CHAPTER 26: DIRECT CURRENT CIRCUITS
• In this chapter, we will learn how to analyze more complicated circuits. Our
focus will specifically be on direct current (dc) circuits, where the direction
of the current is constant in time. Any circuit whose emf is generated by a
battery constitutes a dc circuit.
• Circuits in which the direction of the current oscillates periodically in time
are called alternating current (ac) circuits. All household wiring supply
electrical energy using ac circuits.
Resistors Connected in Series and in Parallel
• Imagine that you are going to use a single battery to light two lightbulbs
(lightbulbs are essentially resistors). There are two possible ways that you
can connect the bulbs – “in series” or “in parallel”
• We say two resistors are connected in series if they are connected directly
to each other with no junction in between
• We say two resistors are connected in parallel if they share two common
junctions; another way of saying this is if two resistors are connected at
both ends, they are in parallel
• In Chapter 24, we learned how to compute the equivalent capacitance for
capacitors in different combinations of series and parallel connections – we
will do the same for resistors using the same 2 principles, conservation of
charge and the fact that the work done by a conservative force between 2
points does not depend on the path taken between the points
Series
• Because there are no junctions (nowhere else for the current to go) and
because of conservation of charge, the same amount of current flows into
one end of a resistor as flows out the other end.
• The current through circuit elements in series is the same.
• The potential difference across each resistor will not be the same (unless
they all have equal resistances)
• Using Ohm’s law we can calculate the potential drop across each resistor
o The potential difference between the points a and x is the potential
drop across R1: ΔV1 = IR1
o The potential difference between the points x and y is the potential
drop across R2: ΔV2 = IR2
o The potential difference between the points y and b is the potential
drop across R3: ΔV3 = IR3
• The potential drop across the entire 3-resistor combination is the sum of the
potential drops across the individual resistors: |ΔVab| = |ΔV1 + ΔV2 + ΔV3| =
I(R1 + R2 + R3)
• Thus, |ΔVab|/I = R1 + R2 + R3. Since the left hand side of the equation is the
potential difference across all 3 resistors divided by the current thru all 3
resistors, it is by definition the equivalent resistance of the 3-resistor series
combination
• For N resistors connected in series, the equivalent resistance is the sum
of the individual resistances: Req = R1 + R2 + R3 + … + RN
• The equivalent resistance is therefore always larger than any individual
resistance
Parallel
• The potential difference between points a and b in the figure above is the
same regardless of the path used to compute it (since the only force doing
work is conservative – work is path-independent). Thus, the 3 resistors have
the same potential difference across them (ΔVab).
• The potential difference across circuit elements in parallel is the same
• The current thru each resistor will not be the same (unless they all have
equal resistances)
• Using Ohm’s law we can calculate the current thru each resistor
o The current flowing thru R1: I1 = ΔVab/R1
o The current flowing thru R2: I2 = ΔVab/R2
o The current flowing thru R3: I3 = ΔVab/R3
• Because of conservation of charge, the current flowing into point a must
equal the sum of the currents in the individual resistors: I = I1 + I2 + I3 =
|ΔVab|(1/R1 + 1/R2 + 1/R3)
• Thus, I/|ΔVab| = 1/R1 + 1/R2 + 1/R3. Since the left hand side of the equation
is the total current divided by the potential difference across all 3 resistors,
it is by definition the reciprocal of the equivalent resistance of the 3-resistor
parallel combination
• For N resistors connected in parallel, the reciprocal of the equivalent
resistance is the sum of the reciprocal of the individual resistances:
1/Req = 1/R1 + 1/R2 + 1/R3 + … + 1/RN
• The equivalent resistance is therefore always less than any individual
resistance
• Note that for the 3-resistor combination above, I1R1 = I2R2 = I3R3. So the
larger the resistance, the smaller the current thru that resistor. In other
words, more current flows thru the path of least resistance.
