Question

igar. 150 3. There is no energy stored in the circuit initially. At t-0, the switch is closed a) Find io for t20. b) Find vo for t20. c) Find i for t20. d) Find i for t20. 18H t0t 30H3 12H 180 V ,42-M2 h2-2M Note: Leg

Answer 1

Given data from above circuit is

R=15$\Omega$,L₁=12H,L₂=30H and M=18H and V=180V

Equivalent inductance=L_eq=(L₁L₂-M²)/(L₁L₂-2M)

Substitute given data in above equation to get equivalent inductance L_eq

L_eq=((12*30)-(18²))/((12*30)-2(18))

L_eq=36/324=0.11H.

(a)

At t=0,inductors will be open because they don't allow sudden changes in current if switch is closed suddenly

Both L₁ and L₂ will be opened and no current flow in the circuit at t=0.

and i_o=0 at t=0,i_o=V/R at t=infinity because both inductors acts as short circuit.Hence,i_o=i_$\oe$=180/15=12A.

Current through L_eq at any time is

i_Leq(t)=i_$\oe$-[i_$\oe$-i₀]e^-Rt/L_eq

Substitute the values to get current through inductor at any time

i_Leq(t)=12-[12-0]e^-15t/0.11.

i_Leq(t)=12-12e^136.6t.

i_Leq(t)=12(1-e^136.6t)

(b).

v₀ is voltage across inductor L₁ and voltage across both inductors is same as both are connected in parallel.

At t=0,Inductors are open.Hence the voltage 180V appears across inductors and v_o=180V at t=0.

at t=$\oe$,Inductors will short.Hence full current flows through inductors and voltage is 0.v_o=0 at t=$\oe$.

Voltage (v_o) across inductor at any time(t)=Ve^-Rt/L

v_o is voltage across inductor.Hence L=12H,R=15$\Omega$ and V=180V.

v_o(t)=180e^-15t/12=180e^1.25t.

c),d)

Both currents i₁,i₂ are 0 at t=0

and i₁ at t=$\oe$ is

i₁=i_o(L₂/(L₁+L₂))=12(30/(12+30))=12(30/42)=8.57A.

i₂=i_o(L₁/(L₁+L₂))=12(12/42)=3.42A.

i₁ at any time is

i₁(t)=i₁(1-e^-Rt/L₁)=8.57(1-e^-15t/12)=8.57(1-e^-1.25t).

i₂ at any time is

i₂(t)=i₂(1-e^-Rt/L₂)=3.42(1-e^-15t/30)=3.42(1-e^-0.5t).