Voltage Across a Resistor Problem

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?

Could someone please help me understand this?

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

frozone45 said:

Could someone please help me understand this?

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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

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

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.

MikeMl said:

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.

Click to expand...

I see. That makes sense. Thank you so much!