To understand the concept of electromotive force and internal resistance; to understand Part H voltmeter. the processes in one-loop circuits; to become familiar with the use of the ammeter and In order for the current in a conductor to exist continuously, the conductor must be part of a loop, that is, a closed path through which the charged particles can move without A voltmeter is connected to the terminals of the battery; the battery is not connected to any other external circuit elements. What is the reading of the voltmeter creating a "build-up." Such build-up, if it occurs, creates its own electric field that cancels Express your answer in volts. Use three significant figures. out the external electric field, ultimately causing the current to stop. AV = 12.0 V However, having a loop, or a closed circuit, is not enough to maintain the current; there must also be a source of energy. Its necessity is fairly obvious: As charged particles move along the circuit, they lose potential energy. In fact, electrostatic forces always Submit Previous Answers push the particles in the direction that leads to a decrease in potential energy. At some point, each charged particle would reach the location in the circuit where it has the lowest possible potential energy. How can such a particle move toward a point where it Correct would have a higher potential energy? Such a move requires that nonelectrostatic forces act upon the charged particle, pushing it toward higher potential energy despite the presence of electrostatic forces. In Part I circuits, such forces exist inside a device commonly known as a battery. In a circuit, the battery serves as the energy source that keeps the charged particles in continuous motion by increasing their potential energy through the action of some kind of nonelectrostatic force. The voltmeter is now removed and a 21.0-ohm resistor is connected to the terminals of the battery. What is the current I through the battery? Express your answer in amperes. Use two significant figures. The amount of work that the battery does on each coulomb of charge that it "pushes through" is called (inappropriately) the electromotive force (pronounced "ee-em-ef" and I = 0.50 A abbreviated emf or denoted by &). Batteries are often referred to as sources of emf (rather than sources of energy, even though they are, fundamentally, sources of energy). The emi of a battery can be calculated using the definition mentioned above: Submit Previous Answers E = W/q. The units of emf are joules per coulomb, that is, volts. The terminals of a battery are often labeled + and - for "higher potential" and "lower Correct potential," respectively. The potential difference between the terminals is called the terminal voltage of the battery. If no current is running through a battery, the terminal voltage is equal to the emf of the battery: A V 1 = E. Part J However, if there is a current in the circuit, the terminal voltage is less than the emf because the battery has its own internal resistance (usually labeled r). When charge q passes through the battery, the battery does the amour of work &q on the charge; however, the charge also "loses" the amount of energy equal to Ir ( I is the current In the situation described in Part I , what is the current I through the 21.0-ohm resistor ? through the circuit); therefore, the increase in potential energy is Eq - qIr, and the Express your answer in amperes. Use two significant figures. terminal voltage is I = 0.50 A AVbat = E - In order to answer the questions that follow, you should first review the meaning of the Submit Previous Answers symbols describing various elements of the circuit, including the ammeter and the voltmeter; you should also know the way the ammeter and the voltmeter must be connected to the rest of the circuit in order to function properly. Correct Since the battery and the external resistor form one loop, the charge that passes through one must pass through another; therefore, the currents must be the same. Note that the internal resistance is usually indicated as a separate resistor drawn next to the "battery" symbol. It is important to keep in mind that this resistor with resistance r is actually inside the battery. In all diagrams, & stands for emf, r for the internal resistance of the battery, and R for Part K the resistance of the external circuit. As usual, we'll assume that the connecting wires have negligible resistance. We will also assume that both the ammeter and the voltmeter are ideal: That is, the ammeter has negligible resistance, and the voltmeter What is the potential difference A V across the 21.0-ohm resistor from Part 1? has a very large resistance. Express your answer in volts. Use three significant figures. > View Available Hint(s) Figure 0 8 ? AV