Liberia Secondary School System (LSSS)
Grade 12, Physics 2nd Marking Period Lecture Note
Instructor: Mr. N. W. Keeta (0770772276/0886419872)
Topic: Direct Current
Sources of direct current
A current is kept in a closed circuit by a source of emf. Among such sources are any devices
(for example, batteries and generators) that increase the potential energy of the circulating
charges. A source of emf can be understood of as a “charge drive” that forces electrons to
move in a direction opposite the electrostatic field inside the source. The emf ‘ɛ’ of a source
is the work done per unit charge; the SI unit of emf is the volt.
Carriers of electric charge
Charge carriers can either be electrons, positive ions (cation), negative ions (anion) or holes.
The flow of a particular electric charge depends on the type of material medium.
Type of material
Charge carriers
metal
electrons (-)
gases
electrons and ions (+)
semi – conductor (like silicon or germanium) electrons (-) and holes (+)
liquid (salt and acid solution)
ions (+ and -)
The battery, represented by the dashed rectangle,
r
consists of a source of emf ɛ in series with an internal
b
a - +
●
●
resistance r. The terminal voltage of the battery,
∆𝑉 = 𝑉𝑏 − 𝑉𝑎
is therefore given by
I
Fig.2.1
I
∆𝑉 = ɛ − 𝐼𝑟 … 2.1
From this expression, we see that ɛ is equal to the
d
c
terminal voltage when the current is zero, called
●
●
the open-circuit voltage. By inspecting Figure 2.1,
R
we find that the terminal voltage ∆V must also equal
the potential difference across the external resistance R, often called the load resistance; that
is,
∆𝑉 = 𝐼𝑅
Combining this relationship with Equation 2.1, we arrive at
ɛ = 𝐼𝑅 + 𝐼𝑟 … 2.2
Solving for the current gives
ɛ
𝐼=
𝑅+𝑟
The preceding equation shows that the current in this simple circuit depends on both the
resistance external (outside) to the battery and the internal (inside) resistance of the battery.
If R is much greater than r, we can neglect r in our analysis (an option we usually select).
ɛ
Resistors in Series
When two or more resistors are connected end to end as in Figure 2.2, they are said to be in
series. The resistors could be simple devices, such as light bulbs or heating elements. When
two resistors R1 and R2 are connected to a battery as in Figure 2.2, the current is the same in
the two resistors because any charge that flows through R1 must also flow through R2.
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A battery made up of cells connected in series has the following characteristics:
(i) The emf of the battery is equal to the sum of the emf of the individual cells.
𝑉𝑒𝑞 = 𝑉1 + 𝑉2 + 𝑉3 + ⋯ + 𝑉𝑛
(ii) The current in each cell and in the external circuit has the same magnitude
throughout.
𝐼𝑇 = 𝐼1 = 𝐼1 = 𝐼3 …
(iii)The internal resistance of the battery is equal to the sum of the internal resistances of
the individual cells.
𝑅𝑒𝑞 = (𝑅1 + 𝑅2 ) … 2.3
Advantage
No exhaustion of cell can occur. Voltage strength increases.
𝐼𝑇 = 𝐼1 = 𝐼1 = 𝐼3 …
Because
the
potential
difference between a and b in
Figure 2.2b equals IR1 and the
potential difference between b
and c equals IR2, the potential
difference between a and c is
c
●
I
R1
b
●
∆V1
R2
∆V1
Fig. 2.2a
a a
● ●
Req = R1 + R2
I
Fig. 2.2b
∆V
∆V
- +
∆𝑉 = 𝐼𝑅1 + 𝐼𝑅2 = 𝐼(𝑅1 + 𝑅2 )
Regardless of how many resistors we have in series, the sum of the potential differences
across the resistors is equal to the total potential difference across the combination.
