 An Introduction to DC Circuits Printer Friendly Version
The term DC means direct current. A DC circuit has current flowing in only one direction and is usually powered by a battery. The principle circuit elements are:

 name variable unit symbol source of emf ε volts V resistor R ohms Ω wires switch ammeter amperes A voltmeter volts V When a switch is closed in a DC circuit, charges in the wire do NOT instantaneously start moving at the speed of light. Closing the switch merely declares a direction of high to low potential - it establishes an electric field in the circuit. Consider the following analogy.
A crowd of people are milling around in the lobby prior to the beginning of a play. Some are standing still, some are circulating to visit friends or find refreshments, others are just randomly strolling to stretch their legs. Then, the bell rings to state that the audience has five minutes before the second half of the play is to begin. At that moment, the crowd now has a preferred direction of motion, out of the lobby (high potential) to their seats (low potential). No one person in the crowd instantaneously pops into his seat, everyone begins to drift in the "circuit's" prescribed direction of flow. Each person's drift velocity is small, although the instantaneous result of the bell is that seats start filling up in preparation for the curtain's rising.
A switch does the same thing as the bell, it tells the charges which direction they are to flow. No one charge moves at the speed of light, all charges begin drifting towards positions of lower potential thus causing a current in the circuit. The average drift velocity of charges in an electric circuit is on the order of a few centimeters per second. As soon as the switch is opened again, the electric field is removed, and the charges once again randomly move in all directions  through the wire canceling out any net flow.

Before practicing with circuits that have one or more resistors, we need to look at some fundamental relationships between volts, ohms, amperes, and watts.

Ohm's Law: V = IR

This law can apply to individual resistors, combinations of resistors or to an entire circuit. The variables in this formula stand for:

 potential difference V volts V = J/C current I amperes A = C/sec resistance R ohms Ω = Jsec/C2

Joule's Law: P = IV

This law relates the power dissipated through a resistor to its current and voltage. The variables in this formula stand for:

 power P watts watt = J/sec current I amperes A = C/sec potential difference V volts V = J/C

By substituting in variations of Ohm's Law, this formula can be rewritten as:

P = IV
P = I(IR) → P = I2R
P = (V/R)V → P = V2/R

These variations allow you flexibilty in how you calculate the power dissipated through a resistor.

Power, which is defined as the rate at which work is done or energy is used, is measured in watts [1 watt = 1 J/sec]. This quantity is CONSERVED in circuits; that is, the power supplied by the battery must be equal to the power consumed by all of the resistors in the circuit. Sometimes problems will ask you to calculate power by asking for "the rate at which heat [i.e. energy] is dissipated through a circuit element."

Rearranging this equation gives us another often used expression, Pt = Energy.

In mechanics, you were sometimes asked to calculate the power required for a constant force to move an object at a constant velocity; for example, the power required by an airplane's engine to maintain a plane's constant forward velocity or the power required to lift a mass on the end of a string at a constant rate. In this instance, power was calculated with the equation P = Fv.

Power = Work/time
Power = Fs/t = F(s/t) = Fv

Resistance of a wire: R = ρ(L/A)

This law relates the resistance of a wire to the material from which it is composed, its resistivity, its length, and its cross-sectional area. The variables in this formula stand for:

 resistance R ohms Ω = Jsec/C2 resistivity ρ rho ρ = Ωm length L meters m area A square meters m2

Details of the relationships in this formula will be discussed in a later lesson called "Filaments and Power."

Combinations of Resistors: Series Circuits

The formulas used to calculate the equivalent resistance, voltage and current through a collection of three resistors wired in series are:

 Req = R1 + R2 + R3 Veq = V1 + V2 + V3 Ieq = I1 = I2 = I3

Note that we will be considering the wires in our examples to have negligible resistance and will not be treating them as resistors; although in real circuits, the length, diameter, and resistivity of the conducting wires are critical components in a circuit.

Refer to the following information for the next three questions. What would be the equivalent resistance of the resistors shown above?

 When the 2A current flows through the resistors, what total voltage will be lost across this combination?

 How much power is dissipated by these resistors?

