Learning Goal: To understand the concept of electromotive force and internal resistance; to understand View Available Hint(s) the processes in one-loop circuits; to become familiar with the use of the ammeter and voltmeter. 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 creating a "build-up." Such build-up, if it occurs, creates its own electric field that cancels out the external electric field, ultimately causing the current to stop. 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 push the particles in the direction at leads to a decrease in potential energy. At some Submit Previous Answers 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 would have a higher potential ener Correct Such a move requires that nonelectrostatic forces act upon the charged particle, In diagrams A and B, the voltmeter readings would actually be quite close to the terminal voltage if the ammeter has a very low resistance, and the voltmeter, a very high one. However, diagram C clearly pushing it toward higher potential energy despite the presence of electrostatic forces. In shows the best way to connect the voltmeter in order to measure the terminal voltage. 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 potent ergy through the action of some kind of nonelectrostatic force. Part F 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 abbreviated emf or denoted by &). Batteries are often referred to as sources of emf Part G (rather than sources of energy, even though they are, fundamentally, sources of energy). The emf of a battery can be calo ing the definition mentioned above: E = W/q. The units of emf are joul per coulomb, that is, volts. The last group of questions refers to a battery that has emf 12.0 volts and internal resistance 3.00 ohms. The terminals of a battery are of eled + and - for "higher potential" and "lower 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 Part H voltage is equal to the emf of the battery: A Vbat = E. However, if there is a current in the circuit, the terminal voltage is less than the emf 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 voltmete because the battery has its own internal resistance (usually labeled r). When charge q passes through the battery, the battery does the amount of work &g on the charge; Express your answer in volts. Use three significant figures. however, the charge also "loses" the amount of energy equal to Ir ( I is the current through the circuit); therefore, the increase in potential energy is Eq - qIr, and the AV = 12.0 V terminal voltage is AVbat = E - IT. Previous Answers In order to answer the questions that follow, you should first review the meaning of the symbols describing various elements of the circuit, including the ammeter and the Correct 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. Note that the internal resistance is usually indicated as a separate resistor drawn next to Part I the "battery" symbol. It is important to keep in mind that this resistor with resistance r is actually inside the battery. 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? In all diagrams, & stands for emf, or for the internal resistance of the battery, and R for the resistance of the external circuit. As usual, we'll assume that the connecting wires Express your answer in amperes. Use two significant figures. 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 has a very large resistance. 6 0 0 0 ? Figure Submit Request Answer Part J Complete previous part(s)