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Circuits. Power. Equivalent resistance in series and parallel circuits. Wheatstone bridge. Terminal voltage of a battery. Kirchhoff’s rules. The potentiometer.

Circuit diagrams. Fig. 1 shows an electrical circuit containing two 1.5 volt batteries, a lamp and a switch in which the current in the form of electrons flow from the negative to the positive terminal of the batteries. Fig. 2 shows a schematic wiring diagram of the circuit. Some of the conventional symbols used in schematic wiring diagrams are shown in Fig. 3.

Convention on the direction of flow of current. Electrons flow from the negative terminal to the positive terminal and thus current flow is from the negative terminal to the positive terminal. The negative terminal is the high potential terminal and the positive terminal is the low potential terminal. However, for many years just what electric current was was not known; it was not realized that electric current consisted in the flow of electrons. For many years it was assumed that electric current flowed from the positive terminal to the negative terminal (opposite from the way it really flows), the positive terminal being the high potential terminal and the negative terminal being the low potential terminal. Flow from the positive terminal to the negative terminal became the accepted convention. Most authors use the old convention and assume that current runs from the positive terminal to the negative terminal, even though it doesn’t. We will henceforth go with convention and assume that electric current flows from the positive terminal to the negative terminal.

Potential energy
drops in a circuit.
A 50 lb steel ball
sitting on top of a 100
foot tower has 5000
ft-lbs of potential
energy relative to the
ground. If it is
dropped, that potential
energy is transformed
to kinetic energy and,
on hitting the ground,
there is no potential
energy left — it has
all been transformed
to another type of
energy. In a similar
way, a one coulomb
test charge moving
from the high
potential terminal of a battery to the low potential terminal by way of a wire conductor does work
and gives up its potential energy as it passes through the circuit. If the circuit contains
resistances R_{1}, R_{2, ... , }R_{n, }it _{ }does work in passing through each of the resistances. When it has
passed through the last resistance, all work has been done, and all the potential energy has been
used up. Let V_{a} be the voltage of the high potential terminal and V_{b} be the voltage of the low
potential terminal. Then as the charge makes its way from the high potential terminal to the low
potential terminal, its potential energy will drop from V_{a} volts to V_{b} volts. If we attach one lead
of a voltmeter to the high potential terminal and check the voltage drop at different points along
the circuit, we will see that the voltage has dropped down to V_{b} at the end of the circuit.

Work done by a moving charge. The work W done, in joules, when a charge of q
coulombs moves through a potential difference of V_{ab} volts in a circuit is

1) W = qV_{ab}

Power. The power, or rate at which work is done over a potential drop of V_{ab} in a circuit, is

**********************

Derivation. From 1)

3) dW = V_{ab }dq

Now

4) i = dq/dt

Substituting 4) into 3) gives

5) dW/dt = i V_{ab}

**********************

Special case when the circuit between points a and b is pure resistance. In the special case when the circuit between points a and b is pure resistance, all of the energy is converted to heat and the power represents the rate at which heat is generated. Here the rate at which heat H is generated is given by

where P is in joules/ sec, i is in amps, V is in volts, and R is in ohms.

Since 1 joule = 0.239 calories, the heat H in calories generated in time t is

7) H = 0.239 i^{2}Rt

Series and parallel connections. An electric circuit may be very simple and consist of one or two electrical devices connected to a power source or it may be very complex and consist of a large number of circuit elements such as multiple emf’s, resistors, capacitors, motors, etc. all connected in more or less complicated ways. A complex circuit is called a network.

Fig. 4 illustrates four different ways in which three resistors having resistances R_{1}, R_{2} and R_{3}
might be connected between the two points a and b.

Series connection. In Fig. 4 (a), the three resistors are said to be connected in series between

points a and b. There is only one path between the two points a and b and the resistors are on that path. Any number of circuit elements such as resistors, cells, motors, etc. are said to be connected in series with one another if they are connected as the resistors in Fig. 4 (a), all being on a single path between points a and b.

Parallel connection. In Fig. 4 (b) the three resistors are said to be connected in parallel between points a and b. Each resistor is on a separate path between points a and b. Any number of circuit elements similarly connected are said to be connected in parallel with one another.

