Please do both problems The following circuit is in a steady

The following circuit is in a steady state and at t= 0 the switch $1 is opened, what will be the current flowing through R2 at t= 0, right after the switch is opened, for vs=18V R1 = 30 and R2 = 6ohm and L=3mH? (a) 9 A (b) 3 A (c) 6 A (d) 4.5 A 14. The voltage and current through a load is: V_L = 10 cosomegat and i_Z = 2sin (omegat + 30degree) The nature of the load is (a) capacitive (b) Inductive

Solution

13)

Before t=0 the circuit is steady state means inductor acts as a short circuit ,

the equivalent circuit considering the fact above is resistors R1 and R2 in parallel with voltage souce Vs

As R1 is in series with inductor L(which acts as short circuit in steady state ) ,current through it is given by

R1=3 ohms, R2=6 ohms Vs=18 V

I_L=I_R1=Vs/R1=18/3=6 A

so I_L=6 A

When switch is opened t=0 the inductor current cannot change instantaneously as previously it is carrying I_L=6 A

So it form the closed circuit with R1 ,R2 in series with L ,so same current circulates through the loop

So current through R₂ I_R2=6A right after switch is opened

I_R2=6A

14)

VL=10 cos(wt) and IL=2 sin(wt+30)=2 cos(90-(wt+30))=2 cos(60-wt)=2cos(wt-60)

following formulaes used above for simplifying IL

sin(t)=cos(90-t) and cos(-t)=cos(t)

now both voltage and current are cosine functions

VL=10 cos(wt)

IL= 2cos(wt-60)

From this current lags voltage by 60 degrees

So the circuit is Inductive