Question

A pure inductance of 318 mH is connected in series with apure resistance of 75 Ohm. The circuit is supplied from 50 Hzsource and the voltage across 75 Ohm resistor is found to be150 V. The supply voltage and phase angle respectively are​

Answer 1

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Answer 2

Given:

Inductance (L) = 318 mH = 318 × 10⁻³ H

Resistance (R) = 75 Ω

Frequency (f) = 50 Hz

Voltage across 75Ω resistor ($V_{R}$) = 150 V

To find:

1. Supply voltage (V)
2. Phase angle (∅)

Explanation:

• We know, in a resistor, that the current and voltage are in the same phase. Whereas, in an inductor, the voltage and current has a phase difference of 90°.
• Hence, we understand that the $V_{R}$ and $V_{L}$ will be perpendicular to each other.

• Hence, the supply voltage can be found out by the law of vector addition or parallelogram rule that is

            $V = \sqrt{(V_{L} )^{2}+V_{R}^{2}+2V_{L} V_{R}cos\alpha }$  

Solution:

Step 1

We know, from Ohm's law.

$V = I R\\I = \frac{V}{R}$

Substituting the values, we get

$I = \frac{150}{75} ; or\\I = 2A$

Step 2

We have,

$L = 318 mH$ and

We know, Inductive reactance

$X_{L}$ = ωL

$X_{L}$ = 2π f L

Substituting the values, we get

$X_{L} = 2 (3.14) (50)(318)10^{-3}$ ; or

$X_{L}$ = 100 Ω (approx)

For voltage across the Inductor, we have

$V_{L} = I X_{L}$ ; or

$V_{L} = 2(100) ; or\\V_{L} = 200 V$

Step 3

Hence, supply voltage will be

$V = \sqrt{(150)^{2}+(200)^{2}+ 2(150)(200)sin(90) }\\sin90 = 1 \\V=\sqrt{22500 + 40000} \\V = \sqrt{62500} ; or\\V = 250 V$

Step 4

We know, phase angle is

tan∅ $= \frac{X_{L} }{R}$ Substituting the values, we get

tan∅  $=\frac{100}{75}$

or ∅ $= tan^{-1}(\frac{4}{3})$

Final answer :

Hence, the supply voltage will be 180 V and the phase angle is $tan^{-1}(\frac{4}{3} )$.