𝑉𝑒𝑞 = 𝑉1 + 𝑉2 + 𝑉3 + ⋯ + 𝑉𝑛
Applying Ohm’s law to this equivalent resistor, we have
∆𝑉 = 𝐼𝑅𝑒𝑞
Equating the preceding two expressions, we have
𝐼𝑅𝑒𝑞 = 𝐼(𝑅1 + 𝑅2 )
Or
𝑅𝑒𝑞 = (𝑅1 + 𝑅2 ) … 2.3
An extension of the preceding analysis shows that the equivalent resistance of three or more
resistors connected in series is
𝑅𝑒𝑞 = 𝑅1 + 𝑅2 + 𝑅3 … 2.4
Therefore, the equivalent resistance of a series combination of resistors is the algebraic
sum of the individual resistances and is always greater than any individual resistance.
Example: Four resistors are arranged as shown in Figure 2.3a. Find (a) the equivalent
resistance of the circuit and (b) the current in the circuit if the closed-circuit terminal voltage
of the battery is 6.0 V. (c) Calculate the electric potential at point A if the potential at the
positive terminal is 6.0 V. (d) Suppose the open circuit
2Ω
voltage, or emf ɛ, is 6.2 V. Calculate the battery’s
4Ω
5Ω
7Ω
internal resistance.
●
R3
Solution:
R1
R2
R4
(𝑎) 𝑅𝑒𝑞 = 𝑅1 + 𝑅2 + 𝑅3 + 𝑅4
A
Fig.2.3a
= 2Ω + 4Ω + 5Ω + 7Ω = 𝟏𝟖. 𝟎Ω
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6.0V
c
●
18.0Ω
∆𝑉
6.0
=
= 𝟎. 𝟑𝟑 𝑨
𝑅𝑒𝑞 18.0
(𝑐) 𝑉1 = 𝐼𝑅1 = (0.33)(2.0) = 0.66𝑉
𝑉𝐴 = ∆𝑉 − 𝑉1 = 6.0 − .66 = 𝟓. 𝟑𝟒𝑽
(𝑏) 𝐼 =
Fig. 2.3b
6.0 V
(𝑑) ∆𝑉 = ɛ + 𝐼𝑟
ɛ − ∆𝑉 6.2 − 6
0.2
𝑟=
=
=
= 𝟎. 𝟔Ω
𝐼
0.33
0.33
Resistors in Parallel
Now consider two resistors connected in parallel, as in Figure 2.4a. In this case the potential
differences across the resistors are the same because each is connected directly across
the battery terminals.
In a parallel combination of cell, all negative terminals are connected together. A battery of
identical cells connected in parallel has the following characteristic.
(i) The total current in a parallel circuit is equal to the sum of the currents in he separate
branches of parallel circuit. IT = I1 + I2 + I3 + etc.
(ii) The potential difference across all branches of a parallel circuit must have the same
magnitude. V = V1 = V2 = V3 = etc.
(iii) The reciprocal of the equivalent resistance is equal to the sum of the reciprocals of
the separate resistance in parallel.
1
1
1
=
+ (𝑃𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝐶𝑖𝑟𝑐𝑢𝑖𝑡) … 2.5
𝑅𝑒𝑞 𝑅1 𝑅2
Advantage of parallel connection
The cell is not easily used up because the cells share the total current generated together.
Disadvantage
The cells must not be left together disconnected to avoid exhaustion arising from short fall in
the strength of one cell, as this is bound to affect others.
The currents are generally not the same.
𝐼𝑒𝑞 = 𝐼1 + 𝐼2 + 𝐼3 + ⋯ + 𝐼𝑛
Because charge is conserved, the current I that enters point
a must equal the total current I1 + I2 leaving that point. The
potential drop must be the same for the two resistors and
must also equal the potential drop across the battery. Ohm’s
law applied to each resistor yields
𝑉
𝑉
𝐼1 =
& 𝐼2 =
𝑅1
𝑅2
Ohm’s law applied to the equivalent resistor in Figure 2.4b b ●
gives
𝐼𝑒𝑞 =
∆𝑉
𝑅𝑒𝑞
1
1
1
=
+
(𝑃𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝐶𝑖𝑟𝑐𝑢𝑖𝑡) … 2.5
𝑅𝑒𝑞 𝑅1 𝑅2
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R2
Fig. 2.4a
∆V
●a
An extension of this analysis to three or more resistors in parallel produces the following
general expression for the equivalent resistance:
1
1
1
1
=
+
+ … 2.6
𝑅𝑒𝑞 𝑅1 𝑅2 𝑅3
From this expression, we see that the inverse of the equivalent
resistance of two or more resistors connected in parallel is
the sum of the inverses of the individual resistances and is
always less than the smallest resistance in the group.