• Note that the result we obtained for the equivalent resistance of resistors in
parallel is the same as what we obtained for the equivalent capacitance of
capacitors in series and vice-versa (see Chapter 24)
**SEE EXAMPLE #1**
Kirchhoff’s Rules
• A junction (sometimes called a node) is a point where three or more
conductors meet, usually where a wire branches
• Kirchhoff’s junction rule: The amount of current flowing into any
junction (positive I) equals the amount of current flowing out of that
junction (negative I). ∑ 𝐼𝐼𝑖𝑖𝑖𝑖 = ∑ 𝐼𝐼𝑜𝑜𝑜𝑜𝑜𝑜 . Or, using the sign convention for
currents at a junction, ∑ 𝐼𝐼 = 0
• This rule is a consequence of conservation of charge (current)
• A loop is any closed conducting path
• Kirchhoff’s loop rule: The sum of the potential differences around any
closed loop equals zero. ∑ ∆𝑉𝑉 = 0
• This rule is a consequence of the fact that the force doing work (the electric
force) on the moving charges is conservative and so ΔV = ΔU/q = 0 on a
closed loop
• In Figure (a) below, points a and b are junctions but c and d are not. Also
shown are the three possible loops to which the loop rule can be applied.
• In Figure (b) below, points a, b, c, and d are junctions but e and f are not.
Also shown are four possible loops to which the loop rule can be applied.
Sign Conventions for the Loop Rule
• First you need to choose a direction for the current in each branch of the
circuit
• Next you apply the loop rule to as many loops as you need to solve the
problem (such as the ones shown in Figures (a) and (b) above). Each loop
corresponds to an equation – you need as many equations as the number of
unknowns.
• Below are the sign conventions for applying the loop rule. The arrow
marked “Travel” is the direction you are traversing a closed loop when
adding up voltages. In all 4 pictures, going from lower to higher potential, as
you go around the loop, is a positive potential difference and going from
higher to lower potential is a negative potential difference. In complicated
circuits it is a good idea to label the terminals on each circuit element with a
+/– sign to indicate whether it’s potential increase or decrease in the
direction of the current flow that you’ve chosen.
**SEE EXAMPLE #2**
**SEE EXAMPLE #3**
Electrical Measuring Instruments
• An ammeter is a device that measures the current through a circuit element
• Because charge flows through circuit elements, an ammeter must be placed
in series with the circuit element through which current is to be measured
• To determine the current in the element, we insert the ammeter in the circuit
just before where the current flows into the element. To do so, we must
break the connection between two elements. Because they are in series, the
ammeter and the element which you are trying to measure have the same
current flowing thru them.
• The resistance of an ideal ammeter is zero so that it can measure the
current without changing it. If its resistance is not zero, or at least negligible,
then the current would be less than what it would be without the presence of
the ammeter.
• A voltmeter is used to measure the potential differences in a circuit
• Because the potential difference is measured across a circuit element, a
voltmeter is connected in parallel with the circuit element whose potential
difference is to be measured
• An ideal voltmeter has infinite resistance so that it can measure the
voltage without changing it
• Because it is in parallel with the resistor, the voltmeter’s resistance must be
very large so that it draws very little current. If it drew a significant amount
of current then less current goes thru the resistor and the potential drop
across it is less than what it would be without the presence of the voltmeter.
• A voltmeter and ammeter can be used in combination to measure resistance
and power in a circuit. Resistance is the current flowing thru an element
divided by the potential difference between its terminals. The rate at which
energy is dissipated in this element is the product of the current flowing thru
it and the potential difference between its terminals.
• Ideally, one would like to measure the current and potential simultaneously
in order to get the resistance, but in practice this is quite difficult
• In Figure (a) above the ammeter correctly measures the current that flows
thru the resistor, however the voltmeter measures the sum of the potential
difference ΔVab = IR across the resistor and the potential difference ΔVbc =
IRA across the ammeter.
• If we instead move one of the terminals of the voltmeter from c to b, then it
will correctly measure the potential difference across the resistor. But now
the ammeter measures the sum of the current in the resistor I and the current
IV in the voltmeter.
**SEE EXAMPLES #4 and #5**
R-C Circuits
• So far in studying circuits we have assumed that the emfs and currents are
steady (constant in time). However, when a capacitor is in the process of
charging or discharging, these quantities do become time-dependent.
• An R-C circuit is a circuit with a resistor and a capacitor (and possibly an
emf source) connected in series
Charging
• The Figures below shows one such R-C circuit that will charge the capacitor.
Here we assume that the battery is an ideal voltage source (no internal
resistance) and also that the wires are ideal conductors (no potential
difference between any 2 points where there is no circuit element).
• Initially the charge on the capacitor is q = 0. I will denote the potential
difference across the resistor as ΔVR and the potential difference across the
capacitor as ΔVC. Since the elements are all connected the current is
everywhere the same.
