CURRENT ELECTRICITY (II) UPDATED NOTES

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Chapter Five

CURRENT ELECTRICITY (II)

ELECTRIC CURRENT AND POTENTIAL DIFFERENCE

Electric Current

 

Using an Ammeter

If this is not the case, use the zero adjusting screw to move it to the correct position,

It is thus connected in series with other components in the circuit so that conventional current enters the ammeter through its positive terminal and exits through the negative terminal.

If the terminals are interchanged, the pointer moves away from thescale in anticlockwise direction. This can damage the instrument.

If 5A scale is selected, the meter can safely read up to a maximum of 5 A.

With such a scale, ten divisions represent 1A.

For scale of2.5 A, ten divisions represent 0.5A.

Potential Difference

The source may be :

From the equation, one volt is equal to one joule per coulomb.
Example 1

In moving a charge of 10 coulombs from point B, 120 joules of work is done. What is the
potential difference between A and B?

Solution

Using a Voltmeter

Points at Same Electric Potential

 

To investigate the current and voltage in a parallel circuit arrangement

Apparatus

Two 1.5 V cells, 3 identical bulbs, 3 ammeters, 4 voltmeters, switch, connecting wires.

Procedure

Observation

Note:

When components are connected in parallel:

 

Example 2

Find the current passing through L1in figure below given that 0.8 A passes through the battery, 0.28 A through L2 and 0.15 A though L3•

To investigate current and voltage in series arrangement

Apparatus

Three ammeters, four voltmeters, three 2.8 V torch bulbs, holder, switch, connecting wires,
two cells.

 

 

Procedure

Switch on the circuit and record the voltmeter readings.

Observations

(i) The reading of current by the ammeters A1and1A2 and A1, is the same.

(ii) The total voltage drops across the bulbs (V 1+ V 2 + V 3) equals the total voltage V 4 acrossthe terminals of the battery.

Note:

The above statements are true even when the bulbs are not identical.

 

 

Conclusion

In a series arrangement,

 

Example 3

In the circuit shown below, what is the potential difference across the bulb and the switchwhen the:
(a) switch is open?

(b) switch is closed?

 

 

 

 

 

(a)     Potential difference across the bulb is zero since no current is flowing through it, while
the p.d across the switch is 1.5 V.

(b) The p.d across the bulb is 1.5 V, since a closed switch is a conductor and has zero voltage.

 

 

OHM’S LAW

The relationship between the voltage across a conductor and the current flowing through it is
summarized in what is referred to as Ohm’s law.

To investigate the relationship between current and voltage across a nichrome wire

Apparatus

Two-metre nichrome wire, 2 dry cells, ammeter, voltmeter, connecting wires, switch, rheostat.

Procedure

Current(A)                
Voltage(v)                

Observation

As current increases, voltage across the coil also increases.

A graph of voltage against current is a straight line, as shown in figure

 

Conclusion

 

Voltage (Volts)

 

Example 4

A current of 4 mA flows through a conductor of resistance 2 00Ω. Calculate the voltage across the conductor.

Solution

Example 5

Calculate the current in amperes flowing through a device of resistance 50Ω when a 10 V
source is connected to it.

Solution

Example 6

In order to start a certain car, a current of 30 A must flow through the starter motor. Calculate
the resistance of the motor given that the battery supplies a voltage of 12 V. Ignore the internal resistance of the battery.

 

Ohmic and Non-ohmic Conductors

 

 

 

Electrical Resistance

 

Factors that affect the Resistance of a Metallic Conductor

Temperature

From the above

Resistance x cross-section area = constant …………………………………(2)

For a given conductor therefore, RA = constant.

Combining equation 1 and equation 2

where  is the resisitivity of the material.

Note:

It is helpful to express all lengths in meters so as to obtain resistivity in ohm-metre units. Every material has its resisitivity

 

Material Resisitivity (Qm) Uses
Silver 1.6 x 10.8 Contacts on some switches
Copper 1.7 x 10.8 Connecting wires
Aluminium 2.8 x 10.8 Power cables
Tungsten 5.5 x 10.8 Lamp filaments
Constantan 49 x 10.8 Resistance boxes, variable resistors
Nichrome 100 x 10-8 Heating elements
Carbon 3000 x 10-8 Radio resistors
Glass 10-8_ 1014  
Polystyrene 1015  

 

Resistors

 

Types of resistors

Fixed resistor

 

 

 

Variable Resistors

These are resistors with a varied range of resistance. They include:

  1. Rheostat

A rheostat is a two-terminal variable resistor represented in electrical circuits by the symbol

Moving the sliding contact along the length of the resistant material varies the resistance
between points A and B.

