# Finding the Thevenin voltage for the circuit.

1. Mar 24, 2014

### johnsy1312

1. For the following circuit, i must find the thevenin's resistance and voltage.

Using R_th = 1/r1 +1/r2 may be effective in finding resistance and V_th = Vout*(R/Rt) may be effective in finding voltage

i have managed to find the thevenin's resistance to be 2.19ohms. I attempted to find the thevenin's voltage by isolating each voltage terminal one at a time to find the voltage and then adding these two voltages together, but i didn't answer correctly.

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2. Mar 25, 2014

### Staff: Mentor

Hi johnsy1312, Welcome to Physics Forums.

In future please retain and use the formatting template provided when starting a new thread.

Can you show your calculations in more detail? Your formula for the Thevenin resistance is not quite right (should be $1/R_{th} = 1/r1 + 1/r2$) but it looks like you made the correct calculation anyway. Be careful of rounding though... it looks to me like you rounded up the final digit for no reason.

How did you arrive at your formula for Vth? Show your calculations.

3. Mar 26, 2014

### johnsy1312

$\frac{}{}$Firstly i isolated the 12V terminal and calculated the voltage:
$V= \frac{18*3}{3+8}=4.91V$
Then i isolated the 18V terminal,
$V= \frac{12*8}{3+8}=3.27V$
V_th = 4.91 + 3.27 = 8.18V

4. Mar 26, 2014

### johnsy1312

$\frac{}{}$Firstly i isolated the 12V terminal and calculated the voltage:
$V= \frac{18*3}{3+8}=4.91V$
Then i isolated the 18V terminal,
$V= \frac{12*8}{3+8}=3.27V$
V_th = 4.91 + 3.27 = 8.18V

5. Mar 26, 2014

### Staff: Mentor

First, when you say isolate a terminal, I presume you mean suppress a voltage source.

Second, check your voltage divider equations; the resistance that appears in the numerator should be the one that you want to find the potential across. For example, with the 12 V source suppressed and the 18 V source driving the circuit, that would be the 8 Ω resistor.

Third, pay close attention to the polarities of the currents and voltages. The polarity of the voltage created by a potential drop across a resistor depends upon the current direction, which in turn depends upon the polarity of the source voltage.

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