The voltage across a 0.01 µF capacitor is 240 sin (1.25 ｘ 10⁴ t - 30°) V. Write the mathematical expression for the current through it. 

Ans (I = 0.03 sin(1.25 ｘ 10⁴ t + 60°) A)



Data Given:

Capacitance of capacitor = C = 0.01 μ F = 0.01 ｘ10⁻⁶ F

Instantaneous Voltage across the  capacitor  `\V_C` = 240 sin (1.25 ｘ 10⁴ t - 30°) 


To Find:


Mathematical Expression for the Current in a capacitor = `\I_C` = ?

Solution:

The general formula for instantaneous Voltage

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

Given Formula is

`\V_C` =  240 sin (1.25 ｘ 10⁴ t - 30°)   ------------(2)


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


Peak Value Voltage = `\V_{max}` = 240 V 

Angular frequency = ယ = 1.25 ｘ 10⁴ rad s⁻¹

ф = 30°



The equation for instantaneous Current `\I_{max}` in capacitor is

`\I_C` = `\I_{max}` sin (ယ t - ф + 90°)  ---------(3)

Where

`\I_{max}` = `\frac {V_{max}}{X_C}` 

but `\X_C` = `\frac {1}{ယC}`

Hence

`\I_{max}` = `\V_{max}` ယC 

putting the `\I_{max}` = `\V_{max}` ယC  in equation (3)

`\I_C` = `\V_{max}` ယC  sin (ယ t - ф + 90°) 

now putting the corresponding values

`\I_C` = 240 V ｘ 1.25 ｘ 10⁴ rad s⁻¹ｘ 0.01 ｘ 10⁻⁶ ｘ sin (1.25 ｘ 10⁴ rad s⁻¹ t - 30° + 90°) 

or 

`\I_C` = 0.03  ｘ sin (1.25 ｘ 10⁴ rad s⁻¹ t + 60°) A

or 

I = 0.03 sin (1.25 ｘ 10⁴ t + 60°) A -------------Ans.
Hence the required mathematical relation for current in a capacitor. 


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