Question

A coil having resistance of 10Ω and inductance of 1H is switched on to a direct voltage of 100 V. Calculate the rate of change of the current (a) at the instant of closing the switch and (b) when t=L/R (c) Also find the steady state value of the current.

Answer 1

The rate of change of current at the instant of closing the switch is 100 A/s. The rate of change of current when t=L/R is also 100 A/s. The steady state value of the current is 10 A.

Step-by-step explanation:

(a) The rate of change of current at the instant of closing the switch can be calculated using the formula:

Rate of change of current = V / L

Where V is the direct voltage applied and L is the inductance of the coil. Substituting the given values:

Rate of change of current = 100 V / 1 H = 100 A/s

(b) The rate of change of current when t = L/R can be calculated by substituting the values of L and R into the same formula:

Rate of change of current = V / L = 100 V / 1 H = 100 A/s

(c) The steady-state value of the current can be found using Ohm's Law:

I = V / R

Where I is the current, V is the voltage, and R is the resistance. Substituting the given values:

Steady-state value of current = 100 V / 10 Ω = 10 A