/sci/ - Science & Math

Because the source emfs are 180 degrees in phase you can get equivelant DC circuit when L1 is at highest tension.

Kirchhoff current law:
[math]I_1=I_2+I_3[/math] (I chose the I_3 direction, does not matter as it will just be negative if chosen wrong)

Kirchhoff voltage law:
Upper loop: [math]U_1=I_1*R_1[/math]
Lower loop: [math]U_2=I_2*R_2[/math]
Immediatly we see that they can be treated as separate entities as do households actually do.
Total loop: [math]U_1+U_2=I_1*R_1+I_2*R_2[/math]
(Absolutely redunant).

Solving [math]I_3[/math]:
[math]I_1=I_2+I_3[/math]
[math]I_3=I_1-I_2[/math]
[math]I_3=U_1 / R_1 - U_2 / R_2[/math]
Since [math]U_1=U_2=U[/math]

[math]I_3=U (1/ R_1 - 1/ R_2)[/math]
and if the lamps have same resistance
[math]R_1=R_2=R[/math]
[math]I_3=U (1/R - 1/R)=0[/math]
and we see that current flows only in the great loop.

This does not mean there would be possibility to short circuit 240 Volts unless you would short circuit simultanously at both resistors and the neutral wire at middle would be broken.