Blog #39 - Calculating Another Circuit Using Thevenin’s Theorem

As part of my problem series in this video, I will be using Loop Analysis Thevenin’s Theorem 

Once I am finished with this series of problems I will be posting them in my Stan store, at this web address…https://stan.store/GVB

Find V0 in this circuit using Thevenin’s Theorem.

Firstly we need to determine the open circuit voltage, V0, 

To begin the process remove the 2 kΩ resistor. 

The open circuit voltage at this point is VTH our Thevenin voltage. The Thevenin resistance is then found with the sources in the network removed (voltage sources short-circuited and current sources open circuit)…let’s find the Thevenin voltage first by letting the voltage drop across the 4 kΩ resistor be V1…and letting the voltage drop across the 3kΩ resistor be V2. Then

VTh = V1 + V2…we now need to determine these voltages. 

We know that the current through the 4KΩ resistor… call it I1…is 4 mA due to the current source. Then clearly, Ohm’s law tells us that…

V1 = I1(4kΩ) = (4mA)(4kΩ) = 16V…to find V2 we need to know I2. and KVL around the loop I2 yields this equation

-12+6k(I2 –I1)+3kI2 = 0…well we know what I1 is so we can plug 4mA into our equation and it becomes

-12 + 6kΩ(I2 – 4mA) + 3kΩI2 = 0…

we know that this is a KVL equation so that all the terms in this equation are going voltages and because we built the equation using thousand Ohm resistors and milliamp currents then I'm going to uncluttered the equation to give us something easier to read…

-12 +     6I2 – 4      +    3I2 = 0

This gives us one equation with one unknown… I2. Therefore we can quite easily solve for the value of I2. Collecting like terms we get 9I2 on the left-hand side of the equation and we'll bring all of our whole numbers to the right hand side of the equation which change from negative numbers to positive numbers and adds up to 36. We can now solve the equation for I2 giving us I2 is equal to… 

4 mA… now V2 is just a voltage drop across the 3kΩ resistor which is 3 kΩ times 4 mA = 12 V. This seems odd…We have a 12 V power supply feeding into a 6kΩ resistor. You’d think there is some voltage drop across the 6kΩ resistor…upon closer examination though we see that the current through the 6 kΩ resistor is I1 - I2 or 4 mA - 4 mA which is equal to zero, so the voltage drop across the 6K on resistor is zero meaning V2 is definitely going to equal the same voltage as the power supply 12 V.

Finally we can solve the equation for VTh which… is 16 V+12 V which…is 28 volts.

The Thevenin equivalent resistance is found by…zeroing all sources and looking into the open circuit terminals to determine the Thevinen resistance RTh. From the network we see that the 6k and 3k Ohm resistors are in parallel and that combination is in series with the 4kΩ resistor. 

Thus RTH =4kΩ+3kΩ||6kΩ = 6kΩ

Therefore, the Thevenin equivalent circuit consists of a 28V source in series with the 6kΩ resistor. If we connect the 2kΩ resistor to this equivalent network we obtain this circuit. V0 is the voltage drop across the 2kΩ resistor which we can find using voltage division…which gives us V0 = 7 Volts

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