Chapter 4: Q8DQ (page 911)
How might a loop of wire carrying a current be used as a compass? Could such a compass distinguish between north and south? Why or why not?
We would have to know the direction of the current in the loop in order to determine where is north and where is south.
Over 30 million students worldwide already upgrade their learning with Vaia!
All the tools & learning materials you need for study success - in one app.
Get started for free
Abulb is connected across the terminals of abattery havingof internal resistance. What percentage of the power of the battery is dissipated across the internal resistance and hence is not available to the bulb?
The circuit shown in Fig. E25.33 contains two batteries, each with an emf and an internal resistance, and two resistors. Find (a) the current in the circuit (magnitude and direction) and (b) the terminal voltage Vabof the 16.0-V battery.
Fig. E25.33
In the circuit shown in Fig. E25.30, the 16.0-V battery is removed and reinserted with the opposite polarity, so that its negative terminal is now next to point a. Find (a) the current in the circuit (magnitude anddirection); (b) the terminal voltage Vbaof the 16.0-V battery; (c) the potential difference Vacof point awith respect to point c. (d) Graph the potential rises and drops in this circuit (see Fig. 25.20).
A heart defibrillator is used to enable the heart to start beating if it has stopped. This is done by passing a large current ofthrough the body atfor a very short time, usuallyabout . (a) What power does the defibrillator deliver to the body, and (b) how much energy is transferred ?
A typical small flashlight contains two batteries, each having an emf of, connected in series with a bulb having resistance. (a) If the internal resistance of the batteries is negligible, what power is delivered to the bulb? (b) If the batteries last forwhat is the total energy delivered to the bulb? (c) The resistance of real batteries increases as they run down. If the initial internal resistance is negligible, what is the combined internal resistance of both batteries when the power to the bulb has decreased to half its initial value? (Assume that the resistance of the bulb is constant. Actually, it will change somewhat when the current through the filament changes, because this changes the temperature of the filament and hence the resistivity of the filament wire.)
What do you think about this solution?
We value your feedback to improve our textbook solutions.
