Answer:
[tex]I_{1}=0.32 A[/tex]
[tex]I_{2}=I_{3}=I_{4}=0.16 A[/tex]
Explanation:
Assuming the circuit is the shown in the attached image, where:
[tex]\Epsilon=12 V[/tex]
[tex]R_{1}=15 \Omega[/tex]
[tex]R_{2}=45 \Omega[/tex]
[tex]R_{3}=20 \Omega[/tex]
[tex]R_{4}=25 \Omega[/tex]
We can use the 2nd Kirchoff Law (or Voltage Law) for each section of the circuit:
1st section:
[tex]-\Epsilon+V_{1}+V_{2}=0[/tex]
2nd section:
[tex]-V_{2}+V_{3}+V_{4}=0[/tex]
Now, according to Ohm's Law, the voltage [tex]V[/tex] is equal to the multiplication of the current [tex]I[/tex] and the resistance [tex]R[/tex]:
[tex]V=IR[/tex]
Then:
1st section:
[tex]-\Epsilon+I_{1}R_{1}+I_{2}R_{2}=0[/tex] (1)
2nd section:
[tex]-I_{2}R_{2}+I_{3}R_{3}+I_{3}R_{4}=0[/tex] (2)
In this case the same current [tex]I_{3}[/tex] that gous trhough [tex]R_{3}[/tex] is the same that goes through [tex]R_{4}[/tex], since they are in series.
And according to 1st Kirchoff Law (or Current Law):
[tex]I_{1}=I_{3}+I_{2}[/tex] (3)
At this point we have a system with three equations. Let's begin by substituting (3) in (1):
[tex]\Epsilon=(I_{3}+I_{2})R_{1}+I_{2}R_{2}[/tex] (4)
[tex]\Epsilon=I_{3}R_{1}+I_{2}R_{1}+I_{2}R_{2}[/tex]
Applying common factor [tex]I_{2}[/tex]:
[tex]\Epsilon=(R_{1}+R_{2})I_{2}+I_{3}R_{1}[/tex] (5)
Isolating [tex]I_{3}[/tex] from (2):
[tex]I_{3}=\frac{R_{2}}{R_{3}+R_{4}}I_{2}[/tex] (6)
Rewriting with the known values:
[tex]I_{3}=\frac{45 \Omega}{20 \Omega+25 \Omega}I_{2}[/tex] (7)
[tex]I_{3}=I_{2}[/tex] (8)
Substituting (8) in (5):
[tex]\Epsilon=(R_{1}+R_{2})I_{3}+I_{3}R_{1}[/tex] (9)
Isolating [tex]I_{3}[/tex]:
[tex]I_{3}=\frac{\Epsilon}{2R_{1}+R_{2}}[/tex] (10)
Rewritting with the known values;
[tex]I_{3}=\frac{12 V}{2(15 \Omega)+45 \Omega}[/tex] (11)
[tex]I_{3}=0.16 A[/tex] (12)
Then: [tex]I_{3}=I_{2}=I_{4}=0.16 A[/tex]
Substituting (12) in (3):
[tex]I_{1}=0.16 A+0.16 A[/tex] (13)
[tex]I_{1}=0.32 A[/tex] (14)