Suppose the ideal battery supplying the power for the resistors in the above example had an emf, or electromotive force, of ε = 40 volts. A graph of the voltage drops across the circuit could be illustrated as follows:  The battery is like a pump, it lifts the charges to a high potential so that they have the energy to flow through the resistors in the circuit. When the circuit is analyzed, the power supplied by the battery will equal the power consumed by the resistors.

R I V      P
5 Ω 2 A 10 V 20 watts
8 Ω 2 A 16 V 32 watts
7 Ω 2 A 14 V 28 watts
80 watts total power consumer
 P = Iε = (2)(40) = 80 watts
battery power supplied

Combinations of Resistors: Parallel Circuits

The formula used to calculate the equivalent current through a collection of resistors wired in parallel is:

Ieq = I1 + I2 + I3

This statement means that when the main line current reaches a parallel configuration of resistors, the current divides. Remember that the current divides proportionally with the larger ratio going along the "path of least resistance."

Refer to the following information for the next three questions. If R1 = 6 Ω and R2 = 6 Ω, how much current flows through each resistor?

 If R1 = 6 Ω and R2 = 12 Ω, how much current flows through each resistor now?

 If R1 = 18 Ω and R2 = 6 Ω, how much current flows through each resistor?

The formula to calculate the equivalent resistance of three resistors wired in parallel is

Req = [(1/R1 ) + (1/R2 ) + (1/R3)]-1

In parallel circuits, this equivalent resistance is always smaller than any of the original resistors. In a parallel circuit, each additional branch increases the size of the main line current that must be drawn from the battery. Thus, the equivalent resistance of the combination must get smaller. Counter-intuitive, yes? Conversely, in series, the equivalent resistance of one or more resistors is always larger than any of the original resistors.

The formula used to determine the equivalent voltage for resistors wired in parallel is:

Veq = V1 = V2 = V3

Remember that voltage, or potential, lost across parallel branches is equivalent since the charges lose the same amount of EPE regardless of which path they "decide" to cross. Using an analogy of ski slopes,
no matter which slope a skier chooses, his loss of PE is the same. Under general usage, more skiers will choose the less difficult slope [path of least resistance]; only the more advanced skiers will choose a more challenging slope [path of higher resistance].

Refer to the following information for the next three questions. If R1 = 18 Ω and R2 = 12 Ω, then what is their equivalent resistance?

 What is the total voltage lost between points X and Y?

 How much current flows through each resistor in the circuit shown above?

Suppose the ideal battery supplying the power for the resistors in the above example had an emf or electromotive force, of ε = 28.8 volts. Since R1 and R2 each occupy individual branches and are in parallel with the battery, each will lose the entire voltage supplied by the battery. A graph of the voltage drops across the circuit could be illustrated as follows:  The battery is like a pump, it lifts the charges to a high potential so that they have the energy to flow through the resistors in the circuit. When the circuit is analyzed, the power supplied by the battery will equal the power consumed by the resistors.

R I V      P
18 Ω 1.6 A 28.8 V 46.08 watts
12 Ω 2.4 A 28.8 V 69.12 watts
115.20 watts total power consumed
 P = Iε = (4)(28.8) = 115.20 watts
battery power supplied

Terminal Voltage of a Real Battery

The formula used to calculate the terminal potential difference of a discharging battery is

TPD = ε - Ir

where I is the main line current drawn through the battery, r is the battery's internal resistance, and ε is the battery's electromotive force, or ideal voltage, when it is not connected to a load and drawing current.

Remember that batteries supply energy through a chemical conversion. As deposits on the electrodes build up, the internal resistance of the battery increases. Eventually, the internal resistance gets to be so large, that the battery is no longer sufficient to supply the energy requirements of the circuit, and it is replaced. Note that a battery's terminal voltage varies with the resistance present in the external circuit which dictates the required currents.

To calculate the terminal potential difference of a charging battery, the formula is

TPD = ε + Ir

If current is being pushed into a battery, it is said to be charging. If current is being drawn from a battery, it is discharging, which is ALWAYS the case when only one battery is present in a circuit. In the diagrams above, the left battery is charging; the right one is discharging. Related Documents