a
●
1
1
1
=
+
𝑅𝑒𝑞 𝑅1 𝑅2
c
●
Fig. 2.4b
Example: Three resistors are connected in parallel as in Figure
∆V
2.5. A potential difference of 18 V is maintained between points
a and b. (a) Find the current in each resistor. (b) Calculate the
power delivered to each resistor and the total power. (c) Find the equivalent resistance of the
circuit. (d) Find the total power delivered to the equivalent resistance.
b
Solution:
●
(𝑎) 𝐼1 =
∆𝑉 18
=
= 𝟔. 𝟎𝑨
𝑅1
3
𝐼2 =
∆𝑉 18
=
= 𝟑. 𝟎𝑨
𝑅2
6
𝐼3 =
∆𝑉 18
=
= 𝟐. 𝟎𝑨
𝑅3
9
I1
3Ω
18V
(𝑏) 𝑃1 = 𝐼12 𝑅1 = (6.0)2 (3.0) = (36)(3) = 𝟏𝟎𝟖𝑾
I2
6Ω
Fig.2.5a
I
●
a
(𝑏) 𝑃2 = 𝐼22 𝑅2 = (3.0)2 (3.0) = (9)(6) = 𝟓𝟒𝑾
(𝑏) 𝑃1 = 𝐼32 𝑅3 = (2.0)2 (3.0) = (4)(9) = 𝟑𝟔𝑾
Sum to get the total power:
(𝑏) 𝑃𝑇 = 𝑃1 + 𝑃2 + 𝑃3 = (108) + (54) + (36) = 𝟏𝟗𝟖𝑾~𝟐. 𝟎 × 𝟏𝟎𝟐 𝑾
(𝑐)
1
1
1
1
1 1 1 (1 × 6) + (3 × 1) + (2 × 1) 6 + 3 + 2 11
=
+
+
= + + =
=
=
𝑅 𝑒𝑞 𝑅1 𝑅2 𝑅3 3 6 9
18
18
18
𝑅𝑒𝑞 =
(𝑑)𝑃 =
18
= 𝟏. 𝟔Ω
11
(18)2
𝑉2
=
= 𝟐. 𝟎 × 𝟏𝟎𝟐 𝑾
𝑅𝑒𝑞
1.6
Example: (Combined Circuit) Four resistors are connected as shown in Figure 2.6 a. (a) Find
the equivalent resistance between points a and c. (b) What is the current in each resistor if a
42-V battery is connected between a and c?
Solution:
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I3
9Ω
(𝑎) 𝑅𝑒𝑞 = 𝑅1 + 𝑅2
= 8.0 + 4.0 = 12.0Ω
1
1
1
1
1
1+2 3
=
+
=
+
=
=
𝑅𝑒𝑞 𝑅3 𝑅4 6.0 3.0
6
6
𝑅𝑒𝑞 = 3.0Ω
𝑅𝑒𝑞 = 𝑅6 + 𝑅7 = 12.0 + 2.0 = 𝟏𝟒. 𝟎Ω
∆𝑉 42
(𝑏) 𝐼 =
=
= 𝟑. 𝟎𝑨
𝑅𝑒𝑞 14
Fig. 2.6
8.0Ω
a
●
4.0Ω
I
6.0Ω
b I1
●
I2
3.0Ω
∆𝑉𝑝𝑎𝑟𝑎 = 𝐼𝑅𝑒𝑞 = (3.0𝐴)(2.0Ω) = 6.0𝑉
𝐼1 =
∆𝑉𝑝𝑎𝑟𝑎 6.0𝑉
∆𝑉𝑝𝑎𝑟𝑎 6.0𝑉
=
= 𝟏. 𝟎𝑨; 𝐼2 =
=
= 𝟐. 𝟎𝑨
𝑅6.0Ω
6.0Ω
𝑅3.0Ω
3.0Ω
Resistance the electric property that impedes (delays) current; for Ohmic materials, it is the
ratio of voltage to current, R = V/I
Resistivity an intrinsic property of a material, independent of its shape or size, directly
proportional to the resistance, denoted by ρ
Factors that affect the resistance of a conductor
✓ Length of conductor (wire) L
As length of the conductor increases, the resistance increases.
