Solved Numerical Problem No. 16.5 ( Unit-16: Alternating Current); HSSC-II (Class-12); Physics (Punjab Curriculum and Text Board, Lahore) With Verified Answers

An inductor of pure inductance 3/π H is connected in series with a resistance of 40 Ω. Find (i) the peak value of the current (ii) the rms value, and (iii) the phase difference between the current and the applied voltage V = 350 sin(100πt). (Ans : (i) 1.16 A, (ii) 0.81 A, (iii) 82 .4° )

Data Given:

Inductance = L = `\frac {3}{ㄫ}` H

Resistance = R =  40 Ω

V = 350 sin (100ㄫt)

To Find:

(i) Maximum value of Current = `\I_0` = ?

(ii) Root Mean Square (rms) current = `\I_{rms}` = ?

(iii) Phase difference  = θ = ?

Solution:

General formula for instantaneous current

V = `\V_0` sin ယ t --------------(1)

Given Formula

V = 350 sin (100ㄫt)  ------------(2)

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

`\V_0` = 350 V

and

Angular frequency = ယ = 100 ㄫ  rad s⁻¹

But ယ =  2ㄫf  So,

2ㄫf  = 100 ㄫ  rad s⁻¹

or

f = 50 Hz 

(i) Maximum value of Current = `\I_0` = ?

Since 

Z = `\frac {V_{rms}}{I_{rms}}` = `\frac {V_0/{sqrt 2}}{I_0/{sqrt 2}}` 

or

Z = `\frac {V_0}{I_0}`

or

`\I_0` = `\frac {V_0}{Z}`

where Z = `\sqrt {R^2 + X_L^2}` So,

`\I_0` = `\frac {V_0}{sqrt {R^2 + X_L^2}}`

`\I_0` = `\frac {V_0}{sqrt {R^2 + X_L^2}}`

where `\X_L` = ယ L = 2ㄫf L So,

`\I_0` = `\frac {V_0}{sqrt {R^2 + (2ㄫf L)^2}}`

by putting values

`\I_0` = `\frac {350 V}{sqrt {(40 Ω)^2 + (2 ｘ ㄫ ｘ 50 hz ｘ 3/ㄫ)^2}}`

or

`\I_0` = `\frac {350 V}{sqrt {1600 Ω^2 + (300 Hz  H)^2}`

`\I_0` = `\frac {350 V}{sqrt {1600 Ω^2 + 90000 Ω^2}`

`\I_0` = `\frac {350 V}{sqrt { 91600 Ω^2}`

`\I_0` = `\frac {350 V}{302.5 Ω}`

`\I_0` = 1.16 A -------------Ans. 1

(ii) Root Mean Square (rms) current = `\I_{rms}` = ?

The equation for Root Mean Square (rms) current  `\I_{rms}` is

`\I_{rms}` = `\frac {I_0}{sqrt 2}`

putting values

`\I_{rms}` = `\frac {1.16 A}{sqrt 2}`

`\I_{rms}` = `\frac {1.16 A}{1.414}`

`\I_{rms}` = 0.82 A---------------Ans(2)

(iii) Phase difference  = θ = ?

The phase difference θ between the current and the applied voltage  can be calculated as

θ = tan⁻¹ (`\frac {ധL}{R}`)

or

θ = tan⁻¹ (`\frac {2ㄫf L}{R}`)

By putting values

θ = tan⁻¹ (`\frac {2ㄫｘ50 Hz ｘ 3/ㄫ H}{40 Ω}`)

θ = tan⁻¹ (`\frac {300 Ω}{40 Ω}`)

θ = tan⁻¹ (7.5)

θ = 82.4° -----------------Ans. 3

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