Parallel circuits, Emfs and kirchoff's laws

AI Thread Summary
To determine the required emf (ε) for a 1.79A current through a 7Ω resistor, the user initially applied Kirchhoff's laws but struggled with sign conventions and equation consistency. After revising their approach, they were advised to clearly define current directions and correctly account for voltage rises and drops in their loop equations. The user reformulated their equations for the outer and left loops, leading to a more consistent solution. This process improved their understanding of circuit analysis and boosted their confidence in applying Kirchhoff's laws. Ultimately, the discussion highlights the importance of careful sign management and systematic analysis in solving circuit problems.
Prodigium
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Homework Statement



YF-26-62.jpg


What must the emf ε in the figure be in order for the current through the 7Ω resistor to be 1.79A ? Each emf source has negligible internal resistance.

Homework Equations



V=IR
Kirchoffs junc and loop rules

The Attempt at a Solution



I put that the outer loop would be 24V +3*I = 7*1.79
and from there solved for I
i got that I would be -5.02A
I then said that because the current I split at the top middle junc the current I{2} down the centre line was I{2}=I-I{1}
and basically used V=IR to solve for ε which i got to be -13.62V



but i can't help but feel I've done it wrong, and in the textbook there is no answers can anyone help? much appreciated
 
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First, check the equation that you wrote for the outer loop. If the current is assumed to be flowing clockwise then there should be voltage DROPS across both the 3.00 and 7.00 Ohm resistors.

You might want to consider using nodal analysis (KCL). After all, there's only one node and you already know it's voltage: 1.79 x 7.00. You'll end up with a single equation with one unknown.
 
Okay so I got (using kirchhoffs lecturer says we have to use kirchhoffs laws)
outer loop :-
24-3I-7I2=0
left hand loop :-
24+ε-3I-2I1=0
right hand loop:-
-ε+2I1-7I2=0

i went on to solve and didnt get a consisted answer... (when put back into equations to check)
are these equations on the right line? I am not the best at circuits.
 
Prodigium said:
Okay so I got (using kirchhoffs lecturer says we have to use kirchhoffs laws)
outer loop :-
24-3I-7I2=0
left hand loop :-
24+ε-3I-2I1=0
right hand loop:-
-ε+2I1-7I2=0

i went on to solve and didnt get a consisted answer... (when put back into equations to check)
are these equations on the right line? I am not the best at circuits.

I see that you're having trouble keeping the signs straight for the voltage rises and drops whe you "walk" around a loop. Your equation for the outer loop looks okay, but the other two have some problems.

Take your figure and draw in the assumed directions of each of the currents (I, I1, I2). Now do the left loop again, walking around the loop clockwise starting at the - terminal of the battery. The next component you come to is the source ε. As you walk over it, will there be a voltage rise or a voltage drop?
 
Thanks so much i have a consistant answer now :D and i am more confident with circuits.
 
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