Download - Chapter 28
Chapter 28
Direct Current Circuits
Introduction
• In this chapter we will look at simple circuits powered by devices that create a constant potential difference across their terminals.
• Most of our circuits will exist in a “Steady State” meaning a constant current in the same direction.
• These are known as “Direct Current” or DC circuits.
28.1 Electromotive Force
• Batteries are often referred to as a source of emf or electromotive force.
• The emf, ε, of a battery is the maximum possible voltage that the battery can provide across its terminals.
• Its not really a force– Think of it as a charge pump.
28.1
• When connected in a circuit, the terminal voltage of a real battery is typically less than its emf.
• This is due to the internal resistance (r) found in a real battery. – Ideal Batteries have zero internal resistance.
IrV
28.1
• Look at a simple circuit, battery and resistor.
28.1
• The current created will equal the terminal voltage divided by the load resistance (R) from Ohm’s Law.
• To find the emf of the battery we need to realize that some of the potential is dropped by the internal resistance and the rest is dropped by the load resistance.
R
VI
28.1
28.1
• The emf is given as
• The current in the circuit is therefore
• And the total power output of the battery is
IrIR
rRI
rIRIP 22 IP
28.1
• Quick Quiz p 860• Examples 28.1 and 28.2
28.2 Resistors in Series and Parallel
• For a series combination of two resistors– The current through each resistor is the same
(All charges must pass through both resistors)
28.2
– The potential difference across both resistors is given as
– So the equivalent resistance for series combinations is given as
21 IRIRV
21 RRIV
...321 RRRReq
21 VVV
28.2
• Quick Quizzes p. 863
28.2
• For a parallel combination of two resistors– The current can split at a junction.– The Sum of currents leaving the junction, will
equal the total entering.
21 III
28.2
– Since both resistors are connected directly across the terminals of the battery, the potential difference across each resistor is the same.
21 VVV
28.2
– To find the equivalent resistance…
21 III
21 R
V
R
V
R
V
eq
21
111
RRReq
28.2
• To Recap Series Combinations:
• To Recap Parallel Combinations
...321 IIII
...321 VVVV
...321 RRRReq
...321 VVVV
...321 IIII ...1111
321 RRRReq
28.2
• Quick Quizzes p. 865• Examples 28.3-28.7
28.3 Kirchoff’s Rules
• Simple Circuits can be analyzed using – Ohm’s Law– Series/Parallel Rules
• If the circuit cannot be reduced to a single loop, we will use Kirchoff’s Rules to analyze.– Junction Rule– Loop Rule
28.3
• Junction Rule- The sum of currents entering any junction in a circuit must equal the sum of currents leaving that junction.
outin II
321 III
28.3
• Loop Rule- The sum of the potential differences across all elements around any closed circuit loop must be zero.
0loop
V
28.3
• When “traveling a loop” the potential difference for resistor elements is –IR with the current, +IR against the current.
28.3
• When “traveling a loop” the potential difference for emf sources is +ε when moving through – to +, -ε when moving from + to –.
28.3
• General Rules– Draw a circuit diagram and label the current with
direction in each branch.– An Equation is needed for each unknown quantity,
usually one for each unknown current.– Series Resistors have the same current.– If a current is solved to be negative, the magnitude
is correct, but direction is opposite.– Charge capacitors act as open branches and current
= 0.
28.3
• Quick Quiz p. 871• Examples 28.8-28.10
28.4 RC Circuits
• One type of DC circuit includes both Resistors and Capacitors.
• An RC circuit will have a current that varies over time.
28.4
• When the switch closes, current will flow charging the capacitor.
• As the capacitor reaches maximum charge, the current approaches zero.
• At any time, the Kirchoff’s Loop rule applies as follows
0 IRC
q
28.4
• The instant the switch is closed, the charge q =0, and the initial current at t = 0 is
• As q increases, I decreases until the capacitor is at its maximum charge and I = 0.
RIo
CQ
28.4
• To develop an expression to analyze the circuit as a function of time, we will begin with an expression for instantaneous currrent (I = dq/dt).
0 IRC
q
RC
q
Rdt
dq
28.4
• We then find a common denominator
• When then combine like terms.
RC
q
RC
C
dt
dq
RC
Cq
dt
dq
dtRC
dqCq
11
28.4
• We then integrate both sides.
• Remember that q = 0 at t = zero.
tqdt
RCdq
Cq 00
11
tq
o RC
tCq
0
)ln(
RC
t
C
Cq
ln
28.4
• Solving for q(t)
• RC is the time constant,τ, to reach 63.2% of Q
RCt
eCtq 1
RCt
eQtq 1
28.4
• We can differentiate this expression with respect to time to get the current expression.
RCt
eCCtq
RCt
eRdt
dqtI
)(
28.4
• Discharging the Capacitor– If we remove the battery and discharge the stored
energy through a resistor.
RCt
Qetq
RCt
eRC
QtI
28.4
• Energy Stored in the Capacitor– We know
– So for a charging RC circuit
– For a discharging RC circuit
C
qU
2
2
21 RCt
eUU o
RCt
eUU o
2
28.4
• Quick Quizzes p 876-77• Examples 28.11-28.14
28.5 Electrical Meters28.6 Household Safety
• Read 879-882– Electrical Meters• Galvanometer• Ammeter• Voltmeter
– Household Safety• Typical Wiring (Live/Neutral)• Shocking/Fatal Currents• 3rd Ground Wire• GFI’s