Voltage Problem: Finding Unknown Voltage in a Circuit with Three Resistors"

[terryds]

Homework Statement

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

Homework Equations

V = I R

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.
Please help

 

[gneill]

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?

 

[terryds]

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?

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

 

[gneill]

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.

 

[terryds]

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.

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

 

[gneill]

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?

 

[terryds]

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?

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

 

[gneill]

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)

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.

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

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).

 

[terryds]

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).

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

 

[gneill]

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

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.