Solved Numerical Problem No.3 ( Unit-15: Alternating Current); HSSC-II (Class-12); Physics (New Book/KPK Text Board, Peshawar) With Verified Answers

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)

Similar Questions:

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)

An alternating current is represented by the equation  I = 20 sin 100 ㄫt. Compute its frequency and the maximum and rms values of current. (Ans :50 Hz, 20 A , 14 A )

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