An inductor-coil of inductance 20 mh having resistance 10 ω is joined to an ideal battery of emf 5.0 v. find the rate of change of the induced emf at t= 0, (b) t = 10 ms and (c) t = 1.0 s.
Answers
L = 20mH; e = 5.0V, R = 10Ω
20mH = 20 × 10⁻³
T = L/R = (20 × 10⁻³) /10
i₀ = 5/10
i = i₀ - i₀e^-t/T²
iR = i₀R - i₀Re^⁻t/T²
a) t = 0
10 × di/dt = d/dt i₀R + 10 × 5/10 × 10/(20 × 10⁻³) × e^-0/10⁻²
= 5/2 × 10⁻³ × 1 = 2.5 × 10⁻³V/s
b) t = 10ms
10ms = 10 × 10⁻³s
Rdi/dt = R × i₀ × 1/Te^-t/T²
dE/dt = 10 × 5/10 × 10/(20 × 10⁻³) × e^-0.01/2 × 10⁻²
= 16. 844V/S
c) at t = 1
dE/dt = Rdi/dt = 5/2 × 10³ × e^10/2 × 10⁻²
= 0.00V/S
L = 20 mH
e =5.0 V
R = 10 Ω
Τ = L/R = 20 x 10^-3 / 10
i base 0 = 5/10
i = I 0(1 – e^–t/τ) ^ 2
i = i base 0 – i base 0 e^-t/τ^2
iR = i0R – i0Re^t / τ^2
a. 10 x di/dt = d/dt i base 0 R + 10 x 5/10 x 10/ 20 x 10^-3 x e^-0 x 10/2 x 10^-2
= 5/2 x 10^-3 x 1
= 5000/2 = 2500
= 2.5 x 10^-3 V/s.
b. Rdi/dt = R x i base 0 x 1/τ x e-t/τ t = 10 ms
= 10 x 10^-3 s dE/dt
= 10 x 5/10 x 10/20 x 10^-3 x e^-0.01 x 10/2 x 10^-2
= 16.844 = 17 V/’
c. For t = 1 s dE / dt = Rdi/dt
= 5/2 10^3 x e^10/2x10^-2 = 0.00 V/s.