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 = 59 , R2 = R6 = 131 R3 = 48 , and R4 = 130 . The positive directions for the currents I1, I2 and I3 are indicated by the directions of the arrows.

1.) 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.

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

Explanation / Answer

1.)
I2 = I1+ I3
I1 = (I2 - I3)

I1*R1 = I3*R3 + V1
(I2 - I3) R1 = I3*R3 + V1
I2R1 - I3R1 = I3 R3 + V1
I2 = (V1 +I3 (R1 + R3)) / R1

I2*R2 + I3*R3 + V1 + I2 *R6 = V2
I2 * (R2 + R6) + I3 *R3 = V2
((V1 +I3 (R1 + R3)) * (R2 + R6)) / R1 + I3*R3 = V2

Substituing Values -
((18 +I3 (59 + 48)) * (131 + 131)) / 59 + I3*48 = 12
Solving for I3
I3 = -0.13 amp

I2 = (V1 +I3 (R1 + R3)) / R1
I2 = (18 - 0.13*(59 + 48))/59
I2 = 0.07 amp

I1 = I2 - I3
I1 = 0.07 -(-0.13 )
I1 = 0.2 amp
Current that flows through the resistor R1, I1 = 0.2 amp

2.)
Va - Vb = I2 * R6
Va - Vb = 0.07 * 131 volt
Va - Vb = 9.2 volt

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