In Fig. 4 (c), resistors R_{2} and R_{3} are connected in parallel and this combination is in series with
resistor R_{1}. In Fig. 4 (d), resistors R_{2} and R_{3} are in series and this combination is in parallel with
R_{1}.

Equivalent resistance. Suppose we have some combination of resistances (combinations
such as those illustrated in the four examples of Fig. 4) situated between two points a and b in a
circuit. It is always possible to find a single resistor that will replace the combination and leave
unchanged both 1) the potential difference V_{ab} between the points a and b, and 2) the line current
i at points a and b. The resistance of this single resistor is called the equivalent resistance of the
combination. For example, if any one of the combinations in Fig. 4 were replaced by its
equivalent resistance R, we could write V_{ab} = Ri where V_{ab} is the potential difference between a
and b and i is the line current at points a and b.

Rules for resistors connected in series. For a set of resistors connected in series:

1. The equivalent resistance R of n resistors R_{1}, R_{2}, ... , R_{n} connected in series is given by

8) R = R_{1} + R_{2} + ... + R_{n}

Thus the equivalent resistance of any number of resistors in series is the sum of their individual resistances.

2. The current in all parts of a series circuit is the same.

3. The total potential drop across several resistors in a series circuit is equal to the sum of the potential drops across the separate resistors.

Def. Conductance. The conductance G of a conductor is defined as the reciprocal of its resistance R:

G = 1/R

The unit of conductance is mhos.

Rules for resistors connected in parallel. For a set of resistors connected in parallel:

1. The equivalent resistance R of n resistors R_{1}, R_{2}, ... , R_{n} connected in parallel is given by

Thus the total conductance of a set of resistors is equal to the sum of their conductances.

2. The total current in the circuit [i.e. the line current at points a and b in Fig. 4 (b)] equals the sum of the currents in the separate branches.

3. The potential drop across several resistors connected in parallel is the same across all branches i.e. the potential difference is the same from point a to point b of Fig. 4 (b), whatever the route.

Problem. Compute the current I through the battery in Fig. 5.

Solution. The 7, 1 and 10 ohms are in series
and their joint resistance is 18 ohms. This 18
ohms is in parallel with 6 ohms and the
equivalent resistance R_{1} is given by

Solving gives R_{1} = 4.5 ohms

The equivalent resistance for the entire circuit is R = 4.5 + 2 + 8 + 0.3 = 14.8 ohms and the battery current is I =V/R = 20/14.8 = 1.35 amp

Measurement of potential differences and
current using the voltmeter and ammeter.
To measure a voltage drop V_{AB} between points A and
B in a circuit, the voltmeter is connected in parallel as
shown in Fig. 6. To measure the current in a circuit, the ammeter is connected in series as shown
in the figure. An ammeter is a very low resistance device and a voltmeter is a very high resistance
device to ensure best accuracy. If very accurate measurements are required, the resistances of the
ammeter and voltmeter must be considered part of the circuit.

Measurement of resistance. We can measure the resistance of a conductor by measuring the potential drop over the conductor with a voltmeter and measuring the current flowing through the conductor with an ammeter as shown in Fig. 6. One then uses Ohm’s law, R = V/ I, to compute the resistance. A much more accurate method involves using the Wheatstone bridge.

Wheatstone bridge method of measuring the resistance of a conductor. The Wheatstone bridge consists of a board a little over a meter in length and about 15 centimeters wide with heavy brass strips mounted along the top and at each end. See Fig. 7. A meter stick is mounted along the bottom with a heavy wire AB of uniform resistance, one meter long, stretched over it as shown in the figure. At the top left are terminals for attaching a resistor of unknown resistance and in the top right are terminals for connecting a resistor of known resistance. One lead of a galvanometer is connected to a terminal at the midpoint of the top brass strip and the other lead is connected to a contact key that slides along the wire on top of the meter stick. (A galvanometer is a very sensitive instrument that measures small currents and indicates their direction of flow.)

