Question 17.1: Turn On the Light GOAL Apply the concept of current. PROBLEM...
Turn On the Light
GOAL Apply the concept of current.
PROBLEM The amount of charge that passes through the filament of a certain lightbulb in 2.00 s is 1.67 C. Find (a) the average current in the lightbulb and (b) the number of electrons that pass through the filament in 5.00 s. (c) If the current is supplied by a 12.0-V battery, what total energy is delivered to the lightbulb filament? What is the average power?
STRATEGY Substitute into Equation 17.1a
I_{aν}={\frac{\Delta Q}{\Delta t}} [17.1a]
for part (a), then multiply the answer by the time given in part (b) to get the total charge that passes in that time. The total charge equals the number N of electrons going through the circuit times the charge per electron. To obtain the energy delivered to the filament, multiply the potential difference, ΔV, by the total charge. Dividing the energy by time yields the average power.
(a) Compute the average current in the lightbulb.
Substitute the charge and time into Equation 17.1a:
I_{aν}={\frac{\Delta Q}{\Delta t}}={\frac{1.67~{C}}{2.00~{s}}}=0.835\,\mathrm{A}
(b) Find the number of electrons passing through the filament in 5.00 s.
The total number N of electrons times the charge per electron equals the total charge, I_{aν} Δt:
(1)\quad N q=I_{aν}\ \Delta t
Substitute and solve for N:
N(1.60\times10^{-19}\,\mathrm{C/electron})=\,(0.835\,\mathrm{A})(5.00\,\mathrm{s})
N={2.61}\times10^{19}\,\mathrm{electrons}
(c) What total energy is delivered to the lightbulb filament? What is the average power?
Multiply the potential difference by the total charge to obtain the energy transferred to the filament:
{\bf(2)}\quad\Delta U=q\Delta V=(1.67\,\mathrm{C})(12.0\,\mathrm{V})\;=\;20.0\mathrm{J}
Divide the energy by the elapsed time to calculate the average power:
P_{aν}={\frac{\Delta U}{\Delta t}}={\frac{20.0~J}{2.00\,\mathrm{s}}}=~10.0~\mathrm{W}
REMARKS It’s important to use units to ensure the correctness of equations such as Equation (1). Notice the enormous number of electrons that pass through a given point in a typical circuit. Magnitudes were used in calculating the energies in Equation (2). Technically, the charge carriers are electrons with negative charge moving from a lower potential to a higher potential, so the change in their energy is \Delta U_{\mathrm{charge}} = qΔV = (-1.67 C)(+12.0 V) = -20.0 J, a loss of energy that is delivered to the filament, \Delta U_{\mathrm{fil}}=-\Delta U_{\mathrm{charge}} = +20.0 J. The energy and power, calculated here using the definitions of Chapter 16, will be further addressed in Section 17.6.