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Ostrovityanka [42]
3 years ago
11

An unknown resistor is connected between the terminals of a 3.00 V battery. Energy is dissipated in the resistor at the rate of

0.560 W. The same resistor is then connected between the terminals of a 1.50 V battery. At what rate is energy now dissipated?
Physics
1 answer:
katrin [286]3 years ago
4 0

Answer:

The energy dissipate at a rate of 11.34 W when the resistor is connected between the terminals of a 1.50 V battery.

Explanation:

the rate at which energy  dissipeted is given by power and we know that when given resistance, R and voltage, V then power is given by:

P = V^2/R

R = V^2/P

  = (3.00)^2/(0.560)

  = 5.04 ohms

then when the resistor is now connected between the terminals of a 1.50 V battery, then the power is given by:

P = V^2/R

  = (1.50)^2/(5.40)

  = 11.34 W.

Therefore, the energy dissipate at a rate of 11.34 W when the resistor is connected between the terminals of a 1.50 V battery.

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