Question

When a series combination of a coil of inductance L and a resistor of resistance R is connected across a 12 V-50 Hz supply, a current of 0.5.A flows through the circuit. The current differs in phase from applied voltage by π / 3 radian. Calculate the value of L and R.

Answer 1

d= volume of the oil drop/ area of the film

 

= 4/3 *pi* (0.025cm)^3/(pi* (10 cm)^2

 

= 2.08X10^-7 cm

Answer 2

Value of inductance 'L' = $12\sqrt{3} H$ and the value of resistance R = 12 Ω.

Explanation:

Given:

            Emf 'E' = 12 V - 50 hz

               Current 'I' = 0.5 A

               Phase difference = $\frac{\pi }{3}$ = $60^{o}$

       

  We know that

                              ⇒ $\frac{E}{\sqrt{R^{2} + L^{2} } }$ = I

       

                             ⇒ $R^{2} + L^{2} = \frac{E}{I}^{2}$           ..........     (1)

            Now, we know, phase difference

                             ⇒  tan $60^{o}$ = $\frac{L}{R}$

                             ⇒  L = $\sqrt{3} R$

Put above equation in equation (1)

                            ⇒  4 $R^{2} = \frac{E}{I}^{2}$

                            ⇒  R = $\sqrt{\frac{12}{0.5\times2} }$

                            ⇒  R = 12 Ω

Now inductance,                

                             ⇒ L = $12\sqrt{3}$ H