A220-v electric heater has two heating coils that can be switched such that either coil can be used independently or the two can be connected in series or parallel, for a total of four possible configurations. if the warmest setting corresponds to 2,000-w power dissipation and the coolest corresponds to 300 w, find the resistance of each coil.

The resistances of both coils are 131.7 Ω and 29.64 Ω.

Explanation:

Since, there are two coils, they can be used independently or in series or parallel. The power is given as:

Power = P = VI

but, from Ohm's Law:

V = IR

I = V/R

therefore,

P = V²/R

R = V²/P

Hence, the resistance (R) and (P) are inversely proportional. Therefore, the maximum value of resistance will give minimum power, that is, 300 W. And the maximum resistance will be in series arrangement, as in series the total resistance gets higher than, any individual resistance.

Therefore,

Rmax = V²/Pmin = R1 + R2

R1 + R2 = (220 V)²/300 W

R1 + R2 = 161.333 Ω     en (1)

Similarly, the minimum resistance will give maximum power. And the minimum resistance will occur in parallel combination. Because equivalent resistance of parallel combination is less than any individual resistance.

Therefore,

(R1 R2)/(R1 + R2) = (220 V)²/2000 W

using eqn (1), we get:

(R1 R2) / 161.333 Ω = 24.2 Ω

R1 R2 = 3904.266 Ω²

R1 = 3904.266 Ω²/R2   eqn (2)

Using this value of R1 in eqn (1), we get:

3904.266/R2 +R2 = 161.333

(R2)² - 161.333 R2 +3904.266 = 0

Solving this quadratic eqn we get two values of R2 as:

R2 = 131.7 Ω     OR     R2 = 29.64 Ω

when ,we substitute these values in eqn (1) to find R1, we get get the same two values as R2, alternatively. This means that the two coils have these resistance, and the order does not matter.

Therefore, the resistance of both coils are found to be 131.7 Ω and 29.64 Ω

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