When the contact is nearer end A, the resistance of the rheostat is lower.

 

  1. Potentiometer

The potentiometer is a variable resistor with three terminals.

Its symbol is shown above.

In potentiometers, a contact is moved to select desired proportions of the total voltage across them.

 

Non-Linear Resistors

The current flowing through these resistors does not change linearly with the changes in the
applied voltage.

Such resistors include the thermistor and light dependent resistor (LDR).

  1. Light-dependent Resistor (LDR)

The resistance of an LDR decreases when it receives light of increasing intensity. Its symbol

 

  1. Thermistor

The thermistor is a temperature-dependent resistor.

Its resistance decreases with increase in temperature.

The electrical symbol of a thermistor is

Thermistors are used in heat-operated circuits.

 

To determine the resistance of a resistor using the voltmeter-ammeter method

 

 

 

 

 

 

 

Procedure

Set up the circuit as shown in the figure 5.24.

With the switch open, record the voltmeter reading V and the corresponding ammeter reading I.
Switch on the current and, by adjusting the variable resistor, record at least five other
values of V and the corresponding I.

Record your results in the table 5.3.

 

 

 

 

 

 

 

 

 

Table 5.3 Compare values of i‘

Plot a graph of V (vertical axis) against I. Note the shape of the graph.
Determine the slope (gradient) of the graph.

Observation

When the switch is open, no current flows through the resistor and therefore both the ammeter
and the voltmeter reading is zero.

When the current through the resistor increases, the voltage across it also increases. An
approximate value of the resistance of the resistor is obtained by dividing the value of the
voltage across the resistor by the corresponding current flowing through it and substituting in the

equationR = i.

The graph of V against I is a straight line whose gradient gives resistance, see figure
5.12. The resistance obtained cannot be accurate since the voltmeter takes some little current,
thus not all of it flows through the resistor.

The Wheatstone Bridge Method

The Wheatstone bridge consists of four resistors and a galvanometer, as shown in figure 5.25.
The operation of the bridge involves making adjustments to one or two of the resistors until
there is no deflection in the galvanometer.

•

 

Fig. 5.25: The Wheatstone bridge

The four resistors K, L, M and N are joined as shown. If K is the unknown resistance, the
values of L, M and N, or the ratio ofM to N must be known. A galvanometer G and a dry cell

are connected as shown.                         .

The variable resistor L (commonly a resistance box) is adjusted until there is no deflection
in G. The bridge is then said to be balanced. No current flows through G at balance and
therefore, the p.d. across BD is zero. At the same time, the potential difference across AB is
then equal to that across AD. Also, the same current I) flows through K and L and current 12

flows through M and N. Then;                                       .

11 = 13 and 12 = 14

Therefore, I) K = 12 M (from V = IR)
Similarly, ~L = I4N.

So, IlL=’I2N

11K _ 12M
IlL – I2N

Therefore, ~ = ~ when the bridge is balanced.

The Wheatstone bridge is more accurate in measuring resistance than the voltmeter-
ammeter method because the value obtained by using the wheatstone bridge method does not
depend on the accuracy of the current-measuring instrument (galvanometer) used.

The Metre Bridge

Figure 5.26 shows a practical form of the Wheatstone bridge known as the metre bridge.

 

Fig. 5.26: Meter bridge

The wire AC of uniform cross-section area and length 1 m with a resistance of several ohms
and made of an alloy such as constantan. The length AD represents resistor M while length CD
represents resistor N. The ratio of M to N is altered by changing the position on the wire of the
movable contact D called ‘jockey’. The other arm of the bridge contains the unknown resistor
K and a known resistor L. The copper strips of low resistance connect the various parts. The
position of D is adjusted until there is no deflection in G. Then;

K = M = resistance of AD

L N resistance of DC

Since the wire is uniform cross-section, its resistance will be proportional to its length and

Hence’ K= AD _ XI
• L DC – X2

Thus K= LXI

, X2

The resistor L should be chosen to give balance points near the centre of the wire. This gives
a more accurate result. After obtaining the balance. K and L should be interchanged and a
second pair of values for XI and X2 obtained. This average of the value eliminates errors due to
non-uniformity of the wire and end corrections. In finding the balance point, the cell key or
switch should be closed before the jockey makes contact with the wire. This is necessary
because of the effect known as ‘self-induction’ in which the currents in the circuit take a short time to grow to their steady values. A high resistance should always be joined in series with
the galvanometer to protect it from damage whilst the balance is being sought.

