Saturday, January 14, 2012

Determine the magnitudes and directions of the currents in each resistor shown in Fig. 19-48. Figure Link..?

Determine the magnitudes and directions of the currents in each resistor shown in Fig. 19-48. The batteries have emfs of E1 = 9.0 V and E2 = 13.0 V and the resistors have values of R1 = 15 ohms, R2 = 20 ohms, and R3 = 28 ohms.

IR1 ________A

IR2 ________A

IR3 ________A

Determine the magnitudes and directions of the currents in each resistor shown in Fig. 19-48. Figure Link..?
E1 = 9, E2 = 13



R1 = 15, R2 = 20, R3 =28

i3 through 28 going up .

i1 is coming down and

i2 = i1 + i3

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13 = 20 i1 + 48 i3

9 = 35 i1 + 20 i3

Solving

Current through R1 = 0.134375A

Current through R3 = 0.21484375A

Current through R2 = 0.34921875

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JUNCTION METHOD SOLUTION:



Point a is between E1 and R1 .

Point b is the junction of R1 and R2

Point c is between E2 and R3 .

Point O is the junction of E1 and E2.

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With reference to O

Potential Ea = 9V

Potential Ec = 13 V

At Junction b

I1 + I3 = I2.

[Ea – Eb] / 15 + [Ec- Eb] / 28 = Eb/20

[9 – Eb] / 15 + [13- Eb] / 28 = Eb/20

28 [9 – Eb] + 15 [13- Eb] = 28 *15* Eb/20

252 – 28Eb + 195 – 15Eb = 21 Eb

252 – 28Eb + 195 – 15Eb = 21 Eb

64Eb = 447 V

Eb = 6.984375 V

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IR2 = 6.984375 / 20 = 0.34921875 A



IR1 = [9 – Eb] / 15 = 0.134375 A



Ir3 = [13- Eb] / 28 = 0.21484375 A
Reply:Use Kirchoff law .



ε1 = I1R1 + I2R2

9 = I1(15) + 20I2



ε2 = I3R3 + I2R2



13 = I3(28) + I2(20)



ε1 - ε2 = I1R1 - I3R3



9 - 13 = I1(15) - I3(28)

-4 = 15I1 - 28I3





9 = I1(15) + 20I2





13 = I3(28) + I2(20)



-4 = 15I1 - 28I3



I1 + I3 = I2



9 = 15I1 + 20(I1 + I3)

9 = 35I1 + 20I3

I1 = ( 9 - 20I3)/35



13 = I3(28) + 20(I1 + I3)

13 = 48I3 + 20I1

13 = 48I3 + 20( 9 - 29I3)/35

55/7 = 220/7I3

I3 = 0.25 A

I1 = 0.1142857143 A

I2 = 0.3642857143 A


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