Chapter

Current and DC Circuits

Blinn College - Physics 1402 - Terry Honan

. - Current and Current Density

Basic Definitions

If Δ Q is the charge that passes through some surface, usually a cross-section of a wire, in the small time Δ t then the current I is defined by

I=((Δ Q)/(Δ t))

Units:  The SI unit for Current is: ampere=(QuantityUnit)=(QuantityUnit)/(QuantityUnit)

Example . - Battery on a Notebook Computer

Solution

({{Q=6.41(QuantityUnit)·(QuantityUnit)=6.41(QuantityUnit)×3600(QuantityUnit)=23000(QuantityUnit)}})

Q=n e   ⟹   n=(Q/e)=(Q/(1.60×(10^(-19))(QuantityUnit)))=1.44×(10^23)

Solution

({{Δ U}})=q Δ V=Q V=461000(QuantityUnit)

. - Resistance

Ohm's  Law

In electrostatics, currents are not allowed. We saw that the electrostatic electric field inside a conductor had to be zero and that implied that the voltage across a conductor also became zero. An electric field in a conductor or a voltage across one will necessarily produce a current.

The conductivity and resistivity are properties of a material.  For an object, like a wire, we can define a quantity called the resistance R by the macroscopic form of Ohm's law

V=I R

R=Resistance

Units:  The SI unit for Resistance is: ohm=(QuantityUnit)=(QuantityUnit)/(QuantityUnit)

V=I R relates the voltage across a resistor to the current through it.  When passing through the resistor in the direction of the current, it is a voltage drop, a decrease in potential. To measure the voltage across a resistor connect the leads of the voltmeter to either side of the resistor.  To measure the current through a resistor connect the ammeter in line with the resistor.

How does the resistance of a wire vary with its dimensions. Take the wire to be a long cylinder with length L and cross-sectional area A.

Double the length of a wire by putting identical wires end to end as shown in (a) below. Keep the current through the long wire the same; the voltage across each short segment will stay the same and the total voltage across the long wire will double. This shows its resistance, the voltage per current, doubles and that resistance is proportional to the length of a wire: R∝L. Next, model doubling the area by placing two wires side-by-side as in (b) below. If we keep the voltage across both the same the the current through each will stay the same and the total current will then double. Since resistance is voltage per current, the resistance is halved; the resistance is inversely proportional to the cross-sectional area: R∝1/A .

Combining these proportionalities we get: R∝L/A. We introduce a material-dependent constant of proportionality ρ, the resistivity. A good conductor has a small resistivity and a poor conductor has a resistivity with a high value.

R=((ρ L)/A).

Units:  The SI unit for resistivity is: (QuantityUnit)·(QuantityUnit)

({{Material, Resistivity-ρ ((QuantityUnit)(QuantityUnit))}, {Copper, 1.68×(10^(-8))}, {Aluminum, 2.65×(10^(-8))}, {Silver, 1.47×(10^(-8))}, {Gold, 2.22×(10^(-8))}, {Glass, (10^5)-(10^8)}})

Example . - Resistance of a Copper Wire

Solution

ρ=1.68×(10^(-8))(QuantityUnit)·(QuantityUnit) ,  L=750(QuantityUnit)  and  d=3×(10^(-3))(QuantityUnit) ⟹ A=π (r^2)=π ((((d/2)))^2)=7.0686×(10^(-6))(((QuantityUnit))^2)

R=((ρ L)/A)=1.78(QuantityUnit)

Solution

V=1.5(QuantityUnit)  and  V=I R  ⟹  I=(V/R)=0.841(QuantityUnit)

Variation of Resistance with Temperature

Resistance in a metal is caused by collisions between the moving electrons with the vibrating atoms.  If there were no vibration in the atoms there would be no collisions and the resistance would be zero.  As the temperature is increased the vibrational motion of the atoms increases and the collisions increase.  Because of this, resistance in metals increases with temperature.

As temperatures approach absolute zero, the vibrational motion of atoms approaches its minimum value consistent with quantum physics. We would expect the resistivity to go to zero as temperatures approach absolute zero. But what we observe in some materials is much more dramatic; below some low critical temperature (Subscript[T, C]), the resistivity abruptly goes to exactly zero. This is known as superconductivity. Superconductivity was first discovered in 1911 in a solid mercury wire at (Subscript[T, C])=4.2(QuantityUnit). In 1986 a new class of materials was discovered; these new materials have critical temperatures greater than 90(QuantityUnit).

Semiconductors have different temperature behavior than metals; the resistivity of pure semiconductors decreases with temperature. This is because, unlike metals, more charge carriers are available for conduction at higher temperatures.

. - Power and DC Voltage Sources

Power in General

Power is generally defined as the rate that work is done or, more generally, the rate that energy is used or provided.

