Question

A coil with an inductance of 2.5 H and a resistance of 15 Ω is suddenly connected to an ideal battery with ε = 140 V. At 0.15 s after the connection is made, what is the rate at which (a) energy is being stored in the magnetic field, (b) thermal energy is appearing in the resistance, and (c) energy is being delivered by the battery?

Answer 1

Answer:

Explanation:

Given that,

Inductance

L = 2.5H

Resistance

R = 15 Ω

EMF of battery

ε = 140 V

Time

t = 0.15s

A. Rate at which Energy stored in the magnetic field

Energy stored can be determined using

U = ½Li²

Then, the rate is

dU/dt = d(½Li²)/dt

Then inductance is constant

Then, dU/dt = Li• di/dt

Where i(t) is give as

i(t) = ε/R (1 —exp(-t/τ))

Where τ is time constant and it is given as

τ = L/R = 2.5/15 = 0.167 seconds

Getting the current I(t)

i(t) = ε/R (1 —exp(-t/τ))

I(t) = 140/15(1—exp(-0.15/0.167)

I(t) = 9.33 × (1—0.407)

I(t) = 9.33 × 0.593

I(t) = 5.54 A

Getting di/dt

di/dt = ε/R (- -1/τ•exp(-t/τ)

di/dt = ε/Rτ exp(-t/τ)

Substituting the given values

di/dt = 140/(15×0.167) exp(-0.15/0.167)

di/dt = 56 × exp(-0.9)

di/dt = 22.78 A/s

Then,

dU/dt = Li• di/dt

dU/dt = 2.5 × 5.54 × 22.78

dU/dt = 315.32 Watts

B. Rate of thermal energy appearing in the resistance

The power dissipated in the resistor can be determined using

P = i²R

We already know the current at t = 0.15s

I = 5.54A and R = 15 ohms

Then,

P = i²R

P = 5.54² × 15

P = 460.37 Watts

C. Rate of energy delivered by the battery

P = iV

P = 5.54 × 140

P = 775.6 Watts