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 = 44 , R2 = R6 = 92 R3 = 73 , and R4 = 122 . The positive directions for the currents I1, I2 and I3 are indicated by the directions of the arrows.

1)

What is V4, the magnitude of the voltage across the resistor R4?V

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?

Last 2 questions are supposed to be tricky so please , pay closer attention and if you are not sure about the answer, Thank you.

Explanation / Answer

here,

The battery voltages are,
V1 = 18 V
V2 = 12 V.

Various resistance are,
R1 = R5 = 44
R2 = R6 = 92
R3 = 73
R4 = 122

Part A:
V4 = V2*R4/(R4+R5)

V4 = 12*122/(122+44)
V4 = 8.819 V

Part B:
Current I3 will be given as:

I3 = ((R5)/( R5*R3 + (R5+R3)*(R2+R6) ) * ((V2 - V1)(1+(R2+R6)/R5))

I3 = ((44)/( (44)(73)+(44+73)(92+92) )) *(12 - 18(1+(92+92)/44))

I3 = -0.145 A

Part C:
I2*R1 - I3*(R1 + R3) = V1

I2 = (V1 + I3*(R1 + R3))/R1
I2 = (18/44) + -0.145*(44+73)/44
I2 = 0.0235 A

Part D:
I1 = I2 - I3
I1 = 0.0235 - (-0.145)
I1 = 0.1685 A

Part E:
Va - Vb = I2 * R2
Del(V) = 0.0235 * 92
del(V) = 2.162 V

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