Two resistors are connected to a 120 V network. When they are connected in series, the current is 3 A

Two resistors are connected to a 120 V network. When they are connected in series, the current is 3 A, and when they are parallel, the total current is 16 A. Find the value of the resistances R1 and R2.

First, determine the total resistance in series connection.

To do this, we divide the direction of the network by the strength of the current that flows in it.

120/3 = 40 ohms.

With a parallel connection, the resistance will be:

120/16 = 7.5 ohms.

We draw up an equation of resistance ratios for serial and parallel connection.

R1 * R2 / (R1 + R2) = 120/16.

Let’s get rid of denominators.

R1 * R2 / 40 = 120/16.

We express the resistance R1.

R1 = (40 / R2) * (120/16) = 300 / R2.

We substitute the obtained value into the formula when the resistances are connected in series.

R1 + R2 = 40.

300 / R2 + R2 = 40.

We get a quadratic equation.

R22 – 40 * R2 + 300 = 0.

By Vieta’s theorem, we obtain:

R2 = 10 Ohm; R1 = 30 ohms.

Answer:

10 and 30 ohms.