Part A Learning Goal: To understand the concept of electromotive force and internal resistance; to understand the processes in one-loop circuits; to become familiar with the use of the ammeter and The next several questions refer to the four diagrams on the left in (Figure 2) shown here labeled A, B, C, and D. 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 Part B 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 Part C 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 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 Part D lowest possible potential energy. How can such a particle move toward a point where it would have a higher potential energy? Such a move requires that nonelectrostatic forces act upon the charged particle, Part E pushing it toward higher potential energy despite the presence of electrostatic forces. In 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 Part F nonelectrostatic force. 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 Part G abbreviated emf or denoted by 2). Batteries are often referred to as sources of emf (rather than sources of energy, even though they are, fundamentally, sources of energy) The emf of a battery can be calculated using the definition mentioned above: The last group of questions refers to a battery that has emf 12.0 volts and internal resistance 3.00 ohms. E = W/q. The units of emt are joules per coulomb, that is, volts. The terminals of a battery are often labeled + and - for "higher potential" and "lower potential," respectively. The potential difference between the terminals is called the Part H 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 Vbat = &. 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 Express your answer in volts. Use three significant figures. passes through the battery, the battery does the amount of work &q on the charge; 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 &q - qIr, and the terminal voltage is O AEd ACE ? AVbat = E - Ir. AV = V 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 voltmeter, you should also know the way the ammeter and the voltmeter must be Submit Request Answer 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 Next Figure the "battery" symbol. It is important to keep in mind that this resistor with resistance r is actually inside the battery. Part1 Complete previous part(s) In all diagrams, & stands for emf, r 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 Part J Complete previous part(s) 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 Part K Complete previous part(s) has a very large resistance. Part L Complete previous part(s)