Voltage Across a Resistor Problem

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frozone45

New Member
Hi everyone, I was having trouble understanding a problem that I have attached. Work for the problem is in the attachment. The problem asks me to find a voltage across a resistor. While doing the problem, I found two different voltages, one positive and one negative. Why does the formula for power absorbed by a resistor give me two voltage values, when only one voltage value is correct?

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ronsimpson

Well-Known Member
I got the same answer using a different method.
Total voltage=10V
Total resistance=14 ohms
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10V/14 ohms = 1V/1.4 ohms.
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10V/14=?V/6 ohms 10V*6 ohms/14 ohms = ?V =4.29V

JimB

Super Moderator
The absolute value of the voltage is correct at 4.29 volts.

The way that you have calculated the voltage, you are taking a square root which you found by doing a power calculation.
From that, mathematically, both +4.29 and -4.29 volts are solutions to the problem.

Electrically, the current could flow in either direction through the 6 Ohm resistor and the power dissipated would be the same.
IN THIS CASE, the conventional current will flow from the positive side of the supply to the negative side.

JimB

frozone45

New Member
I understand now that the negative voltage is not possible due to the conventional current direction. Thank you all so much!

MikeMl

Well-Known Member
Consider this. Label a node as being a reference node (the ground symbol), with respect to which all voltages at the other nodes are measured. In this example, I arbitrarily choose the lower left node to be ground, and labeled the other three a, b, and c.

Obviously, the current in the circuit = I = E/Rt = 10V/(5+3+6) Ω= 10/14A = 0.7143A
It is easy to see that the power in resistor R3 is (I(R3)^2)*R3 = (0.7143^2)*6 = 3.061W

There is no sign ambiguity.

frozone45

New Member
Consider this. Label a node as being a reference node (the ground symbol), with respect to which all voltages at the other nodes are measured. In this example, I arbitrarily choose the lower left node to be ground, and labeled the other three a, b, and c.

View attachment 112974

Obviously, the current in the circuit = I = E/Rt = 10V/(5+3+6) Ω= 10/14A = 0.7143A
It is easy to see that the power in resistor R3 is (I(R3)^2)*R3 = (0.7143^2)*6 = 3.061W

There is no sign ambiguity.
I see. That makes sense. Thank you so much!

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