Assume the resistance values are R 1 = 2,300 , R 2 = 1,200 , R 3 = 4,600 , and R

Question

Assume the resistance values are

R1 = 2,300 , R2 = 1,200 , R3 = 4,600 , and R4 = 5,900 ,

and the battery emfs are

Use Kirchhoff's rules to analyze the circuit in the figure below.

(a) Let I1 be the branch current though R1, I2 be the branch current through R2, and I3 be the branch current through R3. Write Kirchhoff's loop rule relation for a loop that travels through battery 1, resistor 1, and resistor 3. (Use a clockwise current loop when entering your answer. Use any variable or symbol stated above as necessary.)

1 I1R1 I3R3 = 0

(b) Write Kirchhoff's loop rule relation for a loop that travels through battery 2, resistor 2, and resistor 3. (Use a clockwise current loop when entering your answer. Use any variable or symbol stated above as necessary.)

2 + I3R3 I2R2 = 0

(c) Apply Kirchhoff's junction rule to the junction at A to get a relation between the three branch currents. (Use any variable or symbol stated above as necessary.)

I1 = I2+I3

(d) You should now have three equations and three unknowns (I1, I2, and I3). Solve for the three branch currents.

I1= ?

I2= ?

I3= ?

Explanation / Answer

let,

resistance R1=2300 ohms

resistance R2=1200 ohms

resistance R3=4600 ohms

resistance R3=5900 ohms

and

potential V1=1.5 v

potential V2=3 v

by applying Kirchhoff's voltage rule for left side loop,

v1-i1*R1-i3*R3=0

1.5-i1*2300-i3*4600=0 -----(1)

and

by applying Kirchhoff's voltage rule for right side loop,

v2+i2*R2-i3*R3=0

3+i2*1200-i3*4600=0 -----(2)

by applying Kirchhoff's current rule at juction a,

i1=i2+i3 -----(3)

from equation no (1), (2) and (3),

i1=-0.27 mA <------------------

i2=-0.73 mA <----------------

and

i3=0.46 mA <-------------------

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