Question

. A 50 Ω resistor is in parallel with a 100Ω resistor. Current in 50Ω is 7.2A. What is the third resistance to be added in parallel to this circuit to make the total current 12.1A?​

Answer 1

Answer:

Solving for the voltage drop on the first resistor, Vr1:

Vr1 = Ir1 * r1

Vr1 = 7.2 A * 50 Ω

Vr1 = 360 V

The value of 360 V is also the source voltage of r2 and r3.

Solving for the current I2 flowing through resistor r2:

I2 = Vr2 / r2

I2 = 360 V / 100 Ω

I2 = 3.6 A

Solving for the sum of I1 and I2:

Sum of I1 and I2 = I1 + I2

Sum of I1 and I2 = 7.2 A + 3.6 A

Sum of I1 and I2 = 10.8 A

Solving for the current flowing through resistor r3, I3:

I3 = 12.1 A - Sum of I1 and I2

I3 = 12.1 A – 10.8 A

I3 = 1.3 A

Solving for the resistance r3 of the third resistor:

r3 = source voltage / current through r3

r3 = 360 V / 1.3 A

r3 = 276.92 Ω which is rounded to 277 Ω