Question

A coil has 5H inductance and resistance
of 20 ohm in series. An Emf of
100 Volts is applied to it .find the energy stored in inductor in steady state.​

Answer 1

62.5 J is the required answer.

GIVEN:

• Inductance, L = 5 H
• Resistance, R = 20 ohm
• Voltage applied, V = 100 volts

TO FIND:

Energy stored in the inductor, E

FORMULAE:

• According to ohm's law, V = IR
• Energy stored in the inductor, E = (LI²) / 2

SOLUTION:

STEP 1: TO FIND CURRENT

$V = IR \\ \\ 100 = 20 \times I \\ \\ I = \frac{100}{20} \\ \\ I = \frac{10}{2} \\ \\ I = 5 \: ampere$

Current through the inductor, I = 5 A

STEP 2: TO FIND ENERGY

$E = \frac{1}{2} L {I}^{2} \\ \\ E = \frac{1}{2} \times 5 \times {(5)}^{2} \\ \\ E = \frac{1}{2} \times 5 \times 5 \times 5 \\ \\ E = \frac{1}{2} \times 125 \\ \\ E = 62.5 \: joules$

ANSWER:

Energy stored in the inductor is 62.5 J.

OTHER POINTS:

• Inductor is made up of insulated wire wound on a core.

• The inductor is a passive energy storing device.

• It tries to decrease the current when the current is increasing.

• It tries to increase the current when the current is decreasing.

• The SI unit of inductance is henry.