You may have seen the quantity kilowatt-hour written on an electric meter or bill. The kilowatt-hour is a measure of energy -- the electric company bills us for the amount of energy we use, not the amount of power. A kilowatt-hour (kWh) is the amount of energy consumed by a 1kW = 1000W device operating for 1 hour = 3600s, thus 1kWh = (1000W)(3600s) = 3.6×106J.
How much does it cost to watch a complete 21 hour long World Series on a 90.0W television set? Assume that electricity costs $0.0700/kWh.
This problem is easiest to solve by determining how much energy is used to operate the television in kilowatt-hours, not in joules.
Energy = (90.0W)(21h) = 1890Wh = 1.89kWh.
Cost = Energy × Rate = (1.89kWh)($0.0700/kWh) = $0.13 = 13¢
In this chapter we will build on the basic concepts of current and resistance defined in chapter 17. We will start looking at complete electric circuits, learn how to handle combinations of resistors, and look at circuits containing resistors and capacitors.
"Sources of emf" is a phrase that simply means source of voltage. Batteries and electric generators are common sources of voltage.
Although we think of batteries as "perfect" sources of voltage, they aren't. A more accurate picture of a real battery is a perfect voltage source with a small resistance in series. This resistance corresponds to an effective resistance inside the battery, not to a separate device. For purposes of this class, we will treat batteries and generators as ideal sources of voltage, unless specifically stated to the contrary.
We want to be able to treat arbitrary combinations of resistors. As we did with capacitors, we will begin by considering how to handle a pair of resistors, and from this we can build up any arbitrary combination. Once again, there are only two ways to connect two resistors, in series, or in parallel.
We begin by treating the series combination. (Schematic of a battery with two resistors in series, R1 and R2.) We want to determine the value of a single resistor that is equivalent to the two individual resistors, that is, the total voltage drop and current for the single resistor must equal the total voltage drop and current for the pair of resistors. (Schematic of a battery with one equivalent resistor.)
The voltage from the battery must be shared between the resistors; the voltage drop across R1 plus the voltage drop across R2 will equal the battery voltage, V:
V = V1 + V2
The current through each resistor must be equal, I1 = I2.
This is due to conservation of charge -- if the currents are not equal, then the amount of charge arriving at R2, I1Dt, will not equal the amount of charge leaving R2, I2Dt.
The result is that R2 will have a net overall charge (and presumably R1 will have an equal but opposite charge).
But the components of a circuit have no net charge, so this simply cannot occur.
According to Ohm's law, the voltage drops across the resistors is:
V1 = IR1, and V2 = IR2.
Using these relations, the voltage across the battery is:
V = IR1 + IR2 = I(R1 + R2)
We want an equivalent resistance that gives the same current for the given voltage: V = IReq, so we see
Consider a real battery connected to a resistance R.
Let's model the battery as an ideal voltage source, E, with a small series resistor, r.
(Here we use E for voltage to emphasize that this is the ideal voltage source inside a real battery.)
The current in this circuit is given by
I = E/(R + r)
The voltage across the battery terminals equals the voltage that appears across the resistor, R
V = IR = E - Ir
where I get the second form by considering the battery plus its internal resistance.
The effective battery voltage is less than the internal voltage, and decreases as the current it supplies increases.
Batteries give the largest voltage when they are supplying small currents, and if the terminals are shorted, the current from the battery is limited.
This process of modeling real devices by combinations of ideal circuit elements is common in electronics.
A 4.0W resistor, and 8.0W resistor, and a 12W resistor are connected in series with a 24V battery. What are (a) the equivalent resistance and (b) the current in each resistor?
(a) Since the resistors are in series, the equivalent resistance is
Req = (4.0 + 8.0 + 12.0)W = 24.0W.
They are in series, so the same current passes through each, and it equals I = V/Req = 24V/24W = 1A.
Now consider what happens when two resistors are in parallel.
(Draw schematic of a circuit with a battery and two resistors in parallel, also draw circuit with battery and equivalent resistance.)
In this case, the voltage across the resistors is the same and in this case equals the battery voltage, V.
The current from the battery is shared between the resistors, I1 = V/R1 through resistor R1, and I2 = V/R2 through R2.
The total current from the battery equals the sum of the resistor currents:
I = I1 + I2 = V/R1 + V/R2
The same current must flow through the equivalent resistance, I = V/Req.
Equating the currents give V/Req = V/R1 + V/R2, or cancelling the common factor V:
The resistors in P18.2 are connected in parallel across a 24V battery. Find (a) the equivalent resistance and (b) the current in each resistor.
(a) Since the resistances are in parallel, the inverse of the equivalent resistance is the sum of the inverses:
1/Req = 1/4.0 + 1/8.0 + 1/12.0 = (6 + 3 + 2)/24W = 11/24W
Req = (24/11)W = 2.2W
(b) Resistances in parallel have the same voltage across them, but different currents.
We have to solve for each current individually:
I1 = V/R1 = 24V/4.0W = 6.0A
I2 = 24V/8.0W = 3.0A
I3 = 24V/12.0W = 2.0A
The resistance between terminals a and b in Figure P18.13 is 75W. If the resistors labeled R have the same value, determine R.
(Draw diagram.) The resistors R are part of the resistor combination, so if we write out the expression for the equivalent resistance, then we can relate it to R and solve for R. This circuit is a combination of resistors in series and parallel, proceed step by step to find the equivalent resistance. We want to find the "deepest nesting" of resistors, and beginning there, work our way out to the whole circuit.
Begin by combining the series combination of R and the 5.0W resistor, Req1 = R + 5.0W.
Next we can find the equivalent resistor for the parallel combination of the 120, 40, and Req1 resistors:
1/Req2 = 1/120 + 1/40 + 1/(R+5) = (1+3)/120 + 1/(R+5) = 1/30 + 1/(R+5) = (R + 5 + 30)/(30(R+5)) = (R+35)/(30(R+5))
Req2 = (30R+150)/(R+35)
Finally, we sum that equivalent resistance with the second resistor, R, in series with it (the one next to terminal a), and set the result equal to 75W:
Req = 75W = R + Req2 = R + (30R+150)/(R+35) = (R² + 35R + 30R + 150)/(R + 35) = (R² + 65R + 150)/(R + 35)
Therefore, 75 = (R² + 65R + 150)/(R + 35).
To find R, multiply both sides by (R + 35):
75(R + 35) = 75R +2625 = R² + 65R + 150
R² - 10R - 2475 = 0
This quadratic equation has two possible solutions:
R = [10 ± Sqrt{10² + 4(2475)}]/2 = [10 ± Sqrt{10000}]/2 = [10 ± 100]/2 = 5 ± 50 = 55W or -45W
Resistances are positive, therefore the only logical answer is R = 55W.