A circuit is constructed with six resistors and two batteries as shown. The batt

Question

A circuit is constructed with six resistors and two batteries as shown. The battery voltages are V1 = 18 V and V2 = 12 V. The positive terminals are indicated with a + sign, The values for the resistors are: R1 = R5 = 70 ?, R2 = R6 = 147 ? R3 = 99 ?, and R4 = 123 ?. The positive directions for the currents I1, I2 and I3 are indicated by the directions of the arrows.

1) What is I3, the current that flows through the resistor R3? A positive value for the current is defined to be in the direction of the arrow.

2) What is I2, the current that flows through the resistor R2? A positive value for the current is defined to be in the direction of the arrow.

3) What is I1, the current that flows through the resistor R1? A positive value for the current is defined to be in the direction of the arrow.

4) What is V(a) – V(b), the potential difference between the points a and b?

Explanation / Answer

V1 = 18 V and V2 = 12 V. R1 = R5 = 70 , R2 = R6 = 147 R3 = 99 , and R4 = 123 .

From Kirchoff's first law , i2 = i1+i3 -----------( 1)

Apply Kirchoff's second law to left loop you get , V1 = -i1R1 -i3R3

18 = -70i1 -99i3 -----------(2)

Apply Kirchoff second law to middle loop you get ,

V1 - V2 = -i3R3-i2R2-i2R6

18-12 = -99i3 -147i2-147 i2

6 = -99i3 - 294 i2

6 = -99i3 -294(i1+i3)

6 = -393 i3-294 i1 --------( 3)

equation(2) x (294/70) ===>

(294/70) 18 = (294/70) (-70i1) -(294/70) 99i3

75.6 = -294i1 - 415.8 i3 -----------( 4)

equation(3) -equation(4) ==>

6-75.6 = -393 i3+415.8 i3

22.8 i3 = -69.6

(1). i3 = -3.052 A

(2).Substitute i3 value in 6 = -393 i3-294 i1 you get

6 = -393(-3.052) -294 i1

= 1199.68 -294 i1

294 i1 = 1193.68

i1 = 4.06 A

From equation (1 ) , i2 = i1+i3

i2 = 4.06 - 3.052

= 1.008 A

(3).i1 = 4.06 A

(4) the potential difference between the points a and b , = i2 R6 = 1.008 (147) = 148.19 volt

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