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Voltage Across a Resistor Problem

Thread starter #1
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?
 

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ronsimpson

Well-Known Member
Most Helpful Member
#2
I got the same answer using a different method.
Total voltage=10V
Total resistance=14 ohms
-----------------------
10V/14 ohms = 1V/1.4 ohms.
------------------------
10V/14=?V/6 ohms 10V*6 ohms/14 ohms = ?V =4.29V
 

JimB

Super Moderator
Most Helpful Member
#3
Could someone please help me understand this?
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
 

MikeMl

Well-Known Member
Most Helpful Member
#5
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.

4.png

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.
 
Thread starter #6
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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