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.
