Kirchhoff's Rules and potential difference

[AirForceOne]

Homework Statement

Find the potential difference between points a and b. each resistor has R=130 ohms and each battery is 1.5V

Homework Equations

Kirchoff's First Rule: at any junction point, the sum of all currents entering the junction must equal to the sum of all currents leaving the junction

Kirchoff's Second Rule: the sum of the changes in potential around any closed loop of a circuit must be zero

The Attempt at a Solution

Kirchoff's First Rule: since there are no junctions, I just have "I" for the current, going counterclockwise, and I find "I" to be:

I = V/R = 1.5/(4*130) = 0.00289 A

Kirchoff's Second Rule: since there is one loop, I'll set the sum of all the potential changes around the loop to be zero

I redrew the diagram like so:

and when I try to Kirchoff's second rule:

V_ab + V_bc + V_bc + V_cd + V_de + V_ef + V_fa = 0

- 0.375 - 0.375 + 1.5 -0.375 - 0.375 + 1.5 = 1.5

It doesn't add up to zero?

 

[Delphi51]

I = V/R = 1.5/(4*130) = 0.00289 A

Surely it should be 3 volts instead of 1.5.
Final answer 3 Volts, too?

 

[AirForceOne]

Surely it should be 3 volts instead of 1.5.
Final answer 3 Volts, too?

Thanks. So you if you have a series circuit, and you have multiple voltage sources, you just add up the voltage sources for the V in V=IR to find the current?

 

[Delphi51]

True, but probably better to sum the voltages around the loop as you did. You just forgot one of the 1.5's.

My guess of 3V for Vab is wrong! After working out the current I did V=IR to get .75 V on each resistor. But the polarity on them opposes the 1.5 V battery.

 

[zgozvrm]

It may be easier to look at it this way...


This is a single series circuit, therefore, you know that the current, I, is the same through each component.
Since the resistors all have the same value (130[itex]\Omega[/tex]), then the voltage across each resistor is 130I. (Let's say V = 130I).
At any given point (say point A), the sum of voltages around the circuit must be 0.

Considering voltage sources as positive values and voltage drops as negative values, we have (from point A, traveling counter-clockwise around the circuit):

-V + 1.5 - V - V + 1.5 - V = 0
3 - 4V = 0
3 = 4V
V = 3/4 = 0.75

So, the voltage drop across each resistor is, indeed 0.75 volts.

Use the same logic to determine the voltage from point A to point B.