• At t = 0, the switch is closed, which connects the left end of the resistor to
the positive terminal of the battery. The charge on the capacitor cannot
change instantaneously and therefore neither can the potential difference
across it, so the full battery voltage appears across the resistor initially and
ℰ
the current thru it is 𝐼𝐼 = . The current cannot flow thru the capacitor but
𝑅𝑅
instead begins to pile up positive charge on the left plate and negative charge
on the right plate. In a very short time after the switch is closed the current in
the circuit is everywhere the same except in the insulated gap between the
plates of the capacitor.
• As charge accumulates on the capacitor plates, the potential difference
between them increases and is proportional to the increase in charge (ΔVC =
(1/C)*Q). As a result the potential difference across the resistor ΔVR
decreases since, because of the loop rule, their sum has to equal the emf of
the battery (ℰ = ΔVR + ΔVC). Since ΔVR = IR and R is constant, the current
thru the resistor must decrease as well. This in turn decreases both the rate at
which the charge on the plates is increasing and the rate at which ΔVC is
increasing as well. The voltage across the capacitor ΔVC continues to
increase and the current thru the resistor continues to decrease, but both do
so at an ever-decreasing rate.
• In summary, after the switch is closed at t = 0,
1) The voltage across the resistor ΔVR decreases at an ever-decreasing rate
2) The current I in the circuit also decreases at an ever-decreasing rate
3) The voltage across the capacitor ΔVC increases at an ever-decreasing rate
4) The current puts charge Q on the plates at an ever-decreasing rate
• After a long time, the potential difference across the capacitor plates ΔVC is
approximately the same as the emf of the battery ℰ. This means the potential
difference across the resistor (ΔVR) is approximately zero. Therefore, so are
the current through the resistor and the rate of charge buildup on the
capacitor. In the limit 𝑡𝑡 → ∞: I = 0, ΔVR = 0, and ΔVC = ℰ. The
asymptotic value of charge on the plates is Q = Cℰ.
• Let Q denote the charge on the capacitor and I the current in the circuit
which flows in the direction of the conventional current, from the positive
terminal of the battery, around the circuit to its negative terminal. ΔVR = IR
and ΔVC = Q/C and using the loop rule, we have ℰ – IR – Q/C = 0
ℰ
• Thus, the current in the circuit is 𝐼𝐼 = −
𝑅𝑅
ℰ
𝑄𝑄
𝑅𝑅𝑅𝑅
and at t = 0, Q = 0, so the initial
current is as claimed earlier on physical grounds. The capacitor is fully
𝑅𝑅
charged when I = 0 – the amount of "full charge" is Cℰ.
• Differentiating the above equation for current and using (I = dQ/dt) gives:
𝑑𝑑𝐼𝐼
𝑑𝑑𝑑𝑑
=−
1 𝑑𝑑𝑑𝑑
𝑅𝑅𝑅𝑅 𝑑𝑑𝑑𝑑
=−
1
𝑅𝑅𝐶𝐶
𝐼𝐼
𝐼𝐼 𝑑𝑑𝑑𝑑
• Separate variables and integrate both sides: ∫𝐼𝐼
𝐼𝐼
𝑡𝑡
0
𝐼𝐼
=−
1
𝑡𝑡
∫ 𝑑𝑑𝑑𝑑
𝑅𝑅𝑅𝑅 0
• So, ln � � = − . Exponentiate both sides (to invert the logarithm and
𝐼𝐼
𝑅𝑅𝑅𝑅
0
ℰ
solve for I) and we have 𝐼𝐼(𝑡𝑡) = 𝐼𝐼0 𝑒𝑒 −𝑡𝑡/𝑅𝑅𝑅𝑅 = 𝑒𝑒 −𝑡𝑡/𝑅𝑅𝑅𝑅 since 𝐼𝐼0 = 𝐼𝐼(0) =
−𝑡𝑡/𝑅𝑅𝑅𝑅
𝑅𝑅
ℰ
𝑅𝑅
• ∆𝑉𝑉𝑅𝑅 = 𝐼𝐼𝐼𝐼 = ℰ𝑒𝑒
and at t = 0 the potential difference across the resistor
equals the emf of the battery as claimed
• ℰ = ΔVR + ΔVC, so ∆𝑉𝑉𝐶𝐶 = ℰ − ∆𝑉𝑉𝑅𝑅 = ℰ(1 − 𝑒𝑒 −𝑡𝑡/𝑅𝑅𝑅𝑅 )
• Finally, using the definition of capacitance, we can get the charge as
function of time 𝑄𝑄 = 𝐶𝐶∆𝑉𝑉𝐶𝐶 = 𝐶𝐶ℰ(1 − 𝑒𝑒 −𝑡𝑡/𝑅𝑅𝑅𝑅 ) and
• Note that Q(0) = 0 and approaches 𝐶𝐶ℰ as 𝑡𝑡 → ∞ as claimed
ℰ
• Also, I(0) = and approaches 0 as 𝑡𝑡 → ∞ as it should
𝑅𝑅
• See Figures (a) and (b) below on the left
• The quantity τ = RC is called the time constant of the R-C circuit – it has
units of seconds and is the time it takes for the exponential term to fall to 1/e
of its initial value. The smaller τ is, the faster the charge builds up on the
capacitor plates (and the faster the current goes to 0)
• Finally, consider the instantaneous rate of energy transfer in the circuit
(power): P = IV = I2R = V2/R. The battery delivers energy to the circuit at a
rate 𝐼𝐼ℰ, electrical energy is dissipated in the resistor at a rate I2R, and
electrical energy is stored in the capacitor at rate IQ/C. Multiplying the loop
equation by current gives 𝐼𝐼ℰ = I2R + IQ/C. Physically, this means that the
battery supplies power to the circuit and part of it gets dissipated in the
resistor and the rest is stored in the capacitor.
