Electric Power Quiz

PDF page 647: The formula for power may be found by dimensional analysis. Consider the units of power. In the SI system, power is given in watts (W), which is energy per unit time, or J/s

PDF page 647: Recall now that a voltage is the potential energy per unit charge, which means that voltage has units of J/C

PDF page 647: But a Coulomb per second (C/s) is an electric current, which we can see from the definition of electric current, , where Q is the charge in coulombs and t is time in seconds. Thus, equation above tells us that electric power is voltage times current, or

PDF page 651: d. Electric power is proportional to current through the resistor multiplied by the voltage across the resistor.

PDF page 647: Incandescent light bulbs, such as the two shown in Figure 19.20, are essentially resistors that heat up when current flows through them and they get so hot that they emit visible and invisible light. Thus the two light bulbs in the photo can be considered as two different resistors. In a simple circuit such as a light bulb with a voltage applied to it, the resistance determines the current by Ohm’s law, so we can see that current as well as voltage must determine the power.

PDF page 647: Incandescent light bulbs, such as the two shown in Figure 19.20, are essentially resistors that heat up when current flows through them and they get so hot that they emit visible and invisible light. Thus the two light bulbs in the photo can be considered as two different resistors. In a simple circuit such as a light bulb with a voltage applied to it, the resistance determines the current by Ohm’s law, so we can see that current as well as voltage must determine the power.

PDF page 647: But a Coulomb per second (C/s) is an electric current, which we can see from the definition of electric current, , where Q is the charge in coulombs and t is time in seconds. Thus, equation above tells us that electric power is voltage times current, or

PDF page 647: Thus, the power consumed by the circuit is 19.50 Where does this power go? In this circuit, the power goes primarily into heating the resistor in this circuit.

PDF page 648: This gives the power in terms of only the current and the resistance. Thus, by combining Ohm’s law with the equation for electric power, we obtain two more expressions for power: one in terms of voltage and resistance and one in terms of current and resistance. Note that only resistance (not capacitance or anything else), current, and voltage enter into the expressions for electric power. This means that the physical characteristic of a circuit that determines how much power it dissipates is its resistance. Any capacitors in the circuit do not dissipate electric power—on the contrary, capacitors either store electric energy or release electric energy back to the circuit. To clarify how voltage, resistance, current, and power are all related, consider Figure 19.22, which shows the formula wheel. The quantities in the center quarter circle are equal to the quantities in the corresponding outer quarter circle. For example, to express a potential V in terms of power and current, we see from the formula wheel that .

PDF page 648: This gives the power in terms of only the current and the resistance. Thus, by combining Ohm’s law with the equation for electric power, we obtain two more expressions for power: one in terms of voltage and resistance and one in terms of current and resistance. Note that only resistance (not capacitance or anything else), current, and voltage enter into the expressions for electric power. This means that the physical characteristic of a circuit that determines how much power it dissipates is its resistance. Any capacitors in the circuit do not dissipate electric power—on the contrary, capacitors either store electric energy or release electric energy back to the circuit. To clarify how voltage, resistance, current, and power are all related, consider Figure 19.22, which shows the formula wheel. The quantities in the center quarter circle are equal to the quantities in the corresponding outer quarter circle. For example, to express a potential V in terms of power and current, we see from the formula wheel that .

PDF page 648: A typical older incandescent lightbulb was 60 W. Assuming that 120 V is applied across the lightbulb, what is the current through the lightbulb? STRATEGY to We are given the voltage and the power output of a simple circuit containing a lightbulb, so we can use the equation find the current I that flows through the lightbulb. Solution for the current and inserting the given values for voltage and power gives Solving

PDF page 648: Thus, a half ampere flows through the lightbulb when 120 V is applied across it.

PDF page 649: Discussion This is a significant current. Recall that household power is AC and not DC, so the 120 V supplied by household sockets is an alternating power, not a constant power. The 120 V is actually the time-averaged power provided by such sockets. Thus, the average current going through the light bulb over a period of time longer than a few seconds is 0.50 A.

PDF page 652: alternating current electric current whose direction alternates back and forth at regular intervals ampere unit for electric current; one ampere is one ) coulomb per second ( circuit diagram schematic drawing of an electrical circuit including all circuit elements, such as resistors, capacitors, batteries, and so on conventional current flows in the direction that a positive charge would flow if it could move direct current electric current that flows in a single direction electric circuit physical network of paths through which electric current can flow electric current electric charge that is moving electric power rate at which electric energy is transferred in a circuit equivalent resistor resistance of a single resistor that is the

PDF page 649: To warm your boots on cold days, you decide to sew a circuit with some resistors into the insole of your boots. You want 10 W of heat output from the resistors in each insole, and you want to run them from two 9-V batteries (connected in series). What total resistance should you put in each insole? STRATEGY We know the desired power and the voltage (18 V, because we have two 9-V batteries connected in series), so we can use the to find the requisite resistance. equation Solution Solving for the resistance and inserting the given voltage and power, we obtain

PDF page 649: Thus, the total resistance in each insole should be 32 Discussion Let’s see how much current would run through this circuit. We have 18 V applied across a resistance of 32 , so Ohm’s law gives