? voltmeter. in order for the current in a conductor to exist continuously, the conductor must Um e In the situation described in Part I, what is the current I through the 21.0-ohm resistor? of a loop, that is, a closed path through which the charged particles can move without naw_v%m "build-up." Such build-up, if it occurs, creates its own electric field that cancels Express your answer In amperes. Use two significant figures. out the external electric field, ultimately causing the current to stop. I=050 A However, having a loop, or a closed circuit, is not enough to maintain the current; there must also be a source of energy. lts necessity is fairly obvious: As charged particles move along the circuit, they lose potential energy. In fact, electrostatic forces always 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 v Correct would have a higher potential energy? 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. 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 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 v PartK motion by increasing their potential energy through the action of some kind of nonelectrostatic force. 2 The amount of work that the battery does on each coulomb of charge that it "pushes What is the potential difference AV across the 21.0-ohm resistor from Part | 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 (rather than sources of energy, even though they are, fundamentally, sources of energy). P View Available Hint(s) The emf of & batiery can be calculated using the definition mentioned above: & = W /g. The units of emf are joules per coulomb, that is, volts. Express your answer In volt. Use three significant figures. AV =105V The terminals of 2 battery are often labeled + and for \"higher potential" and "lower potential.\" respectively. The potential difierence between the terminals is called the terminal voltage of the battery. If no current is running through a battery, the terminal voltage is equal o the emf of the battery: AV, = Previous Answers However, if there is & current in the circuit, the terminal voltage is less than the emf Comect because the battery has its own internal resistance {usually labeled 7). When charge passes through the battery, the battery does the amount of work 4q on the charge; howeve, the charge aiso "loses" the amount of energy equal to I ( [ is the current through the circuit): therefore, the increase in potential energy is q gIr, and the v PartL terminal voltage is AV = b = Ir. What is the terminal voltage AV of the battery connected to the 21.0-ohm resistor from Part 12 In order to answer the questions that foliow, you should first review the meaning of the symbois describing various elements of the circuit, including the ammeter and the voltmeter; you shouid also know the way the ammeter and the voltmeter must be connected 1o the rest of the circuit in order to function properly. Express your answer in volts. Use three significant figures. View Available Hint(s) Note that the intemal resistance is usually indicated as a separate resistor drawn next to AV = 105 V the \"batiery\" symbol. It is important 1o keep in mind that this resistor with resistance r is aclually inside the batiery y Previous Answers 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 have negligible resistance. We will aiso assume that both the ammeter and the v Correct voltmeter are ideal: That is, the ammeter has negligible resistance, and th ltmet Since the ends of the resistor with 1 3 } e s egligi i ind the voltmeter " with resistance R are attached to the terminals of the battery, the voltage across the resistor is the same as that between the terminals of the battery. | v PartM Figure I 10f2 > How much work W does the battery connected to the 21,0-ohm resistor perform in one m Express your answer In Joules. Use three significant figures. View Avallable Hint(s) ;e