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 = I = ?
(c) Phase angle = ф = ?
(d) Equation for the instantaneous value of current = Iᵢₙₛ = ?
Solution:
(a) Impedance = Z = ?
The formula for Impedance Z is given by
Z = √R2+X2C ------------(1)
But XC is unknown So will first find it .
XC = 1ယC = 12ㄫfC
by putting values
XC = 12x3.1416x50Hzx79.5x10⁻⁶F
XC = 10.02496 Ω
XC = 40 Ω
Now Equation (1)
Z = √(30Ω)2+(40Ω)2
Z = √900Ω2+1600Ω2
Z = √2500Ω2
Z = 50 Ω -----------------Ans. (1)
(b) Current = I = ?
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)
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)
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