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