## We are designing an electric space heater to operate from 120 V. Two heating elements with resistances R1 and R2 are to be used that can be

Question

We are designing an electric space heater to operate from 120 V. Two heating elements with resistances R1 and R2 are to be used that can be operated in parallel, separately, or in series. The highest power is to be 1840 W, and the lowest power is to be 240 W. What values are needed for R1 and R2? What intermediate power setting is available?

in progress 0
2 weeks 2021-11-15T13:10:56+00:00 1 Answer 0 views 0

Step-by-step explanation:

For highest power the resistor R₁ and R₂ must be connected in parallel

effective resistance R eff = 1 / R₁ + 1 / R₂ = R₁ R₂ / ( R₁ + R₂)

P = V² / R eff

1840 W = (120V)² /( R₁ R₂ / ( R₁ + R₂) )

R₁ R₂ / ( R₁ + R₂) = (120V)² / 1840 W = 7.8260

For lowest power, the resistor must be connected in series

Reff = R₁ + R₂

P = V² / Reff

240 W = (120 V)² / ( R₁ + R₂)

R₁ + R₂ = 60

R₁  = 60 – R₂

R₁ R₂ / ( R₁ + R₂) = 7.8260

substitute R₁ into the expression

R₂(60 – R₂)  / ( 60 – R₂  + R₂) = 7.8260

R₂(60 – R₂) / 60 = 7.8260

R₂(60 – R₂) = 7.8260 × 60

R₂(60 – R₂) = 469.6

–  R₂²+ 60R₂ = 469.6

–  R₂² + 60R₂ – 469.6 = 0

multiply the equation by -1 on both side

R₂² – 60R₂ + 469.6 = 0

-b ± √ (b² – 4ac) / 2a

R₂ = 50. 746 Ω or 9.254Ω

R₁  = 60 – R₂ = 60 – 50. 746 Ω  = 9.254 Ω

R₁ = 9.254 Ω, R₂ = 50. 746 Ω

b) intermediate power setting available is the power when each of the voltage is connected to one resistor

P = 120 V² / 9.254Ω = 1556.8 W

P = 120 V² / 50. 746 Ω = 283.8 W