Answer:
0.0173 V
Explanation:
PARAMETERS GIVEN:
Inductance, L = 2 mH = 0.002 H
Time interval, dt = 0.15 s
Change in current, dI = 1.5 - 0.2 = 1.3 A
The magnitude of the induced EMF is given as:
EMF = | -L*dI/dt |
EMF = (0.002 * 1.3) / 0.15
EMF = 0.0173 V
Answer:
0.0173 V
Explanation:
The Emf induced in a coil is given as
E = Ldi/dt......................... Equation 1
Where E = Emf induced in a coil, L = Inductance of the coil, di = change in current, dt = time interval/change in time.
Given: L = 2.0 mH = 2×10⁻³ H, di = 1.5-0.2 = 1.3 A, dt = 0.15 s
Substitute into equation 1
E = 2×10⁻³(1.3/0.15)
E = 1.73×10⁻² V
E = 0.0173 V
Hence the emf induced in the coil = 0.0173 V