✓ Cross – sectional area (A)
The resistance of a conductor is inversely proportional to its cross – sectional area. This
implies that as the cross – sectional area increases the resistance decreases.
✓ Temperature
Increase in temperature of conductor increases the resistance.
✓ Nature of material
The nature of material also determines the effect of temperature on the resistance of a
conductor. From the factors counted: the resistance
𝑳
𝝆𝑳
𝑹 ∝ 𝒊. 𝒆. 𝑹 =
𝑨
𝑨
Where ρ is constant proportionality called resistivity, from the equation
𝝆𝑳
𝑹=
𝑨
Resistivity (ρ) = RA/L. the unit of resistivity is ohm – metre (Ωm)
Example
The resistivity of a given piece of uniform wire of length 2.0 m is 5.4 x 10-7Ωm. If the cross –
sectional area is 9.5 x 10-3cm2, calculate the resistance of the wire.
Given
Unknown
Basic equation
Length of wire, L = 2.0m
R
10-7Ωm
Resistivity, ρ = 5.4 x
Cross – sectional area, A= 9.5x 10-3cm2
= 9.5 x 10-7m2
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𝑹=
𝝆𝑳
𝑨
10000cm2 = 1m2
c
●
If the resistance of the wire = R then 𝑹 =
𝝆𝑳
𝑨
𝟓. 𝟒 × 𝟏𝟎−𝟕 × 𝟐. 𝟎
∴𝑹=
= 𝟏. 𝟏𝟒Ω 𝐚𝐧𝐬.
𝟗. 𝟓 × 𝟏𝟎−𝟕
Heating effects
Energy of an electric current
To maintain an electric current, energy must be supplied by the source of emf. We recognized
that one joule of work is done when one coulomb of charge is moved through a potential
difference of one volt. W = qV
One coulomb of charge transferred in one second constitutes one ampere of current. I = q/t
and q = It therefore W = VIt
Energy conversion in resistance
From Ohm’s law, the potential difference across the resistance of an external circuit is the
product of the current in the circuit and the resistance of the circuit. V = IR substituting this
product for V in equation (V = RI), we have an expression for the energy transferred to the
resistance of a circuit. W = I2Rt
Example
The heating in a coffee maker has a resistance of 25.0Ω. When it is plugged into a 120.0V
circuit and turned on, how long will it take to heat 1.00kg water from room temperature,
20.0℃, to the boiling point, 100.0℃? (Cw =4.19J/g.℃) Assume that no heat is lost to the
surroundings.
Given
Unknown
Basic equation
R = 25.0Ω
t
QLost = QGained
V = 120.0V
Q = I2Rt
m = 1.0kg × 1000 = 1.0 x 103g
Q = mcw∆T
∆T = (100.0 – 20.0) = 80.0℃
I = V/R
cw = 4.19J/g.℃
Solution
Working equation:
𝑽 𝟐
( ) 𝑹𝒕 = 𝒎𝒄𝒘 ∆𝑻
𝑹
Where V2/R2 x Rt is heat lost and mcw∆T is the heat gained. Therefore,
𝑚𝑐𝑤 ∆𝑇𝑅 (1.0 × 103 )(4.19)(80.0)(25.0)
𝑡=
=
= 𝟓𝟖𝟏. 𝟗 𝒐𝒓 𝟓𝟖𝟐𝒔
(120)2
𝑉2
Practice problems
1. A hot plat has a resistance of 30.0Ω. If develops 1.0 × 106J of heat in 60.0 second of
operation. How much current does the hot plate draw? Ans. 23.6A
2. The heating element of an electric teakettle has a resistance of 28.8Ω. It is plugged into a
120V circuit. How many minutes does it take to heat 750g of water from 4.5℃ to 100.0℃ if
there is no heat lost to the surroundings? Ans. 10.0 min
6|Page
Power in electric circuits
Power is defined as the rate of work done. In an electric circuit, it is convenient to think of
power as the rate at which electric energy is sent to the circuit.
