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Chapter Review

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In this chapter we study direct current circuits. The important concepts of Ohm's law and Kirchhoff's rules are covered. Capacitors as circuit elements are also introduced.

21-1   Electric Current

The flow of electric charge constitutes an electric current (I). The current is determined by the net magnitude of charge DQ that flows past a point in time Dt, specifically,

.

As the above definition suggests, the unit of current, called an ampere (A), can be written in terms of charge as 1 A = 1 C/s. However, for technical reasons the ampere is considered to be the fundamental SI unit. From now on, we will adopt this officially correct practice and take the fundamental dimension used for electricity to be [A] representing current.

In this chapter we consider currents that flow through closed paths called electric circuits. Since the flow of current that we study here will be in one direction around the circuit, these are called direct current circuits. A current flows in a circuit when energy is supplied by an electric battery. Batteries have two ends called terminals across which a potential difference exists. This potential difference is called the electromotive force (E), or emf, of the battery. The direction of the flow of current in a circuit is always taken to be the direction in which a positive charge would move. This last statement is true even if the actual charge that flows is negative as is the case in almost all electrical devices.

Practice Quiz

If a current of 2.5 A flows through a wire for 3.0 seconds, how much charge passes through the wire? 0.5 C 7.5 C 5.5 C 1.2 C 0.83 C

21-2   Resistance and Ohm's Law

Typically, current does not flow through a circuit unimpeded. The wires through which the current flows offers some resistance to this flow. The more resistance (R) in the wire, the smaller the current (I) that will flow for a given potential difference (V) across the wire. The relationship between these three quantities is referred to as Ohm's Law:

V = IR.

The resistance R has SI units of V/A; this unit is called an ohm (W): 1 W = 1 V/A.

The amount of resistance in a particular piece of wire depends on the size and shape of the wire as well as the type of material out of which the wire is made. Resistance increases in direct proportion to the length L of the wire and decreases inversely proportional to the cross-sectional area A of the wire. The dependence on the type of material is contained within a measured quantity called the resistivity . Putting this information together gives

.

The resistivity of a material has SI units of .

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Example 21.1   Flashlight:   Assume that the filament in the light bulb of a flashlight is made of a small piece of metal alloy. If this piece of metal has a radius of 0.500 mm, a length of 1.00 cm, and resistivity of 4.39 x 10^-4 , what current runs through this filament if the flashlight operates on two 1.50 volt batteries?

Picture the Problem   The picture shows the circuit for the flashlight.

Strategy   The current can be determined from Ohm's law once the resistance is known. The resistance can be calculated from the given information.

Solution

1. The resistance can be determined from the resistivity:
2. The net potential difference across the resistor is:	E = E1 + E2
3. By Ohm's law we have:
4. The numerical result is:

Insight   Notice that the two batteries oriented with the same polarity provides an emf equal to the sum of the individual emfs.

Physlet Illustration: Ohm's Law
V=  volts R=  W
In this simulation, a resistor is connected to a battery. The voltmeter measures the potential difference across the resistor, and the ammeters measure the currents at two different places in the circuit.  Placing your mouse cursor on top of any circuit element displays the relevant information about that element. Vary the values of the battery voltage (1 volt < V < 50 volts) and resistance (1 W < R < 100 W). Can you verify Ohm's Law?
Hints: How does the voltage across the resistor compare to the battery emf? How does the current that enters the resistor at the top compare to the current that exits the resistor at the bottom? If you double the battery voltage, what happens to the current flowing? If you double the resistance, what happens to the current flowing? Does this verify Ohm's law?

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Practice Quiz

A potential difference V is applied across a resistor of resistance R producing a current I through the resistor. If a potential difference of 2V is applied across the same resistor, what current will flow through it? I/4 I/2 I 2I 4I

A potential difference V is applied across a resistor of resistance R producing a current I through the resistor. If the resistor is replaced by one that has a resistance of 2R, what current will flow through it? I/4 I/2 I 2I 4I

A potential difference V is applied across a resistor, made from a copper wire of length L and radius r, producing a current I through the resistor. If the resistor is replaced by another copper wire of equal length and raduis 2r, what current will flow through it? I/4 I/2 I 2I 4I

21-3   Energy and Power in Electric Circuits

From the definition of electric potential (potential energy per unit charge) we know that energy is transferred when an amount of charge moves through a potential difference. Batteries transfer this energy to the charge and resistors dissipate this energy in the form of heat. The rate at which this energy transfer takes place is the power that is either generated or dissipated by a device in the circuit. In general, when a current I flows as a result of a potential difference V the electrical power used is given by

P = IV,

where you should recall the SI unit of power is the watt, . When this power is being dissipated through a resistance R, we can apply Ohm's law (V = IR) to find other useful relationships for the power. Specifically, it is straightforward to show that

P = IV = I²R = V²/R.

Since power is energy divided by time, then power times time will give energy. A unit of energy that is commonly used to track the household use of electricity is the kilowatt-hour (kWh). This quantity equals the amount of energy used if 1 kW of power is consumed for one hour. Its equivalent energy in joules is the following:

1 kWh = 3.6 x 10⁶ J.

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Example 21.2   Power:   (a) Determine the power requirements of the flashlight in Example 21.1. (b) If the flashlight is left on for a half-hour, how much energy in kilowatt-hours was used?

Picture the Problem   The picture shows the circuit for the bulb in the flashlight.

Strategy   For part (a) we can use the result of Example 21.1 to obtain the power. For part (b) the energy can be found as the power times the time.

Solution
Part (a)

1. Given the current from Example 21.1 the power is:	P = IV = (0.5367 A)(3.00 V) = 1.61 W
Part (b)
1. The energy in kilowatt-hours is:	E = Pt = (0.001610 kW)[0.500 h] = 8.05 x 10-4 kWh

Insight   In part (b) I converted the power to kilowatts and used the time in hours to directly get the result in the desired units.

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Practice Quiz

What is the resistance of a device that consumes 100 W of power when connected to a 120 V source? 144 W 12,000 W 1.2 W 0.83 W 20 W

How much energy is consumed if a 75 W light bulb is left on for 12 hours? 900 kWh 6.3 kWh 0.90 kWh 160 kWh 0.16 kWh

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