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Please help me with this problem I have been set.

enter image description here

So, I have successfully used Kirchhoff's Voltage law to determine the voltages and currents across each of the resistors in question (R2 and R3). However, when I attempt to use Thevenin's Theorem, I don't really know where to start. I understand the logic of Thevenin's theorem, but when I remove 'R2' (changing it to A-B), I'm not sure how to handle the loop to the right, as it isn't your standard 'two voltage-source' network meeting to go through the 'R2' wire, it has an additional branch. This has completely thrown me and I can't seem to get any of my answers to match the answers I got when calculating them using Kirchhoff's laws. I verified my original answers by simulating the circuit on Multi-sim and measuring the current and voltages across R2 and R3, so I know they are correct, however, everything I seem to try using Thevenin's method doesn't match. Please give me some guidance on this. I appreciate your time.

My calculations using Kirchhoff's laws show that R2 has a voltage of 3.012V and a current of 0.00151A. They also show that R3 has a voltage of 3.012V also and a current of 0.0012A.

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The first thing you need to know before proceeding with the Thevenin's equivalent circuit is that the Vth that you find is not the voltage across your resistor (R2 in this case). In fact it is the voltage that appears at the voltmeter when your probes are at nodes A and B and WHEN YOU HAVE NOTHING CONNECTED TO THE TERMINALS A AND B.

Having said this I start with finding the Thevenin's resistance(Rth). For finding Rth, you need to short the voltage sources. Once you short the voltage sources, you end up with something like this circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

Now placing the ohmmeter in the terminals A and B gives you the resistance of 1.386k (5.6k || 2.5k || 7k). This is the required value of Rth.

Now in order to find the Thevenin's voltage, you can short one voltage at a time and carry out the calculation to find the voltage across the terminals AB due to each source individually. Then you can add up the results. I will short the 7V source and find the voltage at AB due to 15V source.

schematic

simulate this circuit

Reducing the parallel combination of 2.5k and 7k (2k is not taken because it is open circuited actually) into single resistor, the equivalent resistance is 1.84k. Simply applying the voltage divider rule gives you the voltage of 3.71V across AB. This is the voltage at AB due to 15V.

Similarly analyzing the circuit by shorting 15V, you will get the voltage of 1.38V across AB due to 7V source.

Adding these two voltages gives you the total voltage of 5.09V at AB which is the required value of Vth.

So the final circuit looks something like this:

schematic

simulate this circuit

If you place your 2k resistor at the terminal AB you will get the result same as your simulation.

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  • \$\begingroup\$ This is extremely useful. Thank you for the detailed break down of each step. Perfectly displayed and now I finally understand how to approach it. I haven't attempted the Superposition theorem for this particular circuit yet, but is there anything specific I should know with regards to how I should handle it? I'm going to assume breaking it down into two circuits again (holding a different voltage source each time) and calculating final values by 'adding' the values of each circuit? Would I be correct in this approach? Again, thank you so much for this explanation. I greatly appreciate it. \$\endgroup\$ – Sc077i3 Apr 10 at 11:21
  • \$\begingroup\$ @Sc077i3 the superposition theorem is just the analysis of circuit using one source at time. When one source is considered at a time, the other source should be opened or shorted. If the other source is current source it is to be opened. If it is voltage source it is to be shorted as in Thevenin's. Now just find the voltage across the resistor and find the current through it. There's no need to remove the resistor in Superposition theorem. Glad to help you!! \$\endgroup\$ – JuneStar_2918 Apr 10 at 13:32
  • \$\begingroup\$ Perfect, thank you 'Fool No. 2918'. You're certainly no fool! Appreciate the help. \$\endgroup\$ – Sc077i3 Apr 10 at 13:50
  • \$\begingroup\$ Hello @Fool No. 2918, I wonder if you could lend a hand again if you would be so kind. So i've done my calculations like you said for the 2k ohm resistor and everything went fine. However, when I have attempted to perform the same for the 2.5k ohm resistor, I have run into problems. I think its got something to do with the way im dividing the voltage. Could you explain this to me please for both the 2k ohm resistor and the 2.5k ohm resistor. Did you use VS * R2/R1+R2 for all scenarios? or will it change? Im confused... \$\endgroup\$ – Sc077i3 Apr 10 at 23:35
  • \$\begingroup\$ Would you please show your calculations? \$\endgroup\$ – JuneStar_2918 Apr 11 at 2:12

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