𝑾
𝑷=
𝒕
Recalling equation (W = VIt), let us solve this expression for Wit.
𝑾
= 𝑽𝑰
𝒕
Combining equation (P = W/t) and (VI = W/t) yields a basic power equation for electric
circuits.
𝑷 = 𝑽𝑰
If I is in ampere and V is in volts, then P is in watts:
𝒋𝒐𝒖𝒍𝒆
𝒄𝒖𝒐𝒍𝒐𝒎𝒃
𝒋𝒐𝒖𝒍𝒆
×
=
= 𝒘𝒂𝒕𝒕
𝒄𝒐𝒖𝒍𝒐𝒎𝒃
𝒔𝒆𝒄𝒐𝒏𝒅
𝒔𝒆𝒄𝒐𝒏𝒅
By Ohm’s law, the potential difference across a load resistance in a circuit is equal to the
product IRL. Substituting for V in equation (P = VI),
PL = I × IRL or PL = I2RL
This shows us that the power expended by a current in a resistance is proportional to the
square of the current in the resistance.
Similarly, the power dissipate in the internal resistance of source of emf is
Pr = I2r
The total power consumed in a circuit must be the sum of the power dissipated within the
source of emf and the power expended in the load of the external circuit.
PT = IT2RT = IT2 (RL + r) Recall that ℰ = IT (RL + r)
Therefore PT = ITℰ
There are instances in which the current in a circuit is unknown or of no interest. Power can
be expressed in terms of voltage and resistance since, by Ohm’s law
𝓔
𝑰𝑻 =
𝑹𝑻
Substituting for IT in equation (PT = ITE), we get
𝓔𝟐
𝑷𝑻 =
𝑹𝑻
And for the internal circuit,
𝑽𝟐
𝑷𝑳 =
𝑹𝑳
Kirchhoff’s Rules and Complex DC Circuits
There are, however, many ways in which resistors can be connected so that the circuits
formed can’t be reduced to a single equivalent resistor. The procedure for analyzing more
complex circuits can be simplified by the use of two simple rules called Kirchhoff’s rules:
i. The sum of the currents entering any junction must equal the sum of the currents
leaving that junction. (This rule is often referred to as the junction rule.)
ii. The sum of the potential differences across all the elements around any closed circuit
loop must be zero. (This rule is usually called the loop rule.)
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I2
The junction rule is a statement of conservation of charge.
Whatever current enters a given point in a circuit must leave that
I1
Fig. 2.7
point because charge can’t build up or disappear at a point. If we
apply this rule to the junction in Figure 18.12a, we get
I3
𝐼1 = 𝐼2 + 𝐼3
When applying Kirchhoff’s rules, you must make two In each diagram, ∆V = Vb – Va
decisions at the beginning of the problem:
and the circuit element is
1. Assign symbols and directions to the currents in all traversed from a to b, left to right.
I
branches of the circuit. Don’t worry about guessing the
a●
●b
direction of a current incorrectly; the resulting answer will be
∆V = – IR
negative, but its magnitude will be correct. (Because the
Fig. 2.8a
equations are linear in the currents, all currents are to the first
I
power.)
a●
●b
2. When applying the loop rule, you must choose a direction for
∆V = + IR
traversing the loop and be consistent in going either clockwise or
Fig. 2.8b
counterclockwise. As you traverse the loop, record voltage drops
ɛ
and increases according to the following rules (summarized in
–
+
Fig. 2.8, where it is assumed that movement is from point a toward a ●
●b
point b):
∆V = + ɛ
(a) If a resistor is traversed in the direction of the current, the change in
Fig.2.8c
electric potential across the resistor is – IR (Fig. 2.8a).
