Question

Q.1 A relay has resistance of 300 ohm and is switched on to 110 V dc supply. If the current reaches
63.2% of its final steady value in
0.002 second, determine
(a) the inductance of the circuit
(b) the final steady state value of the current
(c) the initial
rate of change of current.
​

Answer 1

Given:

R = 300 Ω

V = 110 V

I = 63.2%

T = 0.002 s

To find:

1) inductance l =?

2) steady current =?

3) rate of change of current =?

Explanation:

1)

In the R-L circuit, the time constant is given by

$\displaystyle \tau = \frac{L}{R}$  

⇒ L = T × R

      = 0.002 × 300

  L = 6 H

2)

The final steady state current = $I_m_a_x$

The maximum current flowing through the circuit is the ratio of the applied voltage to the resistance of the conductor.

∴$\displaystyle I _m_a_x = \frac{V}{R}$

          = $\displaystyle \frac{110}{300}$

$\displaystyle \mathbf{I_m_a_x}$ = 0.367 A

3)

The initial rate of change of current is given by the formula

$\displaystyle \frac{dI}{dT} = \frac{d}{dT} \{ \frac{V}{R} \ [ 1 - e^{\frac{-RT}{L}}\ ]\}$

$\displaystyle \frac{dI}{dT} = \frac{d}{dT} \{ \frac{110}{300} \ [ 1 - e^{\frac{-(300)(0.002)}{6}}\ ]\}$

      = $\displaystyle 183.33\ e^{-500t$

Put t = 0 in above equation

$\displaystyle \mathbf{ \frac{dI}{dT} }$ = 183.33 A/s