How can I solve a Thevenin problem involving two batteries?

In summary, the conversation discusses an approach to solving a question related to electronics. The suggested steps include changing a voltage source into a current source, using resistors in parallel, and converting back to a voltage source. The final result is a calculation of the current using thevenin.
  • #1
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Homework Statement



Hi there, I got this question from an old test. I would say I have limited knowledge of electronics and would ask for guidance into what approach to take with this question.

Homework Equations





The Attempt at a Solution



No attempt yet. I thought of breaking it up, in 2, obviously, and then getting the series resistance total, but what then? then I am back at square 1, with a "broken" circuit. What small detail am I missing?
 

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  • #2
1) Change the voltage source into a current source using source conversion.
i.e current source with current = 12V / 4ohm= 3 A. - lookin at the left voltage source.
you will get 3A current source with a resistor of 4 ohm connecting parallel to it.

do for the other side(right) too. 8/3=2.67A

2) Since you then have resistors in parallel.
the effective resistance on the left will then be (4*7)/ (4+7)= 2.545 ohm
right will be 2.25 ohm.

3) Convert them back to voltage source.
LEft: 2.545 *3 = 7.635 V
Right : 2.25 * 2.67= 6V

The resultant voltage will be 7.635 - 6 = 1.635 V.
Sum of all the resistance = 10 + 2.545 + 2=14.545 ohm
Hence without using thevenin,
I = 1.635 / 14.545 = 0.1124A

Using thevenin,

After using source conversion,
Golden rule: Make current source disappear; make voltage source a short circuit, meaning: treat it just as a piece of connecting wire.

thevenin voltage = thevenin resistance * short circuit current.
then calculate the current you wanted by current divider rule.

hope this helps
 
  • #3
1) Change the voltage source into a current source using source conversion.
i.e current source with current = 12V / 4ohm= 3 A. - lookin at the left voltage source.
you will get 3A current source with a resistor of 4 ohm connecting parallel to it.

do for the other side(right) too. 8/3=2.67A

2) Since you then have resistors in parallel.
the effective resistance on the left will then be (4*7)/ (4+7)= 2.545 ohm
right will be 2.25 ohm.

3) Convert them back to voltage source.
LEft: 2.545 *3 = 7.635 V
Right : 2.25 * 2.67= 6V

The resultant voltage will be 7.635 - 6 = 1.635 V.
Sum of all the resistance = 10 + 2.545 + 2=14.545 ohm
Hence without using thevenin,
I = 1.635 / 14.545 = 0.1124A

Using thevenin,

After using source conversion,
Golden rule: Make current source disappear; make voltage source a short circuit, meaning: treat it just as a piece of connecting wire.

thevenin voltage = thevenin resistance * short circuit current.
then calculate the current you wanted by current divider rule.

hope this helps
 

Related to How can I solve a Thevenin problem involving two batteries?

What is the concept of Thevenin with two batteries?

Thevenin with two batteries is a theoretical concept in electricity that involves determining the equivalent circuit of a complex electrical network with two voltage sources (batteries) through the use of Thevenin's theorem. This allows for simplification of the circuit and easier analysis of its behavior.

How do you calculate the Thevenin equivalent voltage with two batteries?

The Thevenin equivalent voltage can be calculated by first removing all components (except the two batteries) connected to the two terminals of interest, and then calculating the open circuit voltage between those terminals. This voltage will be the Thevenin equivalent voltage.

What is the purpose of using Thevenin with two batteries?

Thevenin with two batteries is used to simplify complex electrical networks, making it easier to analyze and understand their behavior. It is also useful in designing and troubleshooting circuits, as it allows for easy calculation of the voltage and current in the circuit.

Can Thevenin with two batteries be applied to any circuit?

No, Thevenin's theorem can only be applied to linear circuits, which means that the components in the circuit must have a linear relationship between voltage and current. Non-linear components, such as diodes and transistors, cannot be analyzed using Thevenin's theorem.

Are there any limitations to using Thevenin with two batteries?

Yes, there are a few limitations to using Thevenin with two batteries. The circuit must be in a steady-state condition, meaning that all transients must have settled. Additionally, Thevenin's theorem assumes that the components in the circuit are independent of each other, which may not always be the case in real-world circuits.

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