Operation of the Wheatstone bridge. Current flows from the cell to binding post C where it divides, part taking the path through the unknown and known resistances at the top and part going through the wire AB. The two paths unite at D and the current returns to the cell. In general, some current will also go through the path containing the galvanometer. With known and unknown resistances in place, one makes a measurement by sliding the contact key along the wire until he finds that point at which no current flows through the galvanometer. This is called balancing the bridge. At this point he can compute the value X of the unknown resistance from the relationship

where L is the meter stick reading in centimeters and L' is 100 - L ( L is the length of the part of the wire from its beginning to the balance point and L' is the length from balance point to end).

A circuit diagram of the Wheatstone bridge is
shown in Fig. 8. Here ρ represents the
resistance of wire AB in ohms /cm. The bridge
will be balanced and no current will flow in path
GK when the potential difference between G
and K is zero. When this happens the potential
difference V_{CG} from C to G is equal to the
potential difference V_{CK} from C to K

11) V_{CG} = V_{GK}

and the potential difference V_{GD} from G to D is equal to the potential difference V_{KD} from K to
D

12) V_{CG} = V_{CK}

From 11) and 12) we get

13) I_{1}X = I_{2}ρL

14) I_{1}R = I_{2}ρL'

Dividing 13) by 14) gives

or

Terminal voltage of a battery or generator. A seat of emf such as a battery or electrical generator will deliver an emf when generating current that is less than its measured terminal voltage when it is not delivering current. This is due to internal resistance which causes internal generation of heat and energy loss when current is flowing. It is customary to represent the internal resistance of a seat of emf by a resistor r which is considered to be in series with a resistanceless source of emf E. Let us find the expressions for the terminal voltage when the cell is discharging and when it is charging.

Case 1. Cell discharging. Fig. 9 (a)
depicts the situation when a cell is
discharging. We can relate the voltage
V_{a} at point a to the voltage V_{b} at point b
by considering the voltage drops and
increases as one proceeds from point b
to point a. There is a voltage drop at the
resistor r and a voltage increase at the
cell so

17) V_{a} = V_{b} - Ir + E

or

18) V_{a} - V_{b} = E - Ir

Case 2. Cell charging. Fig. 9 (b) depicts the situation when a cell is charging. Considering the voltage drops and increases as one proceeds from point a to point b we obtain

19) V_{b} = V_{a} - E - Ir

or

20) V_{a} - V_{b} = E + Ir

In summary, we have

(1) When cell is delivering current (cell discharging):

Terminal voltage = E - Ir

(2) When cell is charging:

Terminal voltage = E + Ir

(3) When no current exists:

Terminal voltage = E

Kirchhoff’s rules. Complicated circuits made up of resistors and sources of emf often cannot be readily resolved into series and parallel combinations of resistors and solved by the method of equivalent resistances. Two rules, first stated by Gustav Robert Kirchhoff (1824 - 1887), enable such problems to be solved systematically.

We first define the terms junction and loop.

Def. Junction. Any point in a network at which three or more conductors are joined.

Def. Loop. Any closed path in an electrical network.

Kirchhoff’s rules are:

Point rule. The algebraic sum of the currents at every junction is zero where a current is regarded as positive if its direction is toward a junction and negative if its direction is away from the junction:

∑i = 0

Loop rule. The algebraic sum of the emf’s in any loop of a network equals the algebraic sum of the Ri products in the same loop:

∑E = ∑Ri

Procedure for solving networks using Kirchhoff’s rules.

Step 1.
Assign an
algebraic
symbol (e.g. i_{1},
i_{2}, ..) and
direction to all
unknown
currents and all
unknown emf’s.
The assumed
directions are
completely
arbitrary. Assign algebraic symbols (e.g. R_{1}, R_{2}, ..) to all unknown resistances.

Note. The solution is carried through on the basis of the assumed directions. If, in the solution of the equations, a negative value is obtained for a current or emf, its correct direction is opposite to that assumed.

Step 2. Create a circuit diagram with all known and unknown quantities and directions shown.

Step 3. Assign letters (e.g. a, b ,c ...) to all junctions.

Step 4. Form the junction equations using the point rule. If there are n junctions, form junction equations for any n-1 of the junctions. (Application of the point rule at the n-th junction won’t lead to an independent relation.)