• The total energy supplied by the battery during the charging of the capacitor
is the total charge delivered (𝐶𝐶ℰ) times the potential difference across the
battery (ℰ), or ℰ 2 𝐶𝐶. The total energy stored in the capacitor is ½QΔV (see
1
1
Chapter 24), or (𝐶𝐶ℰ)(ℰ) = ℰ 2 𝐶𝐶, exactly half the energy delivered by the
2
2
battery. The remaining half is dissipated in the resistor.
Discharging
• The Figures below show another R-C circuit except this time we start with a
fully charged capacitor (Q(0) = Q0) and connect it in series with a resistor
(no battery – the charged capacitor is now the source of emf that drives the
current)
• The capacitor will then discharge thru the resistor and as 𝑡𝑡 → ∞, 𝑄𝑄 → 0
• In Figure (b) below, the current will actually flow in the opposite direction
shown. The potential difference established by the capacitor will cause
positive charges to be pushed from the positive plate, around the circuit in a
CW direction, and onto the negative plate
• Applying the loop rule in a clockwise direction gives Q/C – IR = 0
• Again taking the derivative of the loop equation wrt time and using the
definition of current (I = dQ/dt), we end up with the same expression for
dI/dt as we did for the charging capacitor:
𝑑𝑑𝐼𝐼
𝑑𝑑𝑑𝑑
=−
1
𝑅𝑅𝑅𝑅
𝐼𝐼 except now the initial
current is different. From the loop equation, I(0) = I0 = Q0/RC
• As before, separate variables and integrate… 𝐼𝐼(𝑡𝑡) = 𝐼𝐼0 𝑒𝑒 −𝑡𝑡/𝑅𝑅𝑅𝑅 =
𝑄𝑄0
𝑅𝑅𝑅𝑅
𝑒𝑒 −𝑡𝑡/𝑅𝑅𝑅𝑅
• With no battery, the magnitudes of the potential differences across the two
components are equal: ∆𝑉𝑉𝐶𝐶 = ∆𝑉𝑉𝑅𝑅 = 𝐼𝐼𝐼𝐼 =
𝑄𝑄0
𝐶𝐶
𝑒𝑒 −𝑡𝑡/𝑅𝑅𝑅𝑅 and since 𝑄𝑄 = 𝐶𝐶∆𝑉𝑉𝐶𝐶 ,
𝑄𝑄 = 𝑄𝑄0 𝑒𝑒 −𝑡𝑡/𝑅𝑅𝑅𝑅
• Note that Q(0) = Q0 and approaches 0 as 𝑡𝑡 → ∞ as it should for a
discharging capacitor
• Also, I(0) =
𝑄𝑄0
𝑅𝑅𝑅𝑅
and approaches 0 as 𝑡𝑡 → ∞ as it should
• See Figures (a) and (b) on the right side of the page above. The plot of
current vs time is negative because the book chose the wrong direction for
the current – presumably this was intentional to illustrate the point that if
you choose a direction for I but it is in reality flowing in the opposite
direction, you will get a negative result for I.
**SEE EXAMPLE #6**
**SEE EXAMPLE #7**