circuitspassive-networksresistorstime
So I'm working on an RL circuit and the solution is stated to be that R1 and R3 are in series, but I didn't think that was the case because they don't have the same current through them. Given solution says to add R1 + R3 in series and then use that equivalent resistance and add to R2 as if it were in parallel.
TL;DR: How do I find the equivalent resistance of the circuit below so that I can find the time constant for this RL circuit?
Looks like a trick question. Circuit is shorted, so all the current will flow through the wire. Current through \$L_1\$ is 0. Ran the SPICE sim just to be sure. Results and netlist below :)
I1 0 0 10m
R1 0 0 15
R2 0 N001 47
L1 N001 0 10m
.tran 0 10m 0 1u
.backanno
.end
The two possible solutions are as follows.
The first diagram is Ra in parallel with R2. Then R3 is in series with the result. Finally, R1 is in parallel with the result.
Ra || R2 = 0.99
0.99 + R3 = 10.99
10.99 || R1 = 5.15
Notice that in the second diagram the voltage of the supply is lower. This is called a Thevenin equivalent.
For a Thevenin equivalent first, remove the load and calculate the voltage across where the load was. In this case, R2 and R3 form a resistive divider.
V1 * (R2 / (R3 + R2)) = 0.090V
Then second, short the supply (turn it off or reduce it to zero for a voltage supply and calculate the equivalent resistance. With the supply sorted R3 is completely bypassed. This leaves R2 in parallel with R1.
R1 || R2 = 0.9 ohms.
Finally, take the voltage and resistance and set it up in the manner shown.
simulate this circuit – Schematic created using CircuitLab
To keep current the same though Ra first you need to calculate what the current in Ra is. Then find the resistance needed to keep the same current. First Ra and R2 are in parallel.
R2 || Ra = .99
Then the result is in series with R3. (R1 is ignored because it is directly across the supply and does not affect Ra)
R3 + 0.99 = 10.99
The current is then V1 / 10.99 = 0.090A
Then you do a current divider to see what part of the current is flowing through Ra and what part is flowing through R2.
0.090 * (R2 / (Ra + R2)) = 9.09e-5
Now set up an equasion with X equal to the needed resistance. The total voltage drop in the circuit should be 1V.
9.09e-5 * 1000 + 9.09e-5 * X = 1
Solve for X to get 10121 ohms.
Best Answer
If you are trying to find the R-L time constant (L/R numerically) then the constant current generator has infinite impedance and can be ignored. This leaves R1 and R3 in series (R1+R3) in parallel with R2 hence the effective resistance of the circuit is: -
R2 || (R1 + R3)
Or do it using Thevenin's theorum; a current source in parallel with a resistor (R1) is equivalent to a voltage source in series with that resistor and this of course makes R1 in series with R3.