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

Question

Joyce Solomon

Physics

4 March, 07:27

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

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Dax Campbell · Answer 1

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