πΌ
1
= 0. 001363 π΄ + 0. 001223 π΄
πΌ
1
= 0. 002586 π΄
Equation 5:
πΌ
2
= (πΌ
3
Β· 14. 468) + 0. 0638 π΄
πΌ
2
= (β 0. 003957 π΄ Β· 14. 468) + 0. 0638 π΄
πΌ
2
= β 0. 05725 π΄ + 0. 0638 π΄
πΌ
2
= 0. 00655 π΄
Do not forget that we have supposed the directions of the currents, in the final answer if we get a
positive answer, like and the direction we have supposed is true, but if we get a negative answer,
πΌ
1
πΌ
2
like the direction we have supposed is wrong, so we must reverse it.
πΌ
3
The equation will be changed to:
πΌ
1
= πΌ
2
+ πΌ
3
πΌ
2
= πΌ
1
+ πΌ
3
Now, it is easy to calculate the voltage on the resistors using Ohm's law:
π = πΌ Β· π
The voltage on 2.2 kΞ©
π = πΌ
1
Β· 2. 2 πΞ©
π = 0. 002586 Β· 2200 Ξ©
π = 5. 7 π
The voltage on 680 Ξ©
π = πΌ
3
Β· 680 Ξ©
π = 0. 003957 Β· 680 Ξ©
π = 2. 7 π
The voltage on 47 Ξ©
π = πΌ
2
Β· 47 Ξ©
π = 0. 00655 π΄ Β· 47 Ξ©
π = 0. 3 π
30
Mastering the Art of Measurement (DM401B Smart Digital Multimeter) r.01 www.plusivo.com