formula for the instantaneous value of current  Iᵢₙₛ is

Iᵢₙₛ Imax sin (ယt + ф)

where ယ = 2ㄫf and Imax = 2 Irms, So

Iᵢₙₛ 2 Irms sin (2ㄫf t + ф)

putting values

Iᵢₙₛ 2 2A sin (2x3.1416x 50 Hz) t + 53.13°)

Iᵢₙₛ 2.828 sin (314 t + 53.13°) A ----------------Ans. (4)




************************************

Shortcut Links For 


1. Website for School and College Level Physics   
2. Website for School and College Level Mathematics  
3. Website for Single National Curriculum Pakistan - All Subjects Notes 

© 2022-Onwards by Academic Skills and Knowledge (ASK    

Note:  Write me in the comment box below for any query and also Share this information with your class-fellows and friends.