Example 10

In an experiment to determine the resistance of a nichrome wire using the metre bridge, the
balance point was found to be at 38 em mark. If the value of the resistance in the right hand
gap needed to balance the bridge was 25 Q, calculate the value of the unknown resistor.

Resistor Networks

Resistors Connected in Series

Figure 5.28 shows three resistors connected in series.

 

 

 

 

Fig. 5.28: Resistorsin series

Since this is a series arrangement, V T = VI + Vz+ VyThe same current I flows through each of
the resistors. Using Ohm’s law and the fact that same current flows through the resistors;

I~ = I(RI+ R2+ R3)

Diving through by I;

Thereforfor resistors connected in series, the equivalent resistance is equal to the sum of individual resistances.

Example 11

Three resistors of 2.5 Q, 12 Q, and 3.5 Qare connected in series. What single resistor can
replace them in a circuit?

Example 12

Figure 5.29 shows three resistors in series connected to a power source. A current of 2 A flows
through the circuit.

 

 

Calculate:

(a) the voltage drop across each resistor.
(b) the voltage across the source.

(c) the total resistance in the circuit.

 

Resistors Connected in Parallel

Fig. 5.30 shows three resistorsRj, R, and R3 connected in parallel.

 

 

 

 

 

Example 13

The circuit diagram in figure 5.31 shows four resistors in parallel connected across a3 V supply.

 

 

Calculate:

(a) the effective resistance.

(b) the current through the 80 resistor.

Solution

 

(b) Current through the 8 0 resistor

 

 

Example 14

Two resistors

of 300 and 70 0 are connected in parallel. Calculate their equiva.ent resistance.

Example 15

Several 150 Q resistors are to be connected so that a current of 2 A flows from a 50 Y source.
How many resistors are required and how should they be connected?

Solution

Example 16

Calculate the current through each resistor in the figure 5.32.

 

 

Series – Parallel Arrangement

To find the effective resistance of a series – parallel arrangement, the network is systematically
reduced into a single resistor.

Example 17

Determine

the equivalent

resistance for the resistors in figure 5.33.

Since the 30 ,Qand 70 ,Q resistors are in parallel, the two can be replaced by a single one whose
value is;

The 21 ,Qresistor is now in series with the 19,Q resistor, see figure 5.34 (b). The two resistors
can be replaced by a single resistor RAC= 19 + 21 = 40 ,Q, see figure 5.34 (c).

Example 21

Calculate the effective resistance in the figure 5.35.

 

Solution

The reduction begins by combining the 5 Q and 7 Q resistors, which are in series, to get 12 Q.
The circuit is then re-drawn as in figure 5.36 (a). The 12 Q resistor in parallel with the 8 Q

resistor may be replaced ~c == fft! == 4.8 Q. The circuit is re-drawn as in figure 5.36 (b).

 

Finally.the 4.8 Q is in series with the 4 Q resistor. giving an equivalent resistance of;
RAC== 4 +4.8

== 8.8Q

Example 19

Two resistors of 6 Q and 3 Q in parallel are connected in series to a 4 Q resistor and a cell of
e.m.f. 1.5 V. Calculate:

(a)     the equivalent resistance of the circuit.

(b) the current through each of the resistors and the p.d. across each.

Solution

(b) Total current flows through the 4 Q resistor.

Now 16Q + 13Q = 0.25 A…………….. :           (1)

But, voltage across 6 Q = voltage across-3Q

6 X 16Q = 3 X 13Q

13Q = 216Q …………………….•… ~       ·i •••••• (2)

Substituting (2) in (1);
16Q + 216Q = 0.25 A

3 x 16Q= 0.25

16Q = 0.0833 A
Substituting in (2);

13Q = 2 x 0.0833

= 0.167 A

V6Q = 0.0833 x 6
=,0.5V

This is also the voltage across the 3 Q resistor, since they are in parallel.

