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Here is the problem:

Here is the problem

In the above figure I want help on finding the potential difference between X and Y. It is getting quite confusing due to the battery in the middle. I found the current in both the loops using Kirchhoff's Voltage law but then I'm confused on the proper method to find the potential drop between the two loops.

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I suggest you insert two extra points A and B on both ends of the battery (on the bottom right of the left square and the top left of the right square). Can you calculate the potential difference between X and A, A and B, and finally B and Y? –  Gerben Oct 21 '11 at 12:46
    
Can you be more specific about what exactly you're getting confused about? –  David Z Oct 21 '11 at 14:05
    
@DavidZaslavsky I found the current in both the loops using Kirchhoff's Voltage law but then I'm confused on the proper method to find the potential drop between the points X and Y may be can you explain me the procedure as well as find the potential drop between X and Y since there is a 4V battery in middle through which no current flows I wonder how to calculate the potential drop between X and Y+ –  Praveen Gowda I V Oct 21 '11 at 14:59
    
@PraveenGowdaIV - There is no complete circuit between nodes A and B, so no current flows. The potential across the battery is...the potential across the battery; 4V. –  Kevin Vermeer Oct 21 '11 at 15:33

2 Answers 2

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Consider the loop to the left to be loop A and that to the right to be loop B According to Kirchhoff's voltage law find out the current in loop A and loop B

current in loop A comes out to be 0.4A in the anticlockwise direction

current in loop B comes out to be 0.5A in the clockwise direction

while applying KVL do not consider the 4V battery connecting the two loops since no current flows through it.

Now when it comes to finding the potential difference between the two points X and Y assume you connect a voltmeter between the points X and Y

then the circuit becomes as shown in the diagram below(you can also observe that i have marked different points on the loop with alphabets so that it is easier to explain in the later steps)

voltmeter connected

now when it comes to finding the potential difference you can either consider the potential drop recorded across the path XABPQY or the path XABPRY(you can later verify that both the paths give the same answer)

Now i am going to choose the path XABPRY since this is more simpler

in this path the potential drop is recorded across the 2V battery, 4V battery in the middle and the 3ohm resistor in the second loop

calculations:

so total p.d recorded is going to be +2 volts(by 2V battery), +4V(by the 4V battery) and +3*0.5 V(1.5V) (potential drop across the 3 ohm resistor - taken with positive sign because the assumed direction is opposite to the direction of current)

so the total potential drop between X and Y comes out to be 7.5 V

you will get the same answer if you go along the path XABPQY

Hope it helps.

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That schematic is quite confusing; draw it with higher voltages at the top, lower voltages at the bottom, and signals flowing from left to right per standard procedure.

It's also helpful to label components and nodes so that you can solve the equation symbolically. I haven't labeled some of the nodes here because they can be represented by the names of the voltages or because they're irrelevant to the problem.

Redrawing it this way will probably make it much easier to understand.

enter image description here

I've redrawn it this way using LTspice, with which you can validate your answer.

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