𝒫=((Δ Energy)/(Δ t))

When a charge Q is moved across a potential difference ΔV the potential energy difference is Δ U=Q Δ V.  It follows that when an small charge Δ Q moves across a voltage of V the infinitesimal energy change is Δ U=V Δ Q.  Writing 𝒫=Δ U/Δ t and using I=Δ Q/Δ t gives

𝒫=V I.

Power Dissipated in a Resistor

Ohm's law V=I R relates the voltage drop across a resistor to the current through it. Using it we can write equivalent expressions for the power dissipated in a resistor.

𝒫=V I=(I^2)R=((V^2)/R)

The energy lost to resistance is dissipated as heat.  This is called Joule heating.

Terminal Voltage

Treat every DC voltage source as an ideal voltage source with EMF (electromotive force)  ℰ in series with its internal resistance r.  The voltage across the terminals (Subscript[V, t]) of the source is then

(Subscript[V, t])=ℰ-I r.

When there is no load,  I=0, the terminal voltage (Subscript[V, t]) is the same as the EMF ℰ.  With a load the terminal voltage drops.

Example . - Terminal Voltage

Solution

ℰ=1.486(QuantityUnit)

I=0.250(QuantityUnit)   and  (Subscript[V, t])=1.454(QuantityUnit)

(Subscript[V, t])=ℰ-I r   ⟹   r=((ℰ-(Subscript[V, t]))/I)=0.128(QuantityUnit)

Circuit Diagrams and Nodes

A real-world wire has resistance.  When we draw circuit diagrams we always consider the wires to be perfect conductors.  Since R=0, the voltage drop across a wire is zero.  A wire in a circuit diagram is a point of constant voltage; this is what we call a node.  The most effective way to analyze complex circuit diagrams is in terms of nodes and the circuit elements (voltage sources, resistors, capacitors, etc.) connected between nodes.

When it is necessary to consider the real-world resistance of a wire one can simply view it as an ideal conductor with a resistor with ρ ℓ/A of resistance placed in line.

Voltages in circuits are always differences.  If we choose some node to be zero voltage then we can assign a voltage to each node in a circuit.  A point of zero voltage in a circuit is called a ground.

. - Combinations of Resistors

Any combination of resistors with one wire in and one wire out can be reduced to its equivalent resistance.  If the combination were placed inside some black box then outside the box the combination would look like a single resistor, which we call its equivalent resistance.  For series and parallel resistor combinations, there are simple formulas for finding these equivalent resistances.

Series

Resistors are in series when all the current through one passes through the others; there is no branching between them.  The total voltage is the sum of the voltages.

I=(Subscript[I, 1])=(Subscript[I, 2])=…   and   V=(Subscript[V, 1])+(Subscript[V, 2])+…

Using V=I R gives I (Subscript[R, eq])=I (Subscript[R, 1])+I (Subscript[R, 2])+….  The equivalent resistance of series resistors is given by

(Subscript[R, eq])=(Subscript[R, 1])+(Subscript[R, 2])+….

Parallel

Resistors are in parallel when the voltage across the one is the same as the voltage across the others.  Resistors are in parallel when they are connected between the same two nodes, where a node is a point of constant voltage in a circuit.  The current branches and the total current is the sum of the currents.

V=(Subscript[V, 1])=(Subscript[V, 2])=…   and   I=(Subscript[I, 1])+(Subscript[I, 2])+…

Using I=V/R gives V/(Subscript[R, eq])=V/(Subscript[R, 1])+V/(Subscript[R, 2])+….  The equivalent resistance of series resistors is given by

(1/(Subscript[R, eq]))=(1/(Subscript[R, 1]))+(1/(Subscript[R, 2]))+….

Example . - Identifying Nodes

Solution

(Subscript[R, eq])=((((1/(Subscript[R, 1]))+(1/(Subscript[R, 2]))+(1/(Subscript[R, 3]))))^(-1))

Example . - A Resistor Circuit

Solution

(1/(Subscript[R, ∥]))=(1/(12(QuantityUnit)))+(1/(24(QuantityUnit)))=(1/(8(QuantityUnit)))   ⟹  (Subscript[R, ∥])=8(QuantityUnit)

(Subscript[R, eq])=4(QuantityUnit)+8(QuantityUnit)=12(QuantityUnit)

(Subscript[I, 4(QuantityUnit)])=(Subscript[I, battery])=((Subscript[V, battery])/(Subscript[R, eq]))=((36(QuantityUnit))/(12(QuantityUnit)))=3(QuantityUnit)

(Subscript[V, 4(QuantityUnit)])=(Subscript[I, 4(QuantityUnit)])4(QuantityUnit)=12(QuantityUnit)

