Engineering Finding the Thevenin voltage for the circuit.

AI Thread Summary
The discussion focuses on finding the Thevenin voltage and resistance for a given circuit. The calculated Thevenin resistance is 2.19 ohms, but there are concerns about the accuracy of the formula used, which should be 1/R_th = 1/r1 + 1/r2. The participant attempted to calculate the Thevenin voltage by isolating each voltage source and summing the results, arriving at 8.18V, but was advised to double-check the voltage divider equations and the polarity of voltages. Clarifications on isolating terminals and the correct application of voltage divider principles were also provided. Accurate calculations and attention to detail are emphasized for correct results.
johnsy1312
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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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Hi johnsy1312, Welcome to Physics Forums.

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

johnsy1312 said:
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.

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.
 
\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
 
gneill said:
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.


\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
 
johnsy1312 said:
\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

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.

attachment.php?attachmentid=68013&stc=1&d=1395828469.gif
 

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