The emf induced in the second coil is given by:
V = -M(di/dt)
V = emf, M = mutual indutance, di/dt = change of current in the first coil over time
The current in the first coil is given by:
i = i₀
i₀ = 5.0A, a = 2.0×10³s⁻¹
i = 5.0e^(-2.0×10³t)
Calculate di/dt by differentiating i with respect to t.
di/dt = -1.0×10⁴e^(-2.0×10³t)
Calculate a general formula for V. Givens:
M = 32×10⁻³H, di/dt = -1.0×10⁴e^(-2.0×10³t)
Plug in and solve for V:
V = -32×10⁻³(-1.0×10⁴e^(-2.0×10³t))
V = 320e^(-2.0×10³t)
We want to find the induced emf right after the current starts to decay. Plug in t = 0s:
V = 320e^(-2.0×10³(0))
V = 320e^0
V = 320 volts
We want to find the induced emf at t = 1.0×10⁻³s:
V = 320e^(-2.0×10³(1.0×10⁻³))
V = 43 volts
Answer:
the money that would be saved is $13.14.
Explanation:
Given;
power consumed by the light bulb, P = 100 W = 0.1 kW
time of running the bulb, t = 3 hours for 365 days = 1,095 hours
cost rate of power consumption, C = $0.12 per kWh
Energy consumed by the light bulb for the given days;
E = Pt
E = 0.1 kW x 1,095 hr
E = 109.5 kWh
Cost of energy consumed = 109.5 kWh x $0.12 / kWh
= $13.14
Therefore, the money that would be saved is $13.14.
Here’s my work to your question. I used Newton’s Second Law and a kinematics equation to arrive at the answer.