(Subscript[𝒫, 4(QuantityUnit)])=(Subscript[I, 4(QuantityUnit)]) (Subscript[V, 4(QuantityUnit)])=36(QuantityUnit)

(Subscript[V, 12Ω])=(Subscript[V, 24Ω])=(Subscript[V, battery])-(Subscript[V, 4Ω])=36(QuantityUnit)-12(QuantityUnit)=24(QuantityUnit)

(Subscript[I, 12(QuantityUnit)])=((Subscript[V, 12(QuantityUnit)])/(12(QuantityUnit)))=((24(QuantityUnit))/(12(QuantityUnit)))=2(QuantityUnit)   and   (Subscript[I, 24(QuantityUnit)])=((Subscript[V, 24(QuantityUnit)])/(24(QuantityUnit)))=((24(QuantityUnit))/(24(QuantityUnit)))=1(QuantityUnit)

(Subscript[𝒫, 12(QuantityUnit)])=(Subscript[I, 12(QuantityUnit)])(Subscript[V, 12(QuantityUnit)])=48(QuantityUnit)   and   (Subscript[𝒫, 24(QuantityUnit)])=(Subscript[I, 24(QuantityUnit)])(Subscript[V, 24(QuantityUnit)])=48(QuantityUnit)

. - Combinations of Capacitors

As we saw for resistors, any network of capacitors can be reduced to an equivalent capacitance.  For capacitors its charge plays the role the current played in resistors.  (Recall that I=Δ Q/Δ t.)

The voltage to charge relation for a capacitor is

V=(Q/C).

Series

In the case of series resistors the charge on each capacitor is the same and both are the same as the charge on the equivalent.  The voltages add.

Q=(Subscript[Q, 1])=(Subscript[Q, 2])=…   and   V=(Subscript[V, 1])+(Subscript[V, 2])+…

Using the voltage to charge relation gives  Q/(Subscript[C, eq])=Q/(Subscript[C, 1])+Q/(Subscript[C, 2])+… which gives the expression for equivalent capacitance

(1/(Subscript[C, eq]))=(1/(Subscript[C, 1]))+(1/(Subscript[C, 2]))+….

Parallel

For parallel resistors the voltages are equal and the charges add.

V=(Subscript[V, 1])=(Subscript[V, 2])=…   and   Q=(Subscript[Q, 1])+(Subscript[Q, 2])+…

Using Q=C V gives (Subscript[C, eq])V=(Subscript[C, 1])V+(Subscript[C, 2])V+…  giving

(Subscript[C, eq])=(Subscript[C, 1])+(Subscript[C, 2])+… .

Note that the series and parallel formulas for capacitors are reversed relative to their resistor counterparts.

Example . - Equivalent Capacitance

Solution

3(QuantityUnit)+3(QuantityUnit)+9(QuantityUnit)=15(QuantityUnit)   and   2(QuantityUnit)+8(QuantityUnit)=10(QuantityUnit)

((((1/(15(QuantityUnit)))+(1/(10(QuantityUnit)))))^(-1))=6(QuantityUnit)

(Subscript[C, eq])=6(QuantityUnit)+4(QuantityUnit)=10(QuantityUnit)

. - Kirchhoff’s Rules

Kirchhoff's rules are used to solve for the currents in the case of a circuit involving many resistors and DC voltage sources.  A junction is a point in the circuit where three or more wires meet; if there are just two wires it is just a bend in the wire and not a junction.  For every branch in the circuit we can define a current.  Kirchhoff's rules gives a set of linear equations in the currents.  It is not essential to choose the proper direction for the currents, and in fact one typically doesn't know the current directions until a solution is found.  If the chosen current direction is wrong then that current will be negative when the solution is found.

Junction Rule

At every junction in a circuit the total current in is equal to the total current out.

∑(Subscript[I, in])=∑(Subscript[I, out]).

In every case (at least where the circuit is one connected piece) the junction rule equations will not be independent; there will always be one equation more than is needed.  Summing all the equations gives ∑I=∑I which is equivalent to 0=0.  (This is because every current leaves one junction and enters another.)  Because of this any one junction rule equation is the negative of the sum of the others.  To get an independent set of equations one must delete one (any one) of the equations.

Loop Rule

Around every closed loop in a circuit the sum of all the voltage gains is zero.

∑ ΔV=0

The sign conventions are:

When moving through a resistor in the direction of the current:  ΔV=-I R.
When moving through a resistor opposite the current:  ΔV=+I R.

When moving  through a DC source from - to + terminals:  ΔV=+ℰ.
When moving  through a DC source from + to - terminals:  ΔV=-ℰ.

To avoid non-independent equations consider only the smallest loops.

