Finding the potential difference in a circuit

In summary: However, due to the impossible nature of the circuit described, it is not possible to accurately determine the potential difference using either path.
  • #1
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Homework Statement
For the circuit below, calculate the potential difference between points a and b. The current in the 2.00 Ω resistor is 0.909 A, and the current in the 4.00 Ω resistor is 1.636 A.
Relevant Equations
loop rule
junction rule
1675383984537.png

The solution chooses the centre wire to determine the potential difference, where Va−(0.909 A)(2.00 Ω)=Vb and Vb - Va = -1.82

If I choose the top wire (passing through the 12 V battery and 4 Ω resistor), Va - 12 + (1.636 A)(4.00 Ω)=Vb, and Vb - Va is different (= -5.46 V). Why would this path not work?

Thank you.
 
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  • #2
I think the problem is that the situation they describe is impossible, as it breaks Kirchhoff's circuit law for voltage. Given the currents they've specified for the top two resistors, the PDs across the three elements in the upper circuit are 12, 0.909 x 2 and 1.636 x 4 and, whatever signs we give to either of those last two, we cannot get the three PDs to add to zero as Kirchhoff requires.

I suspect the question just contains a misprint, which makes it unsolvable.

For an impossible circuit, we should not be surprised if it gives different measurements when approached in different ways.

EDIT: In fact, they should not specify any currents in the circuit. We can calculate all currents using just the voltages of the two cells and the three resistances. Use Kirchhoff's laws. The current they specify for the 2 Ohm resistor is correct but that for the 4 Ohm resistor is not. It needs to be way more than that.
 
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  • #3
andrewkirk said:
The current they specify for the 2 Ohm resistor is correct but that for the 4 Ohm resistor is not.
Yes. The 1.636 A is the current in the 6 Ohm resistor.
 
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  • #4
andrewkirk said:
I think the problem is that the situation they describe is impossible, as it breaks Kirchhoff's circuit law for voltage. Given the currents they've specified for the top two resistors, the PDs across the three elements in the upper circuit are 12, 0.909 x 2 and 1.636 x 4 and, whatever signs we give to either of those last two, we cannot get the three PDs to add to zero as Kirchhoff requires.

I suspect the question just contains a misprint, which makes it unsolvable.

For an impossible circuit, we should not be surprised if it gives different measurements when approached in different ways.

EDIT: In fact, they should not specify any currents in the circuit. We can calculate all currents using just the voltages of the two cells and the three resistances. Use Kirchhoff's laws. The current they specify for the 2 Ohm resistor is correct but that for the 4 Ohm resistor is not. It needs to be way more than that.
Okay, thank you. Ignoring the actual values, could either path be used to get the same potential difference?
 
  • #5
jolly_math said:
Okay, thank you. Ignoring the actual values, could either path be used to get the same potential difference?
yes
 
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