A voltage of 110 V is maintained on a coil with a resistance of 24.6 Ohms. Determine the EMF of self-induction in the coil

A voltage of 110 V is maintained on a coil with a resistance of 24.6 Ohms. Determine the EMF of self-induction in the coil if its inductance is 10 mH and the current disappears within 6 ms.

Given:

R = 24.6 Ohm;

U = 110V;

L = 10 mH;

Δt = 6 ms = 0.006 s.

Find:

E (EMF of self-induction in the coil) -?

To solve the problem, we first of all apply the formula:

Ec = -L * ΔI / Δt.

where L is the inductance (10 mH = 0.01 H);

ΔI – current strength;

Δt – time (6 ms = 0.006 s).

To calculate the current strength, we apply the formula:

I = U / R;

where U is the voltage (110 V);

R – resistance (24.6 ohms).

We apply the formula and get:

I = 110 / 24.6 = 4.47 A.

Let’s find the EMF of self-induction in the coil:

E = -0.01 * 4.47 / 0.006 = -0.01 * 745 = -7.45.