Two circuit problems?

First problem:

You got wrong because you assumed that the current was the same before and after the heaters were connected in series. Actually it changed because the resistance of the circuit had changed. Only the voltage across the terminal of the battery did not depend on the circuit.

Eliminating I between the formulae P = UI and P = RI², we get:

P = U² / R . . . . . . . . . . . . .[P² / P = (UI)² / RI² = U² / R]

hence

R = U² / P

The resistance of each heater is:

R1 = U² / P1

R2 = U² / P2

The resistance of 2 heaters in series is

R(total) = R1 + R2

. . . . . = U² (1/P1 + 1/P2)

The total power is:

P(total) = U² / R(total) = U² / U² (1/P1 + 1/P2).= P1P2 / (P1 + P2)

P(total) = 340×235/(340 + 235) = 139 W

(The total power is less than the power of each heater used separately because of the amperage drop due to the increase in resistance of the circuit)

Second problem.

Applying Ohm's law in 2 situations, we get:

(Eq 1): 14 = 2 (R1 + R2)

(Eq 2): 14 = 9 R1R2 / (R1 + R2)

From Eq 1: R1 + R2 = 14/2 = 7

From Eq 2: R1R2 = 14 (R1 + R2) / 9; hence R1R2 = 14×7/9 = 98/9

We have to find two positive numbers R1 and R2 whose sum is S = 7 and whose product is P = 98/9.

R1 and R2 are positive roots, if they exist, of the quadratic equation

x² - Sx + P = 0

Let's plug in the values of S and P:

x² - 7x + 98/9 = 0

9x² - 63x + 98 = 0

Δ = 63² - 4×9×98 = 441 = 21²

R1 = (63 − 21) / 18 = 42/18 = 2.33 ohms

R2 = (63 + 21) / 18 = 84/18 = 4.67 ohms

First question: V^2/R1 = 340 V^2/R2 = 235 thus

R1 = V^2 / 340 R2 = V^2 / 235

In series the power will be: P = V^2/ (R1 + R2)

Substitute for R1 and R2 from line two into line 3 and you will see that the V^2 terms cancel giving you:

P = 1 /((1/340) + (1/235)) = 138.95 watts.

Second question you got the equations right, just keep working on the algebra and you'll get it. You will have to solve the first equation for either R1 or R2 and then substitute in the second eqn. It will get a little messy, but with care you can do it. Good luck.

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