# Voltage problem

1. Mar 27, 2016

### terryds

1. The problem statement, all variables and given/known data

Three resistors (R1 = 3 ohm, R2 = 4 ohm, R3 = 5 ohm) is arranged like the picture above.
The voltage at point 1 is 24 Volt and the voltage at point 2 is 10 Volt.
The currents that go in R1 and R2 are 2 Amperes.
The voltage at point 3 is ... Volt

A. 10
B. 12
C. 16
D. 18
E. 24

2. Relevant equations
V = I R

3. The attempt at a solution

Umm...
I think there are weird things, such as I think that the problem does not obey the Kirchhoff 1 Law.
If the current 1 is 2 ampere into the junction, so I think current 2 and current 3 will share. (The problem says that The current 2 is 2 ampere which means that current 3 is 0 ampere (huh?) )

Voltage at R1 is I1 * R1 = 2 * 3 = 6 Volt
Voltage at R2 is I2 * R2 = 2 * 4 = 8 Volt
Voltage at R3 is I3* R3 = 0 * 5 = 0 Volt ( I get the current is zero because the Kirchhoff 1 law which states that current into junction = current leave junction)

But, then I think that the voltage at point 2 and 3 must be the same because it's parallel connected, right?
So, I think it's 10 Volt.
But, really, I'm still confused.

2. Mar 27, 2016

### Staff: Mentor

There are no parallel or series connections shown: No resistors share connections at both ends, so there are no parallel connections; The only junction shown has three connections, so that rules out series connections.

What can you say about the potential at the junction?

3. Mar 27, 2016

### terryds

Potential at the junction?
Hmm... I have no idea...
Maybe it is the same as the voltage in R1 = I1 * R1 = 6 Volt?

4. Mar 27, 2016

### Staff: Mentor

When a current flows through a resistor there is a potential drop across the resistor. If the potential at point 1 is 24 V, then what is the potential at the junction? DO a "KVL walk" from point 1 to point 0.

5. Mar 27, 2016

### terryds

Oh yeah, I get the idea...
The voltage at the junction is 24 - 6 = 18 Volt
Then?

6. Mar 27, 2016

### Staff: Mentor

Confirm that the 2 A flowing through R2 satisfies KVL for point 2. If it does, what can you conclude about the current through R3?

7. Mar 27, 2016

### terryds

So V2 is 2 * 4 = 8 Volt
It means that V3 is 18 - 8 = 10 Volt (I'm not sure about this since V3 is at another branch)

The current is V3/R3 = 10/5 = 2 ampere ??

But, I really doubt it. Since by KCL, it should be zero ampere since I1 = I2 + I3 so I3 = I1-I2 = 2-2 = 0 ampere..
I'm confused

8. Mar 27, 2016

### Staff: Mentor

No. It means that point 2 is at 10 V: 18V - 8V = 10V , which is what was given in the problem statement, so it's confirmed that the currents and potentials given for points 1 and 2 are good, and that the center junction is at 18 V.
And I3 is zero, which is a perfectly good value. What does that make the potential drop across R3? Then what's the potential at point 3 (KVL walk from the junction to point 3).

9. Mar 27, 2016

### terryds

Since the voltage at the resistor is zero, it must be 18 volt at point 3, right?

10. Mar 27, 2016

### Staff: Mentor

That's right.

So to sum up, from the given information and KCL you know that the current through the R3 path is zero. By a KVL walk from one of the known potentials (we used point 1 at 24 V) you know the potential at the junction is 18 V. Then the potential at point 3 is determined by a KVL walk from the junction to point 3. Since that current is zero there's no drop, so it's also at 18 V.