Different loop getting different answers?

Thread Starter

hitmen

Joined Sep 21, 2008
159
I need help with regard to this qns. I am supposed to use Thevenin theorem to find V aross the 8kΩ resistor.

Note that the voltage across the voltage source is 12V.

I broke the circuit after the 4kΩ resistor.
I am now trying to find V ab open circuit.

Since no current flows through 4kΩ, 2mA flows through 2kΩ resisor.

I across 6kΩ is (3k)/(9k) * 2mA = 2/3 mA
I across 3kΩ is 4/3 mA using voltage divider rule.

PD across 3kΩ is 4V.
PD across 6kΩ is 4V.

Here is my qn. If I take the loop across the 12V voltage src, my V ab open circuit will be 4V + 4V + 12V = 20V.

However, if I take the loop across the 6kΩ resistor, V ab open circuit is 4V + 4V = 8V.

Where have I gone wrong?

This problem is adapted from Basic Engineering Circuit Analysis 8th Edition.
 

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mik3

Joined Feb 4, 2008
4,843
The voltage across ab is 8 volts. If you find the thevenin voltage across ab with the voltage divider circuit of the 3K and the 6K resistors then you find 8 volts and a thevenin resistance of 4K.
 

Thread Starter

hitmen

Joined Sep 21, 2008
159
Err.. I am not concerned with the answer. I am more concerned with why I use 2 loops and get different answers. I get 8V for 1 loop and 20V for another.

Can we use pd divider rule when there is a voltmeter in the circuit?
 

mik3

Joined Feb 4, 2008
4,843
Err.. I am not concerned with the answer. I am more concerned with why I use 2 loops and get different answers. I get 8V for 1 loop and 20V for another.

Can we use pd divider rule when there is a voltmeter in the circuit?
We assume the voltmeter has a very high input resistance (open circuit) so we can use the voltage divider rule.

I cant see where did u find the 4+4+12 voltages. Resistor 2K does not determine the thevenin voltage. Think of it.
 

Thread Starter

hitmen

Joined Sep 21, 2008
159
I see. Do you mean to imply that we do not need to add the 12V voltmeter when we determine Vab open circuit. If so, then I understand where I have gone wrong.
 
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