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 = XC = ?

2. Impedance = Z = ?

3. Maximum Current = Imax = ?



Solution:


The general formula for instantaneous Voltage

VC = Vmax sin ယ t --------------(1)

Given Formula is

VC =  100 sin (314t) ------------(2)


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


Peak Value Voltage = Vmax = 100 V 

Angular frequency = ယ = 314 rad s⁻¹



1. Capacitive Reactance = XC = ?

Since the capacitive reactance XC is given by:

XC  = 1ယC 

by putting values

XC  = 1314rads⁻¹ｘ100ｘ10⁻⁶F

XC  = 10.0314 Ω

XC  = 31.8 Ω -----------------Ans. (1)



2. Impedance = Z = ?

Z = √R2+X2C

Z = (10Ω

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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