Question

When an inductor L and a resistor R in series are connected across a 12 V, 50 Hz supply, a current of 0-5 A flows in the circuit. The current differs in phase from applied voltage by π/3 radian. Calculate the value of R.

Answer 1

hope this would help you

Answer 2

Answer: The value of R is 12 ohm.

Explanation:

Given that,

Voltage $E_{v} = 12 V$

Frequency f = 50 Hz

Current I = 0.5 A

Phase difference $x = 60^{0}$

We know that,

For LR circuit,

$I = \dfrac{E_{v}}{\sqrt{R^{2}+XL^{2}}}$

$\dfrac{E_{v}}{I}={\sqrt{R^{2}+XL^{2}}}$

Squaring on both sides

$(\dfrac{E_{v}}{{I}})^{2}={R^{2}+XL^{2}$

$(\dfrac{12}{0.5})^{2}={R^{2}+XL^{2}$......(I)

Now,

$tan\ x = \dfrac{XL}{R}$

$XL = R\ tan60^{0}$

$XL = R\sqrt{3}$.....(II)

Put the value of XL in equation (I) from equation (II)

$(\dfrac{12}{0.5})^{2}={R^{2}+(\sqrt{3}R)^{2}$

$R = \dfrac{1}{2}\times\dfrac{12}{0.5}$

$R = 12 \Omega$

Hence, The value of R is 12 ohm.