Given data from above circuit is
R=15,L1=12H,L2=30H and M=18H and V=180V
Equivalent inductance=Leq=(L1L2-M2)/(L1L2-2M)
Substitute given data in above equation to get equivalent inductance Leq
Leq=((12*30)-(182))/((12*30)-2(18))
Leq=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 io=0 at t=0,io=V/R at t=infinity because both inductors acts as short circuit.Hence,io=i=180/15=12A.
Current through Leq at any time is
iLeq(t)=i-[i-i0]e-Rt/Leq
Substitute the values to get current through inductor at any time
iLeq(t)=12-[12-0]e-15t/0.11.
iLeq(t)=12-12e136.6t.
iLeq(t)=12(1-e136.6t)
(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 vo=180V at t=0.
at t=,Inductors will short.Hence full current flows through inductors and voltage is 0.vo=0 at t=.
Voltage (vo) across inductor at any time(t)=Ve-Rt/L
vo is voltage across inductor.Hence L=12H,R=15 and V=180V.
vo(t)=180e-15t/12=180e1.25t.
c),d)
Both currents i1,i2 are 0 at t=0
and i1 at t= is
i1=io(L2/(L1+L2))=12(30/(12+30))=12(30/42)=8.57A.
i2=io(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).
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