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) = `\V_{rms}` = 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 = `\sqrt {R^2 + X_C^2}` ------------(1)

But `\X_C` is unknown So will first find it .

`\X_C`  = `\frac {1}{ယ C}` = `\frac {1}{2ㄫf C}`

by putting values

`\X_C`  = `\frac {1}{2 ｘ 3.1416 ｘ 50 Hz ｘ 79.5 ｘ10⁻⁶ F }`

`\X_C`  = `\frac {1}{0.02496}` Ω

`\X_C`  = 40 Ω 

Now Equation (1)

Z = `\sqrt {(30 Ω)^2 + (40 Ω)^2}`

Z = `\sqrt {900 Ω^2 + 1600 Ω^2}`

Z = `\sqrt {2500 Ω^2}`

Z = 50 Ω -----------------Ans. (1)


(b) Current = I = ?

Since

`\I_{rms}` = `\frac {V_{rms}}{Z}`

putting values

`\I_{rms}` = `\frac {100 V}{50 Ω}`

`\I_{rms}` = 2 A -----------------Ans. (2)



(c) Phase angle = ф = ?

Phase angle can be calculated as

ф = tan⁻¹(`\frac {X_C}{R}`)

ф = tan⁻¹(`\frac {40 Ω}{30 Ω}`)

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 = `\I_{max}` sin (ယt + ф)

where ယ = 2ㄫf and `\I_{max}` = `\sqrt 2` `\I_{rms}`, So

Iᵢₙₛ = `\sqrt 2` `\I_{rms}` sin (2ㄫf t + ф)

putting values

Iᵢₙₛ = `\sqrt 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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