In the circuit below, the voltage source is a 5 V power supply. The inductor is
ID: 1410855 • Letter: I
Question
In the circuit below, the voltage source is a 5 V power supply. The inductor is considered to be ideal (it has no internal resistance) and large (say, 1 henry). R_1 = 2 Ohm and R_2 =8 Ohm. Assume the switch had been closed for a long tune. Use Kirchhoff's rules to determine the voltage drop across R_2. V_2 = At the instant just after the switch was opened, what is the current though the inductor? I_L = At the instant just after the switch was opened, what is the current though R_2? I_2 = At the instant just after the switch was opened, what the current though R_1? I_1 = At the instant just after the switch was opened, what the magnitude of the voltage drop across R_1? V_1 = Suppose R_1 were replaced with a 20 Ohm resistor and the switch had been closed for a long tune. What would the magnitude of the voltage drop across R_1 be just after the instant in which the switch was opened? With R_1 = 20 Ohm. V_1 =Explanation / Answer
Hi,
In this case we have an RL circuit. When the switch is closed, then the current flows through the circuit and it can be charged by the battery. When the switch is opened, the current should not be able to flow, but if we ignore this fact we could say that the circuit will enter a discharging process.
The relations for both process are as follows:
I(t) = V/R (1 - exp(-t/A)) (charging process)
I(t) = V/R exp(-t/A) (discharging process)
A = L/R and R = R1 + R2
V = IR (Ohm's Law, which I assume as valid)
1) If the switch has been closed for a long time, we can consider that t is infinity and that the circuit has reached its maximum current, therefore:
I(infinity) = V/R ::::::: I = 5 V / 10 = 0.5 A
As the current is the same due to the resistances being in series, we have the following:
V2 = IR2 = (0.5 A)*8 = 4 V
2) Once the switch is opened, the discharging process begins, therefore we can consider this moment as the time equal to zero, so:
I(0) = V/R = 0.5 A
The current through the inductor should be the same as the current shown above, as well as the same through any element (device) of the circuit as all of them are in series, therefore:
I1 = I2 = IL = 0.5 A
The voltage drop across R1 at t = 0 is therefore:
V1 = IR1 = (0.5 A)*2 = 1 V
3. We have changed the value of R1, so there is a new maximum current through the circuit once a long time has passed. Again we consider that the time is equal to infinity:
I = V/R = 5 V / 28 = 0.18 A
If then the switch is opened, the circuit enters a discharging process, but as we consider the instant where t=0, the current remains the same, so the voltage drop through R1 is:
V1 = IR1 = (0.18 A)*(20 ) = 3.6 V
I hope it helps.