ɛ
(b) If a resistor is traversed in the direction opposite the current, the
+
–
a
change in electric potential across the resistor is + IR (Fig. 2.8b).
●
●b
(c) If a source of emf is traversed in the direction of the emf (from – to +
∆V = – ɛ
on the terminals), the change in electric potential is + ɛ (Fig. 2.8c).
Fig.2.8d
(d) If a source of emf is traversed in the direction opposite the emf (from
+ to – on the terminals), the change in electric potential is – ɛ (Fig.
2.8d).
Example: Find the currents in the circuit shown in Figure 2.9 by using Kirchhoff’s rules.
Solution:
Apply the junction rule to point c. I1 is directed into the
5.0 Ω
junction, I2 and I3 are directed out of the junction.
𝐼1 = 𝐼2 + 𝐼3
I2
4.0 Ω
c●
Select the bottom loop and traverse it clockwise starting
●d
at point a, generating an equation with the loop rule:
I3
∑ ∆𝑉 = ∆𝑉𝑏𝑎𝑡 + ∆𝑉4.0Ω + ∆𝑉9.0Ω = 0
I1
9.0 Ω
6.0𝑉 − (4.0Ω)𝐼1 − (9.0Ω)𝐼3 = 0
Select the top loop and traverse it clockwise from point c.
+ –
●
●
a
b
Notice the gain across the 9.0 Ω resistor because it is
6.0 V Fig. 2.9
traversed against the direction of the current!
∑ ∆𝑉 = ∆𝑉5.0Ω + ∆𝑉9.0Ω = 0
−(5.0Ω)𝐼2 + (9.0Ω)𝐼3 = 0
Rewrite the three equations, rearranging terms and dropping units for the moment, for
convenience:
(1) 𝐼1 = 𝐼2 + 𝐼3
(2) 4.0𝐼1 + 9.0𝐼3 = 6.0𝑉
(3) − 5.0𝐼2 + 9.0𝐼3 = 0
Solve Equation (3) for I2 and substitute into Equation (1):
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𝐼2 = 1.8𝐼3
𝐼1 = 𝐼2 + 𝐼3 = 1.8𝐼3 + 𝐼3 = 2.8𝐼3
Substitute the latter expression into Equation (2) and solve for I3:
4.0(2.8𝐼3 ) + 9.0𝐼3 = 6.0 → 𝐼3 = 0.297 ~ 𝟎. 𝟑𝟎 𝑨
Substitute I3 back into Equation (3) to get I2:
−5.0𝐼2 + 9.0(0.30) = 0 → 𝐼2 = 𝟎. 𝟓𝟒 𝑨
Substitute I3 into Equation (2) to get I1:
4.0𝐼1 + 9.0(0.30A) = 6.0 → 𝐼1 = 0.825 ~ 𝟎. 𝟖𝟑 𝑨
Example: 1. A battery has an emf of 12.0 V and an internal resistance of 0.05 Ω. Its
terminals are connected to a load resistance of 3.00 Ω. (a) Find the current in the circuit and
the terminal voltage of the battery. (b) Calculate the power delivered to the load resistor, the
power delivered to the internal resistance of the battery, and the power delivered by the
battery.
