# The light bulb and rheostat are connected in series and connected to a current source.

The light bulb and rheostat are connected in series and connected to a current source. The voltage at the terminals of the bulb is 40V, the resistance of the rheostat is 60 Ohm. The external circuit draws power 120 W. Find the current in the circuit.

Given:
Ul = 40 V
Rр = 60 Ohm
Ptot = 120 W
I = Il = Iр – for serial connection
Ptot = I ^ 2 * (Rl + Rp) – for serial connection
120 = I ^ 2 * (Rl + 60)
Rl = Ul / I – according to Ohm’s law
Rl = 40 / I
120 = I ^ 2 * (40 / I + 60)
120 = 60 * I ^ 2 + 40 * I
60 * I ^ 2 + 40 * I-120 = 0
3 * I ^ 2 + 2 * I-6 = 0
Let’s solve this quadratic * equation:
D = 2 ^ 2-4 * 3 * (- 6) = 4 + 72 = 76
I = (- 2 ± √76) / 2 * 3 = (- 2 ± √4 * 19) / 2 * 3 = 2 * (- 1 ± √19) / 2 * 3 = (- 1 ± √19) / 3
I = (- 1/3) ± (√19 / 9) = – 0.333 ± 1.453
Interested in a positive solution to this equation
I = 1.12 A – circuit current One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities.