The a.c. voltage across a 0.5 µF capacitor is 16 sin (2 ｘ 10³ t) V. Find (a) the capacitive reactance (b) the peak value of current through the capacitor. Ans (1000Ω, 16mA) 


Data Given:

Capacitance of capacitor = C = 0.5 μF = 0.5 ｘ10⁻⁶ F

Instantaneous Voltage across the  capacitor  `\V_C` = 240 sin (2 ｘ 10³ t) 


To Find:

Capacitance Reactance = `\X_C` = ?

Peak Value of Current = `\I_{max}` = ?

Solution:

The general formula for instantaneous Voltage

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

Given Formula is

`\V_C` =  16 sin (2 ｘ 10³ t) ------------(2)


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


Peak Value Voltage = `\V_{max}` = 16 V 

Angular frequency = ယ = 2 ｘ 10³ rad s⁻¹

The equation for capacitance reactance `\X_C` is given by:

`\X_C`  = `\frac {1}{ယ C}`

putting values

`\X_C`  = `\frac {1}{2 ｘ 10³ rad s⁻¹ ｘ 0.5 ｘ10⁻⁶ F}`

`\X_C`  = `\frac {1}{0.001 rad s⁻¹ F}`

`\X_C`  = 1000 Ω ---------------Ans(1)



To find Peak Value of Current `\I_{max}`  we have

`\I_{max}` = `\frac {V_{max}}{X_C}`

putting values

`\I_{max}` = `\frac {16 V}{1000 Ω}`

`\I_{max}` = 0.016 A

or

`\I_{max}` = 16 ｘ 10⁻³ A

`\I_{max}` = 16 mA ------------------Ans.(2)





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