Step 5. Imagine the network to be broken up into a number of simple loops, like pieces of a jigsaw puzzle. Make sure every circuit element is included in at least one loop. Apply the loop equation to each of the loops.

When applying the loop rule, some direction around the loop (i.e. clockwise or counterclockwise) must be chosen as the positive direction. All currents and emf’s in this direction are regarded as positive, all in the opposite direction are regarded as negative.

Problem.
Given the network shown in Fig. 10 where the indicated emf’s E_{1}, E_{2}, ... and resistances R_{1}, R_{2}, ...
are assumed to be known quantities. Write the junction equations and loop equations.

Solution. First we need to assign algebraic symbols and directions to all currents. Fig. 11 shows the assignment we have made.

Letters have been assigned to the various junctions. There are four junctions: a, b, c, and d.

We now form the junction equations:

Junction a: i_{1} + i_{2} - i_{3} = 0

Junction b: -i_{1} - i_{4} - i_{6} = 0

Junction c: i_{4} + i_{5} - i_{2} = 0

Let us now break the network up into loops and form the loop equations. We will consider the clockwise direction positive for each loop. The loop equations are:

1] -E_{1} - E_{5} = i_{1}R_{1} + i_{1}r_{1} - i_{2}r_{5} - i_{4}R_{4} + i_{1}R_{3}

2] E_{2} + E_{5} = i_{3}r_{2} + i_{3}R_{2} + i_{3}R_{6} + i_{5}R_{5} + i_{2}r_{5}

3] E_{4} = i_{4}R_{4} - i_{5}R_{5} - i_{6}r_{4} - i_{6}R_{7}

We now have six independent equations to be solved for six unknown currents.

What are the advantages of parallel wiring as opposed to serial wiring? In house wiring, receptacles and appliances are wired in parallel. See Fig. 12. Why? One must consider that everything in a series circuit receives the same amount of current. In parallel wiring this isn’t the case. In parallel wiring each appliance will receive whatever current it requires. An electric light draws a different amount of current than a stove. It needs to be on its own circuit.

The potentiometer. The potentiometer is an instrument that is used to measure the emf of a seat of emf (such as a cell) without drawing any current from the seat. If a voltmeter is used for this purpose, the reading will be less than the emf because the voltmeter draws current from the seat of emf and the reading includes the voltage drop due to the internal resistance of the seat.

A schematic diagram of a potentiometer is
shown in Fig. 13. In the upper loop of the
figure, a resistance wire ab of uniform
resistance is connected to a cell of emf E. In
the lower loop, a cell X of unknown emf,
whose emf E_{x} is to be measured, is connected in series with a galvanometer and a sliding contact
C. Determining the value of an unknown emf E_{x} involves moving the contact C to a position
where there is no current flowing through the galvanometer. Assume that at the point where no
current is flowing, the contact C is at the value C = x. We can then write

(upper loop) V_{ax} = iR_{ax}

(lower loop) V_{ax} = E_{x}

i.e. when no current is flowing through the galvanometer

21) E_{x} = iR_{ax}

In order to use a potentiometer it is necessary to first calibrate it with the aid of a standard cell
such as the Weston normal cell whose emf has previously been determined. Let us replace the
cell X of unknown resistance with our standard cell W_{S} with emf E_{S }and let us move contact C so
as to determine the balance point of no current. Assume it occurs at the point C = s. Then we
have

22) E_{S} = iR_{as}

Dividing 21) by 22) gives

23) E_{x} = E_{S} (R_{ax}/R_{as})

If l_{s} is the length of the slide wire from a to s and l_{x} is the length of the slide wire from a to x,
then

24) R_{ax}/R_{as} = l_{x} / l_{s}

Substituting 24) into 23) gives

25) E_{x} = E_{S} (l_{x} / l_{s})

The measurement of the emf has thus been reduced to the measurement of two lengths along a uniform slide wire.

References

1. Dull, Metcalfe, Brooks. Modern Physics.

2. Sears, Zemansky. University Physics

3. Semat, Katz. Physics.

4. Freeman. Physics Made Simple.

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