# 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.