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 = 48 , R2 = R6 = 153 R3 = 117 , and R4 = 95 . 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.

-the answer is not 0.103

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.

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

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

Explanation / Answer

by using Kirchhoff's current branch law
I1 + I2 = I3

by using Kirchhoff's voltage loop law, left loop, traced clockwise:

V1 - I1 R1 + I2 R2 - V2 - I1 R4 = 0

By using Kirchhoff's voltage loop law, right loop, traced clockwise:
V2 - I2 R2 - I3 R3 = 0

Three equations, three unknowns (I1, I2, I3). Solve system, algebra not displayed.

Resulting expressions:
I1 = ((V1-V2)R3 + R2V1)/(R2R1 + R2R4 + R1R3 + R2R3 + R4R3)
I2 = (V2(R1 + R4 + R3) - R3V1)/(R2R1 + R2R4 + R1R3 + R2R3 + R4R3)
I3 = (V1R2 + V2(R1+R4))/(R2R1 + R2R4 + R1R3 + R2R3 + R4R3)

Data:
V1:=18 V & V2:= 12 V
R1:=R5=48 Ohms; R2:=R6=153 Ohms; R3:=117 Ohms; R4:= 95 Ohms;

plug in the values in the above formula and you will get

I1, I2 &I3

voltage drop

since loop resistances are in series so R2+R3+R6 = 153+153+117 =423omhs

I=V/R = (V1+V2)/R= 30/423 =0.71A

voltage drop between points a and b

V= IR = 0.71x4153 =108.63V

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