20.4 Electric power and energy  (Page 4/7)

Why do incandescent lightbulbs grow dim late in their lives, particularly just before their filaments break?

The power dissipated in a resistor is given by $P=V^2/R$ , which means power decreases if resistance increases. Yet this power is also given by $P=I^{2}R$ , which means power increases if resistance increases. Explain why there is no contradiction here.

What is the power of a $1.00\times {\text{10}}^{\text{2}}\text{MV}$ lightning bolt having a current of ${\mathrm{2.00\; \times \; 10}}^{\text{4}}\text{A}$ ?

What power is supplied to the starter motor of a large truck that draws 250 A of current from a 24.0-V battery hookup?

A charge of 4.00 C of charge passes through a pocket calculator’s solar cells in 4.00 h. What is the power output, given the calculator’s voltage output is 3.00 V? (See [link] .)

How many watts does a flashlight that has $6.00\times {\text{10}}^{\text{2}}\text{C}$ pass through it in 0.500 h use if its voltage is 3.00 V?

Find the power dissipated in each of these extension cords: (a) an extension cord having a $0\text{.}\text{0600}\text{-}\Omega$ resistance and through which 5.00 A is flowing; (b) a cheaper cord utilizing thinner wire and with a resistance of $0\text{.}\text{300}\Omega .$

Verify that the units of a volt-ampere are watts, as implied by the equation $P=\text{IV}$ .

Show that the units $1{\text{V}}^2/\Omega =1\text{W}$ , as implied by the equation $P=V^2/R$ .

Show that the units $1{\text{A}}^2\cdot \Omega =1\text{W}$ , as implied by the equation $P=I^{2}R$ .

Verify the energy unit equivalence that $1\text{kW}\cdot \text{h = 3}\text{.}\text{60}\times {\text{10}}^6\text{J}$ .

Electrons in an X-ray tube are accelerated through $1.00\times {\text{10}}^{\text{2}}\text{kV}$ and directed toward a target to produce X-rays. Calculate the power of the electron beam in this tube if it has a current of 15.0 mA.

An electric water heater consumes 5.00 kW for 2.00 h per day. What is the cost of running it for one year if electricity costs $\text{12.0 cents}\text{/kW}\cdot \text{h}$ ? See [link] .