The circuit in the figure is composed of two batteries (e 1 = 9 V and e 2 = 5 V)

Question

The circuit in the figure is composed of two batteries (e1 = 9 V and e2 = 5 V) and four resistors (R1 = 110 ?, R2 = 40 ?, R3 = 30 ?, and R4 = 50 ?, as shown.

1) What is the current I1 which flows through R1? I1= ______ A

2) What is the current I3 that flows through R3? I3= ______ A

The circuit in the figure is composed of two batteries (e1 = 9 V and e2 = 5 V) and four resistors (R1 = 110 ?, R2 = 40 ?, R3 = 30 ?, and R4 = 50 ?, as shown.

1) What is the current I1 which flows through R1? I1= ______ A

2) What is the current I3 that flows through R3? I3= ______ A

Explanation / Answer

given R1=110ohm

R2=40 ohm

R3=30 ohm

R4=50 ohm

e1=9V

e2=5V

current in loop 1is I1

current in loop is I3

let current in R2 is I2.

according to the kirchhof's law

in loop 1

-e1+(I1*R1)+(R2*I2)+e2+(I1*R4)=0 .... (eq 1)

in loop 2

-e2-(R2*I2)+(I3*R3) =0 .....(eq 2)

I1= I2+I3 ....(eq 3)

put the value of I1 in eq 1 from the eq 3

-e1+(I2(R1+R2+R4))+e2+(I3(R1+R4)) =0.....(eq 4)

put the given values in eq2 &eq4

(200*I2)+(160*I3)=4

(50*I2)+(40*I3)=1.....(eq5)

(30*I3)-(40*I2) =5......(eq 6)

by solving eq 5& eq6

I3=29/310 A=0.093 A

I2=-0.054

put the values of I2&I3 in eq 3

I1=-0.054+0.093

I1= 0.038 A

so current in R1 is I1=0.038A

current in R3 is I3 =0.093 A

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