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
