a circuit with 3 resisters {9 ohms 10 ohms and 12 ohms} connected in parallel with the source 150v\50HZ.use kitchoffs law to find the current of the circuit and the current across each resister?
R1 = 9 Ohms.
R2 = 10 Ohms.
R3 = 12 Ohms.
1/Rt = 1/R1 + 1/R2 + 1/R3,
1/Rt = 1/9 + 1/10 + 1/12 = 0.294444,
Rt = 3.4 Ohms.
It = E/Rt = 150/3.4 = 44.1A.
I1 + I2 + I3 = It = 44.1A.
Let I3 = I,
I2 = (12/10)*I = 1.2I.
I1 = (12/9)*I = 1.33I.
1.33I + 1.2I + I = 44.1,
3.53I = 44.1,
I = I3 = 12.49A.
1.2I = I2 = 1.2*12.49 = 14.99A.
1.33I = I1 = 1.33*12.49 = 16.61A.
It = I1 + I2 + I3,
It = 16.61 + 14.99 + 12.49 = 44.1A.