is a property that determines the amount of current for a given source voltage, whereas a is a device that controls the current in a circuit. Resistance, resistor Resistance, battery Resistor, resistance
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- Consider the circuit below. The battery has an emf of = 30.00 V and an internal resistance of r = 1,00 . (a) Find the equivalent resistance of the circuit and the current out of the battery. (b) Find the current through each resistor, (c) Find die potential drop across each resistor, (d) Find the power dissipated by each resistor, (e) Find the total power supplied by the batteries.A battery with an internal resistance of 10.0 produces an open circuit voltage of 12.0 V. A variable load resistance with a range from 0 to 30.0 is connected across the battery. (Note: A battery has a resistance that depends on the condition of its chemicals and that increases as the battery ages. This internal resistance can be represented in a simple circuit diagram as a resistor in series with the battery.) (a) Graph the power dissipated in the load resistor as a function of the load resistance. (b) With your graph, demonstrate the following important theorem: The power delivered to a load is a maximum if the load resistance equals the internal resistance of the source.Three identical 60.0-W, 120-V lightbulbs are connected across a 120-V power source as shown in Figure P28.72. Assuming the resistance of each lightbulb is constant (even though in reality the resistance might increase markedly with current), find (a) the total power supplied by the power source and (b) the potential difference across each lightbulb.
- Construe! Your Own Problem Consider a rechargeable lithium cell that is to be used to power a camcorder. Construct a problem in which you calculate the internal resistance of the cell during normal operation. Also, calculate the minimum voltage output of a battery charger to be used to recharge your lithium cell. Among the things to be considered are the emf and useful terminal voltage of a lithium cell and the current it should be able to supply to a camcorder.Integrated Concepts A 12.0-V emf automobile battery has a terminal voltage of 16.0 V when being charged by a current of 10.0 A. (a) What is the battery’s internal resistance? (b) What power is dissipated inside the battery? (c) At what rate (in °C/min ) will its temperature increase if its mass is 20.0 kg and it has a specific heat of 0.300 kcal/kg. °C, assuming no heat escapes?Some camera flashes use flash tubes that requite a high voltage. They obtain a high voltage by charging capacitors in parallel and then internally changing the connections of the capacitors to place diem in series. Consider a circuit that uses four AAA batteries connected in series to charge six 10-mF capacitors through an equivalent resistance of 100 . The connections are thenswitched internally to place the capacitors in series. The capacitors discharge through a lamp with a resistance of 100 . (a) What is the RC time constant and the initialcurrent out of the batteries while they are connected in parallel? (b) How long does it take for the capacitors to charge to 90% of the terminal voltages of the batteries? (c) What is the RC time constant and the initial current of the capacitors connected in series assuming it discharges at 90% of full charge? (d) How long does it rake the current to decrease to 10% of the initial value?
- Figure 21.55 shows how a bleeder resistor is used to discharge a capacitor after an electronic device is shut off allowing a person to work on the electronics with less risk of shock, (a) What is the time constant? (b) How long will it take to reduce the voltage on the capacitor to 0.250% (5% of 5%) of its full value once discharge begins? (c) If the capacitor is charged to a voltage V0through a 100-O resistance, calculate the time it takes to rise to 0.865V0(This is about two time constants.)Figure P18.26 shows a voltage divider, a circuit used to obtain a desired voltage Vout from a source voltage . Determine the required value of R2 if = 5.00 V, Vout = 1.50 V and R1 = 1.00 103 (Hint: Use Kirchhoff's loop rule, substituting Vout = IR2, to find the current. Then solve Ohms law for R2. Figure P18.26Check Your Understanding In considering the following schematic and the power supplied and consumed by a circuit, will a voltage source always provide power to the circuit, or can a voltage source consume power?
- The rather simple circuit shown below is known as a voltage divider. The symbol consisting of three horizontal lines is represents “ground” and can be defined as the point where the potential is zero. The voltage divider is widely used in circuits and a single voltage source can be used to provide reduced voltage to a load resistor as shown in the second part of the figure, (a) What is the output voltage Vout of circuit (a) in terms of R1,R2,andVin (b) What is the output voltage Vout of circuit (b) in terms of R1,R2,RLandVinCheck Your Understanding How would you use a river and two waterfalls to model a parallel configuration of two resistors? How does this analogy break down?An ideal voltmeter connected across a certain fresh 9-V battery reads 9.30 V, and an ideal ammeter briefly connected across the same battery reads 3.70 A. We say the battery has an open-circuit voltage of 9.30 V and a short-circuit current of 3.70 A. Model the battery as a source of emf in series with an internal resistance r as in Figure 27.1a. Determine both (a) and (b) r. An experimenter connects two of these identical batteries together as shown in Figure P27.45. Find (c) the open-circuit voltage and (d) the short-circuit current of the pair of connected batteries. (e) The experimenter connects a 12.0- resistor between the exposed terminals of the connected batteries. Find the current in the resistor. (f) Find the power delivered to the resistor. (g) The experimenter connects a second identical resistor in parallel with the first. Find the power delivered to each resistor. (h) Because the same pair of batteries is connected across both resistors as was connected across the single resistor, why is the power in part (g) not the same as that in part (f)? Figure P27.45