Potential difference between two points on a circuit

AI Thread Summary
The discussion revolves around calculating the potential difference between points B and D in a circuit with a 3-volt supply and resistors of 2 ohms. Using Kirchhoff's laws, participants analyze the circuit's symmetry, likening it to a Wheatstone bridge. They derive the currents through various branches, ultimately determining that the potential difference between B and D is zero. The calculations confirm that the voltage drop across the resistors leads to equal potentials at points B and D. The conclusion affirms the circuit's characteristics as similar to a Wheatstone bridge, resulting in no potential difference.
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Homework Statement



n76vt.png


The circuit is shown as above.
If ε = 3 volt and each resistor has 2 ohm resistance, then what's the potential difference between point B and point D ?

A. 4
B. 3
C. 2
D. 1
E. 0

Homework Equations



V = I R
R series = R1 + R2 + ..
1/ R parallel = 1/R1 + 1/R2 + ...

The Attempt at a Solution


[/B]
The circuit is very weird. I don't even know how to start.
 
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Are you familiar with Kirchhoff's laws?
 
cnh1995 said:
Are you familiar with Kirchhoff's laws?

Yes, of course.. KVL and KCL..
Okay, I'll use KVL for the two triangles

Triangle ADC

4IR = ε
I = ε/4R = 3/8 ampere

The current in triangle ABC also 3/8 ampere, right ? (The resistance are equal also the emf)

But, what's the potential difference between point B and D?
 
Assume current through diagonal AC as i and current through ABC as i1. So, current through ADC is i-i1. Using these currents, write KVL and you'll get values for i and i1.
Actually, this problem can be solved by observation only, using the symmetry of the circuit.
 
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cnh1995 said:
Assume current through diagonal AC as i and current through ABC as i1. So, current through ADC is i-i1. Using these currents, write KVL and you'll get values for i and i1.

##Triangle\ ABC \\
2I_1R+2IR - \varepsilon = 0 \\
4I_1 + 4I = 3\\
Triangle\ ADC \\
2((I-I_1)R) + 2IR - \varepsilon = 0 \\
4IR-2I_1R = 3 \\
8I - 4I_1 = 3 \\##

Solving the equation, I get i = 0.5 A, and i1 = 0.25 A, and it means that i2 = 0.25 A
How to determine the potential difference between B and D ??

cnh1995 said:
Actually, this problem can be solved by observation only, using yhe symmetry of the circuit.

Yeah, I also see that it's kinda like the Wheatstone bridge, isn't it ??
But, eventually, the current through diagonal AC is not zero (not as I thought it would be in Wheatstone bridge)..
Please help me
 
terryds said:
Solving the equation, I get i = 0.5 A, and i1 = 0.25 A, and it means that i2 = 0.25 A
Good!
terryds said:
How to determine the potential difference between B and D ??
Let's say point A is at a potential Va and point B is at potential Vb. Voltage drop across the resistor between A and B is Va-Vb. What is the value of this voltage drop? You know the currents now.
 
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cnh1995 said:
Good!

Let's say point A is at a potential Va. You know the currents now. What is the potential of point B in terms of Va and known current and resistor values? (Hint:Ohm's law).

Vb = 3 - 0.5 * 2 - 0.25 * 2 = 1.5 V
Vd = 3 - 0.5 * 2 - 0.25 *2 = 1.5 V
The potential difference is zero, right ?

Anyway, is this circuit a Wheatstone bridge?
 
terryds said:
The potential difference is zero, right ?
Right.
terryds said:
Anyway, is this circuit a Wheatstone bridge?
Yes. Supply is connected between A and C and output is taken between B and D.
 
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