A capacitor with a resistance of 10 Ω and a capacitance value of 100 µF is connected to a supply voltage as V₍ₜ₎ = 100 sin (314t). Calculate the capacitive reactance, circuit impedance, and peak current flowing into the capacitor. Ans (31.85 Ω, 33.4 Ω and 3 A )



Data Given:

Resistance of the capacitor = R =  10 Ω

Capacitance of capacitor = C = 100 μF = 100 ｘ10⁻⁶ F

The equation for supply voltage V₍ₜ₎ = 100 sin (314t)


To Find:

1. Capacitive Reactance = `\X_C` = ?

2. Impedance = Z = ?

3. Maximum Current = `\I_{max}` = ?



Solution:


The general formula for instantaneous Voltage

`\V_C` = `\V_{max}` sin ယ t --------------(1)

Given Formula is

`\V_C` =  100 sin (314t) ------------(2)


By Comparing both equations (1) and (2) we have the following values


Peak Value Voltage = `\V_{max}` = 100 V 

Angular frequency = ယ = 314 rad s⁻¹



1. Capacitive Reactance = `\X_C` = ?

Since the capacitive reactance `\X_C` is given by:

`\X_C`  = `\frac {1}{ယ C}` 

by putting values

`\X_C`  = `\frac {1}{314 rad s⁻¹ｘ 100 ｘ10⁻⁶ F }`

`\X_C`  = `\frac {1}{0.0314}` Ω

`\X_C`  = 31.8 Ω -----------------Ans. (1)



2. Impedance = Z = ?

Z = `\sqrt {R^2 + X_C^2}`

Z = `\sqrt {(10 Ω)^2 + (31.8 Ω)^2}`

Z = `\sqrt {100 Ω^2 + 1011.24 Ω^2}`

Z = `\sqrt {1111.24 Ω^2}`

Z = 33.335 Ω -----------------Ans. (2)


3. Maximum Current = `\I_{max}` = ?

the relation for maximum current flowing in the capacitor is 
 
`\I_{max}` = `\frac {V_{max}}{X_C}` 

putting values

`\I_{max}` = `\frac {100 V}{33.335 Ω}`

`\I_{max}` = 2.9998 A

or 

`\I_{max}` = 3 A -------------Ans. (3)




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