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A resistor with R1=25.0 ohm is connected to a battery that has internal resistance and electrical energy is dissipated by R1 at rate of 36.0 W . If a second resistor with R2= 15.0 ohm is connected in series with R1 , what is total rate at which electrical energy is dissipated by the two resistors

Respuesta :

Use the power and R1 to find the voltage of the battery
P = V²/R
V = √(PR) = √(36.0 W)(25.0 ohm) = 30 V

Now find the equivalent resistance of the new configuration
R_equivalent = R1 + R2 = 40 ohm

Now find the power (energy rate)
P = V²R = (30 V)²/(40 ohm) = 22.5 W
onyebuchinnaji

The total rate at which electrical energy is dissipated by the two resistors is 48 W.

The given parameters;

  • First resistor, R₁ = 25 ohm
  • Second resistor, R₂ = 15 ohm
  • power dissipated in the first resistor, P₁ = 36 W

In series circuit arrangement, the current flowing in each resistor is the same.

The current flowing in the first resistor is calculated as follows;

P = I²R₁

[tex]I^2 = \frac{P}{R_1} \\\\I = \sqrt{\frac{P}{R_1}} \\\\I = \sqrt{\frac{36}{25}}\\\\I = 1.2 \ A[/tex]

The equivalent resistance of the series arrangement is calculated as;

R = R₁ + R₂

R = 25 + 15

R = 40 ohms

The total rate at which electrical energy is dissipated by the two resistors;

P = I²R

P = 1.2(40)

P = 48 W

Thus, the total rate at which electrical energy is dissipated by the two resistors is 48 W.

Learn more here: https://brainly.com/question/2937340

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