Kirchhoff's Rules. Solving for currents in circuit with two batteries

  • Thread starter Painguy
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  • #1
Painguy
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Homework Statement



zAhzBia.png


What are the sizes and directions of the currents through
resistors (a) R2 and (b) R3 in Fig. 27-42, where each of the three resistances is 4.0 Ω?

Homework Equations


ΔV=IR


The Attempt at a Solution


Equations:
1) ΔVB2 -(i1)R1 -(i2)R2 -ΔVB1=0
5V - 4(i1) - 4(i2) =0
4(i2)=5V -4(i1)

2) ΔVB2 -(i1)R1 -(i3)(R3)=0
10V -4(i1) -4(i3) =0
-4(i3)=-10V + 4(i1)

not sure about this one
3) ΔVB1 -(i2)R2 -(i3)(R3)=0
5V + 4(i2) -4(i3) =0

4) i1=i2 + i3

5V +5V -4(i1) -10V +4(i1)=0
0=0

I'm not really sure what to do here.
 
Last edited:

Answers and Replies

  • #2
ehild
Homework Helper
15,543
1,915

Homework Statement



zAhzBia.png


What are the sizes and directions of the currents through
resistors (a) R2 and (b) R3 in Fig. 27-42, where each of the three resistances is 4.0 Ω?

Homework Equations


ΔV=IR


The Attempt at a Solution


Equations:
1) ΔVB2 -(i1)R1 -(i2)R2 -ΔVB1=0
5V - 4(i1) - 4(i2) =0
4(i2)=5V -4(i1)

2) ΔVB2 -(i1)R1 -(i3)(R3)=0
10V -4(i1) -4(i3) =0
-4(i3)=-10V + 4(i1)

not sure about this one
3) ΔVB1 -(i2)R2 -(i3)(R3)=0
5V + 4(i2) -4(i3) =0


4) i1=i2 + i3

[STRIKE]5V +5V -4(i1) -10V +4(i1)=0
0=0[/STRIKE]

I'm not really sure what to do here.

The blue equation is not independent from the previous ones. Use eq.4 to find i1.

ehild
 
  • #3
rude man
Homework Helper
Insights Author
Gold Member
8,031
866
I also suggest superposition:
solve for the currents with one battery shorted, solve with the other battery shorted, then add the results.
 

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