Solution:
𝜀
12
12
(𝑎𝑖) 𝐼 =
=
=
= 𝟑. 𝟗𝟑 𝑨
𝑅 + 𝑟 3.0 + 0.05 3.05
(𝑎𝑖𝑖) ∆𝑉 = 𝜀 − 𝐼𝑟 = 12.0𝑉 − (3.93𝐴)(0.05) = 𝟏𝟏. 𝟖 𝑽
To check this result, we can calculate the voltage across the load resistance R:
∆𝑉 = 𝐼𝑅 = (3.93𝐴)(3.0Ω) = 𝟏𝟏. 𝟖𝑽
(bi) The power delivered to the load resistor is
𝑃𝑅 = 𝐼 2 𝑅 = (3.93𝐴)2 (3.0Ω) = 𝟒𝟔. 𝟑𝑾
(bii) The power delivered to the internal resistance is
𝑃𝑟 = 𝐼 2 𝑟 = (3.93𝐴)2 (0.05Ω) = 𝟎. 𝟕𝟕𝟐𝑾
Hence, the power delivered by the battery is the sum of these quantities, or 47.1 W. You
should check this result, using the expression
𝑃 = 𝐼𝜀
2. What If? As a battery ages, its internal resistance increases. Suppose the internal
resistance of this battery increases to 2.00 Ω toward the end of its useful life. How does this
change the ability of the battery to deliver energy? Let us connect the same 3.00 Ω load
resistor to the battery. The current in the battery now is
𝜀
12
12
𝐼=
=
=
= 𝟐. 𝟒𝟎 𝑨
𝑅 + 𝑟 3.0 + 2.00 5.00
and the terminal voltage is
(𝑎𝑖𝑖) ∆𝑉 = 𝜀 − 𝐼𝑟 = 12.0𝑉 − (2.40𝐴)(2.00) = 𝟕. 𝟐 𝑽
Notice that the terminal voltage is only 60% of the emf. The powers delivered to the load
resistor and internal resistances are
𝑃𝑅 = 𝐼 2 𝑅 = (2.4𝐴)2 (3.0Ω) = 𝟏𝟕. 𝟑𝑾; 𝑃𝑟 = 𝐼 2 𝑟 = (2.4𝐴)2 (2.0Ω) = 𝟏𝟏. 𝟓𝑾
Notice that 40% of the power from the battery is delivered to the internal resistance. In part
(B), this percentage is 1.6%. Consequently, even though the emf remains fixed, the increasing
internal resistance significantly reduces the ability of the battery to deliver energy.
Assignment questions (20Pts)
Direction: Please show diagrams in all steps why solving for the equivalent resistance as
points will be allocated for each diagram that correctly describes the solution.
1. Two resistors connected in series have an equivalent resistance of 690 Ω. When they are
connected in parallel, their equivalent resistance is 150 Ω. Find the resistance of each resistor.
9|Page
2. Consider the combination of resistors shown in Figure A2.1 (a) Find the equivalent
resistance between point a and b. (b) If a voltage of 35.0 V is applied between points a and b,
find the current in each resistor.
4.0 Ω
12.0 Ω
5.0
Ω
a
6.0 Ω
Fig. A2.1
3. (a) Find the equivalent resistance between
points a and b in Figure A2.2. (b) Calculate the
4.0 Ω
current in each resistor if a potential difference of
34.0 V is applied between points a and b.
b
8.0 Ω
7.0 Ω
9.0 Ω
Fig. A2.2
10.0 Ω
4. Consider the circuit shown in Figure A2.3. (a)
●
Calculate the equivalent resistance of the 10.0- Ω a
and 5.00 -Ω resistors connected in parallel . (b)
Using the result of part (a), calculate the combined
resistance of the 10.0-Ω, 5.00-Ω, and 4.00-Ω resistors. (c) Calculate the equivalent resistance
of the combined resistance found in part (b) and the parallel 3.00-Ω resistor. (d) Combine the
equivalent resistance found in part (c) with the 2.00-Ω resistor. (e) Calculate the total current
in the circuit. (f) What is the voltage drop across the 2.00-Ω resistor? (g) Subtracting the
result of part (f) from the battery voltage, find the voltage across the 3.00-Ω resistor. (h)
Calculate the current in the 3.00-Ω resistor.
10.0 Ω
4.0 Ω
5.0 Ω
2.0 Ω
3.0 Ω
Fig. A2.3
+
-
8.0 V
10 | P a g e
●
b