Can symmetry help me solve this circuit? Circuit diagram question

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The discussion centers on solving a circuit problem involving symmetry to simplify analysis. Participants confirm that the circuit is symmetric, allowing for simplification by connecting nodes of the same voltage. The consensus is that loop equations are unnecessary for this specific case, as the symmetry provides a clear path to the solution. Users are encouraged to identify equivalent nodes to facilitate circuit simplification.

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Homework Statement


I am trying to find the current across Rg. I do not understand the answer provided

Homework Equations

The Attempt at a Solution


Am I supposed to add the two voltage sources and then continue?
 

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No, you can't simply add them at this point. You'll need to write loop equations.
 
I can't read the writing on the circuit but it might be symmetrical left to right. If it is that may simplify things.
 
CWatters said:
I can't read the writing on the circuit but it might be symmetrical left to right. If it is that may simplify things.

yes it is symmetric
 
In that case you can simplify the circuit. Remember that nodes that are the same voltage can be connected together with a wire without affecting it in any way. Try that. Meanwhile I'll check the given answer is correct.
 
I agree with the given answer.

No need to write loop equations to get that answer.

If you get stuck let us know but see the hint in the previous post.
 
CWatters said:
I agree with the given answer.

No need to write loop equations to get that answer.

If you get stuck let us know but see the hint in the previous post.

I'm a little stuck in trying to simplify the circuit using the hint. I'm not sure set of nodes to connect
 
The left and right hand branches are the same and connect to the same points on the centre branch. Here is a quick sketch with the left hand branch moved over to the right hand side. Electrically the circuit is identical to the original.

Can you see (using symmetry) which points are at the same voltage?

IMG.jpg
 

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