In the circuit shown in the figure (Figure 1) , E = 56.0 V , R1 = 45.0 , R2 = 26
ID: 1795131 • Letter: I
Question
In the circuit shown in the figure (Figure 1) , E = 56.0 V , R1 = 45.0 , R2 = 26.0 , and L = 0.350 H .
Part A
Switch S is closed. At some time t afterward the current in the inductor is increasing at a rate of di/dt=50.0A/s. At this instant, what is the current i1 through R1?
Part B
Switch S is closed. At some time t afterward the current in the inductor is increasing at a rate of di/dt=50.0A/s. At this instant, what is the current i2 through R2?
Part C
After the switch has been closed a long time, it is opened again. Just after it is opened, what is the current through R1?
2 SaExplanation / Answer
When dI/dt = 50 , the voltage drop across it will be
V = L (dI/dt ) = 50 *.35 = 17.5 V
FOr part one , the voltage across R1 is 56v
I1 = V/R = 56/45 = 1.244 A
For the part 2
The voltage drop is V - Vl = 56- 17.5 = 38.5
hence the current will be
I2 = 38.5/26 = 1.48 A
The current in case will be generated due to stored energy in the inductor.
Hence it will be
I= 17.5/45+26 =0.246 A