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How do I find the Thévenin voltage in this diagram?

In parallel the voltage remains the same, so should it be 10 volts across AB?

I am given that since \$R_3\$ and \$R_4\$ are not connected on one end, they do not carry current. Therefore, they cannot have a voltage drop. The voltage present between points A and B is the voltage drop across \$R_2\$.

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The Thévenin equivalent consists of a single voltage source in series with a single resistor, together between points A and B. To find the voltage of the voltage source and the resistor value you consider two different load situations.

1.
No load at all, like it's drawn. The circuit consists of voltage source, R1, R2 and R5. There's no current through R3 or R4. We calculate the current: \$\dfrac{V+}{R1+R2+R5}=\dfrac{10V}{3k + 4k + 3k}=1mA\$. Then the voltage over R2 is 1mA * 4k = 4V, and since there's no voltage drop over R3 or R4 that's also the voltage between A and B.
In the Thévenin equivalent, when A-B is open there won't flow any current, so no voltage drop over the internal resistor. If we want 4V between A and B, the voltage source has to be 4V.

2.
Short-circuit A and B. Now R2 is parallel to the series resistance of R3 and R4. We need to know the equivalent of these (call it R6): \$\dfrac{1}{R6}=\dfrac{1}{R2}+\dfrac{1}{R3+R4}=\dfrac{1}{4k}+\dfrac{1}{6k}=\dfrac{0.417}{1k}\$ so \$R6=\dfrac{1k}{0.417}=2k4\$.
Again we calculate the current: \$\dfrac{V+}{R1+R6+R5}=\dfrac{10V}{3k + 2k4 + 3k}=1.19mA\$. The voltage over R6 is \$10V - 1.19mA \times (R1 + R5) = 2.85V\$, hence the current through R3 and R4 (and the short-circuit A-B) is \$\dfrac{2.85V}{R3 + R4}=\dfrac{2.85V}{6k}=476\mu A\$.
Our Thévenin circuit had a voltage source of 4V. To have 476\$\mu\$A through a short-circuited A-B the internal resistance has to be \$\dfrac{4V}{476\mu A}=8k4\$.

And that's our solution:

Equivalent voltage = 4V,
Equivalent series resistance = 8k4

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  • \$\begingroup\$ @Federico - True, but I find this makes more sense :-) \$\endgroup\$ – stevenvh Jun 27 '11 at 12:59
  • \$\begingroup\$ @ stevenh: Agreed with everything, but found your answer notation confusing. I thought, by "8k4," you meant 80k. I see now that you meant 8.4K. \$\endgroup\$ – Vintage Jun 27 '11 at 18:59
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    \$\begingroup\$ @Vintage - The scaling prefix/infix notation was covered in this answer. \$\endgroup\$ – stevenvh Aug 9 '11 at 12:58
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For Rth,first short the power supply of 10V, then calculate resistance.
R1 is in serie with R5, 3k+3k=6k, the result is in parallel with R4 , (6k x 4k )/(6k+4K)=2k4, then that is in serie with R3 and R4.
2k4+3k+3k=8k4.

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