Calculating EMF Using Kirchhoff's Rules

[PirateFan308]

Homework Statement

What must the emf ε in the figure be in order for the current through the 7.00 ohm resistor to be 1.75A? Each emf source has negligible internal resistance.

Homework Equations

Kirchhoff's Rule:
\sum I=0 junction rule

\sum V=0 loop rule

The Attempt at a Solution

I have drawn my current directions as shown in the attachment below.

By the junction rule, I know that I_2 = I_3+I_1 and it is given that I_2=1.75A.

I drew a loop clockwise around the entire thing and came up with the following equation:

24V-(7 ohm)(1.75A)-(I_1)(3 ohm) = 0

so that I have

I_1 = (24V-12.25V)/3= 3.92A

and then

I_3 = I_2 - I_1 = 2.17

Drawing loop 2 clockwise within the right inner loop I have the following equation:

-(7 ohm)(I_2) - (I_3)(2 ohm) + ε = 0

solving for ε, I get

ε = (12.25V)+(4.34V) = 16.59V

Also, if I draw a clockwise loop inside the right inner loop, I have the following equation:

-ε + (I_3)(2 ohm) - (I_1)(3 ohm) + 24 V = 0

which simplifies to

ε = 16.58V

I can find no mistakes (and I have checked over my work a few times) but it is the wrong answer. Could anyone please help me? Thanks!

 

[SammyS]

Homework Statement

What must the emf ε in the figure be in order for the current through the 7.00 ohm resistor to be 1.75A? Each emf source has negligible internal resistance.

Homework Equations

Kirchhoff's Rule:
\sum I=0 junction rule

\sum V=0 loop rule

The Attempt at a Solution

I have drawn my current directions as shown in the attachment below.

By the junction rule, I know that I_2 = I_3+I_1 and it is given that I_2=1.75A.

I drew a loop clockwise around the entire thing and came up with the following equation:

24V-(7 ohm)(1.75A)-(I_1)(3 ohm) = 0

so that I have

I_1 = (24V-12.25V)/3= 3.92A

and then

I_3 = I_2 - I_1 = 2.17

That should be I_3 = I_2 - I_1 = -2.17\text{A}

Drawing loop 2 clockwise within the right inner loop I have the following equation:

-(7 ohm)(I_2) - (I_3)(2 ohm) + ε = 0

solving for ε, I get

ε = (12.25V)+(4.34V) = 16.59V

Also, if I draw a clockwise loop inside the right inner loop, I have the following equation:

-ε + (I_3)(2 ohm) - (I_1)(3 ohm) + 24 V = 0

which simplifies to

ε = 16.58V

I can find no mistakes (and I have checked over my work a few times) but it is the wrong answer. Could anyone please help me? Thanks!

 

[PirateFan308]

Thank you soo much! I can't believe I missed that - I was freaking out thinking I had the completely wrong idea. Thanks!