Answer to Question #100546 in Electrical Engineering for Hardi Amiru Sulemana

Question

Two coils A and B are so placed that 80% of the total Flux produced by one links the other. A has 2000 turns and B has 3000 turns. When the current in A is changing at a rate of 500A/s,the Flux linking it,is changing at the rate of 1m wb/s, calculate,
a)the corresponding EMF induced in each coil
b) the self inductance of each coil
c) the mutual inductance of each coil
d) the effective inductance of the two coils in series

Answer 1

a) The corresponding EMF induced in each coil:

$$V_1=-N_1\frac{d\Phi}{dt}=-2000\cdot1\cdot10^{-3}=-2\text{ V}.\\ \space\\ V_2=-kN_2\frac{d\Phi}{dt}=-0.8\cdot3000\cdot1\cdot10^{-3}=-2.4\text{ V}.$$

b) Self-inductance can be calculated from Faraday's law as well:

$$V_1=-L_1\frac{dI_1}{dt},\\ \space\\ L_1=-\frac{V_1}{dI_1/dt}=\frac{2}{500}=4\cdot10^{-3}\text{ H}.$$

For the second coil, since

$$kN_2\frac{d\Phi}{dt}=L_2\frac{dI_2}{dt},\\ \space\\ k\frac{dI_2}{dt}=\frac{dI_1}{dt},\\ \space\\ L_2=-\frac{V_2}{kdI_1/dt}=\frac{2.4}{0.8\cdot500}=6\cdot10^{-3}\text{ H}.$$

c) The mutual inductance:

$$M=k\sqrt{L_1L_2}=\\=0.8\sqrt{4\cdot10^{-3}\cdot6\cdot10^{-3}}=3.9\cdot10^{-3}\text{ H}.$$

d) The effective inductance of the two coils in series cumulatively coupled:

$$L_{\text{cef}}=L_1+L_2+2M=13.9\cdot10^{-3}\text{ H}.$$

Differentially coupled:

$$L_{\text{def}}=L_1+L_2-2M=6.1\cdot10^{-3}\text{ H}.$$