Suppose R13 R22 R32 E13V E 13 V and I5A Use Kirchhoffs v

Suppose R1=3? , R2=2? , R3=2? , E=13V E 13 V and I=5A .

Use Kirchhoff\'s voltage law to find an equation for the voltage drop around the outer loop (loop containing the resistors R1 , R2 , R3 and the cell E ).

......i1+...... i2= ......

Find a relationship between the current source I=5A and the loop currents.

......i1+...... i2= ......

Find the loop currents i1 , i2 (positive for clockwise direction). i

i1= .....A,

i2= .....A

Use Kirchhoff\'s voltage law to find the voltage gain across the current source I .

Voltage gain across I is....... V

Suppose I I is adjustable so that it is no longer 5A . What should I be changed to so that there is no current through the cell E ?

The current through I should be changed to .........A

Solution

R1=3 , R2=2 , R3=2 ; E = 13V

Applying Kirchoff voltage Law:

E = i1*R1+ I2(R2 +R3)

13= I1*3 + 5I2 -------(1)

Applying Kirchoff\'s current law at node:

I1 +I =I2

5 = I2 -I1 -------( 2)

I1 = I2 -5

Usin equation 1 and 2 to get I1 and I2:

13 =3I1 + 5I2

13 = 3(I2 -5) +5I2

13 = 3I2 - 15 +5I2

28 = 8I2

I2 = 7/2 = 3.5A

I1 = I2 -5 = 3.5 -5 = -1.5 A

Voltage Drop Across current source I : Va + E - I1R1 = Vb

Va -Vb = I1R1 -E = -1.5*3 -13 = - 17.5 V

Now we have I1 = I2 -I

To have I1 = : I = I2

E = I2R2 + I3R3

13 = I2(5)

I2 = 13/5 = 2.6 Amp

I = 2.6 Amp