Example . - A Two-loop Circuit

Solution

(Subscript[I, 2])=(Subscript[I, 1])+(Subscript[I, 3])   (Junction 1)

(Subscript[I, 1])+(Subscript[I, 3])=(Subscript[I, 2])  (Junction 2)

0=8-30 (Subscript[I, 1])+10 (Subscript[I, 3])   (Loop 1)

0=6-10 (Subscript[I, 3])-20 (Subscript[I, 2])   (Loop 2)

0=6-10 (Subscript[I, 3])-20 ((Subscript[I, 1])+(Subscript[I, 3]))  ⟹  0=6-20(Subscript[I, 1])-30(Subscript[I, 3])  ⟹  20(Subscript[I, 1])+30(Subscript[I, 3])=6

30 (Subscript[I, 1])-10 (Subscript[I, 3])=8

({{3×(30 (Subscript[I, 1])-10 (Subscript[I, 3]))=8×3}, { 20(Subscript[I, 1])+30(Subscript[I, 3])=6}})

110(Subscript[I, 1])=30  ⟹ (Subscript[I, 1])=(30/110)(QuantityUnit)=0.27273(QuantityUnit)

30 (Subscript[I, 1])-10 (Subscript[I, 3])=8  ⟹  (Subscript[I, 3])=((30 (Subscript[I, 1])-8)/10)=0.01818(QuantityUnit)

(Subscript[I, 2])=(Subscript[I, 1])+(Subscript[I, 3])=0.29091(QuantityUnit)

(Subscript[I, 1])=0.273(QuantityUnit), (Subscript[I, 2])=0.291(QuantityUnit)  and   (Subscript[I, 3])=0.0182(QuantityUnit)

. - RC Circuits

When the switch is thrown to the Charging position the current flows from the battery to charge the capacitor.  In the Discharging position the charge flows from the capacitor and its energy is dissipated in the resistor.  The charge on  the capacitor is related to the current in the wire by

I=((Δ Q)/(Δ t)) for small Δ t

Note that when the capacitor is discharging the charge is decreasing and thus the current is negative.

Discharging

For the discharging case, applying Kirchhoff’s loop rule to the circuit gives:

0=I R+(Q/C).

Writing the current in terms of the charge gives.

0=R ((Δ Q)/(Δ t))+(1/C)Q   or   ((Δ Q)/(Δ t))=-(Q/τ)

where τ is defined as the time constant.

τ=R C

Note that τ is a time.

Take the initial charge to be (Subscript[Q, 0]). Solving the equation for t is a calculus problem. We will just write down the solution.

Q(t)=(Subscript[Q, 0]) (e^(-t/τ)).

This is an exponential decay. After on time constant τ the charge drops to 1/e of its initial value. A small time constant means a rapid decay and a large time constant means a slow decay.

The current is negative, since the charge is decreasing, and is given by

I(t)=-((Subscript[Q, 0])/τ) (e^(-t/τ)).

Note that with this exponential decay, the charge on a capacitor never goes fully to zero but after several time constants it becomes very small.

Charging

The loop rule for the charging case gives:

0=ℰ-I R-(Q/C)

and we can rewrite this as:

ℰ=R ((Δ Q)/(Δ t))+(1/C)Q

Using the same time constant τ we can write down the solution where we take the initial charge, at time zero, to be zero.

Q(t)=(Subscript[Q, ∞]) (1-(e^(-t/τ)))  where  (Subscript[Q, ∞])=ℰ C.

The current is now positive and is exponentially decaying.

I(t)=(ℰ/R) (e^(-t/τ))

Example . - An RC Circuit

Solution

c=50×(10^(-6))(QuantityUnit),  ({{ℰ=12(QuantityUnit)  and  R=300(QuantityUnit)}})

τ=R C=0.015(QuantityUnit)

Q(t)=ℰ C (1-(e^(-t/τ)))    ⟹  Q(20(QuantityUnit))=4.42×(10^(-4))(QuantityUnit)

I(t)=(ℰ/R) (e^(-t/τ))  ⟹   I(20(QuantityUnit))=0.0105(QuantityUnit)

Solution

Q(0)=(Subscript[Q, 0])=(Subscript[Q, ∞])=ℰ C=6.0×(10^(-4))(QuantityUnit)

I(0)=-((Subscript[Q, 0])/τ) (e^(-t/τ))=-((Subscript[Q, 0])/τ) (e^0)=0.040(QuantityUnit)

Solution

0.020(Subscript[Q, 0])=Q(t)=(Subscript[Q, 0]) (e^(-t/τ))   ⟹   0.020= (e^(-t/τ))

-t/τ=ln(0.020)=-3.91202   ⟹   t=3.91202 τ=0.587(QuantityUnit)