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 = 42 , R2 = R6 = 126 R3 = 111 , and R4 = 79 . The positive directions for the currents I1, I2 and I3 are indicated by the directions of the arrows.

2)

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.A

3)

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.A

4)

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.A

5)

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

V

Explanation / Answer

consider loop 1 containing resistors R1 and R3 and voltage source V1:

-I3R3 -V1 -I1R1 = 0

-111*I3 - 18 - 42*I1 = 0

so, I1 = - [111*I3 +18]/42

consider loop 2 contaning R2, R3 and R6 and V1 and V2:

-I3R3 -V1 -I3R6 +V2 -I2R2 = 0

also, I2 =I1 + I3

so, -I3R3 -V1 -I3R6 +V2 - I1R2 -I3R2 = 0

-I3R3 -I3R6 + [R3I3 +V1]/R1 -I3R2 = V1 - V2

-I3 (R3 +R6 +R2 ) +R3I3/R1 +V1/R1 = V1 - V2

so, I3 = - {[R1(V2 - V1) - V1]/[R1R3 + R1R6 +R1R2 +R3]}

susbtituing values , we get: I3 = 15.2 mA

I1 = - [111*I3 +18]/42 = 0.47A

I2 = I1 + I3 = 0.47 + 0.00152 = 0.472A

Vab = Vb - Va = -I3R6 = 0

15.2*10-3*126 = 0.19V

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