How Does Kirchhoff's Loop Rule Apply to a Two Battery, Three Resistor Circuit?

[helpme12345]

I'm not sure where to begin. The problem asks to find the magnitude of the direction of the current in a two battery system with three resistors. However, it appears the junction rule cannot be applied. The loop rule may be implemented but I don't know how to get the current. The link to the problem is below.

http://negatrons.homestead.com/files/Kirchhoff.jpg

 

[Adrian Baker]

Add the two cell pd's together for a start - just because they aren't next to each other doesn't mean that you can't add them up (they are both connected the same way round.

Then find the total resistance and you are nearly there.

Also, post these in the homework help section in future!

 

[deltabourne]

Magnitude of the direction of the current? I didn't know direction had magnitude.

 

[Doc Al]

I'm not sure where to begin. The problem asks to find the magnitude of the direction of the current in a two battery system with three resistors. However, it appears the junction rule cannot be applied. The loop rule may be implemented but I don't know how to get the current.

It's a simple series circuit. Combine the batteries and the resistors and apply Ohm's law.

 

[robphy]

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While it may be really simple, let's apply the loop rule... to see that the rule really works [and so I can try to draw a circuit ].

(Assume that we traverse the loop clockwise, which defines our choice of "positive current" [akin to choosing an axis].)
$$\begin{align*}<br /> 0&=(V_B-V_A) + (V_D-V_B) + (V_E-V_D)+ (V_C-V_E)+ (V_A-V_E)\\<br /> &=(-I_1R_1)+(V_1)+(-I_2R_2)+(V_2)+(-I_3R_3)\\<br /> \intertext{in series, the currents are equal}<br /> &=(-IR_1)+(V_1)+(-IR_2)+(V_2)+(-IR_3)\\<br /> &=-I(R_1+R_2+R_3)+(V_1+V_2)\\<br /> \end{align*}$$
Now, solve for the current $I$.

I think the original poster meant "magnitude and direction" of the current.