Alternatively; .

,To calculate current ,through either 6,Q or 3 Q .resistor, U1e p.d, across them.must be

found first. Thus, p.d. across 6Q& 3′ Q’ + p.d. across 4· Q ~ voltage’ ofthesupply

 

 

 

Voltage across 412 resistor == ‘0.25 x 4
= 1 V

Hence, the voltage across (6Q and 3Q) is;

V = 1.5-1
V=0.5V

 

CUrrent through 6 Q = 05
6
  = 8.33 x 1O-2A
Current through 3 Q = 0.5
  3
  = 1.67 X 10-1 A

 

 

 

 

Exampk20

Four resistors of 5 0, 13 0, 3 0 and 6 0 are connected to 6 V supply, as shown in figure 5.38.

 

Fig. 5.38
Calculate:

(a) the current through the 13 0 resistor.
(b) total current in the circuit.

(c) voltage V AB and V BO.

Solution

(a)   From figure 5.38 (a), current from the supply divides into two parts at junction A. Part of
it flows through the 5 0 resistor and the rest through 13 0 resistor. The current
through the 13 0 divides into two at junction B, some flowing through the 3 0 and the
rest through the 6 ,Q. The 3 ,Qand 6 ,Q resistors are in parallel and equivalent to a 2 ,Q
resistor, which would then be in series with the, 13 ,Q resistor. The circuitin figure 5.38
(a) is thus reduced to the one in figure 5.38 (b).

The 13 0 and 2 O’inseries”f<:nw a resistor that is in parallel with the 5 ,Q resistor.

The p.d. across the 13 ,Qand 2 ,Q resistor, is equal to the supply of 6 V.

6

Current through 13.Q = 13 + 2

= 6

15

= O.4A

 

 

 

 

(b)   Total current = current through 13 .Q resistor + current through 5 .Q resistor
Current through 13 .Q = 0.4 A

Current through 5,Q

p.d. across 5,Q

=         5Q

= 6

:5

= 1.2A

Therefore, current in the circuit = 0.4 + 1.2
= 1.6A

(c) VAB= 0.4×13

= 5.2V

VBD= VAC– VAB

= 1.2 x 5 – 5.2
= 6.0-5.2

= O.8V

Alternatively;
VBD= 0.4 x 2
= O.8V

Example 20

Figure 5.39 shows a current of 0.8 A passing through an arrangement of four resistors.

 

Find the current through the 10 Q resistor.
Solution

The network of resistors can be replaced by two resistors of 40 Q and 60 Q in parallel.

C

n _ p.d. between P and R

urrentoug             ;lot. –         (30 + 10)

P.d. between P and R = 0.8 X RE, where RE is equivalent resistance for the whole network.
R = 40×60

E        60+40

= 24Q

P.d. across P and R = 0.8 x 24
= 19.2

Therefore, current through 10

= 0.48A

 

 

ELECTROMOTIVE FORCE AND INTERNAL RESISTANCE

The function of a cell in a circuit is to supply electrical energy. By definition.the electromotive
force (e.m.f.) of a cell is the potential difference between its terminals when no charge is

flowing out of the cell (cell in open circuit).                                                    .

Figure 5.40 shows a circuit that may be used to demonstrate the difference between
e.m.f. of a cell and terminal voltage. The reading of the voltmeter when the switch is open is
the e.m.f. of the cell.

Once a cell supplies current to an external circuit, the potential difference across it
drops by a value referred to as ‘lost voltage’. This loss in voltage is due to the internal resistance
of the cell. The potential difference across the cell when the circuit is closed.is referred to
as the terminal voltage of the cell.

 

 

Fig. 5.40: Em.f. ofa cell

A cell or any source of e.m. f. is made up of materials that are not perfect conductors of electricity.
They therefore offer some resistance tothe flow of current that they generate. This resistance
is usually low and is called the internal resistance of the cell or battery.

Relationship between E.M.F. and Internal Resistance

If a resistor R is connected in series with a cell as shown in figure 5.41, the internal resistance
of the cell r is considered to be connected in series with the external resistor R.

 

 

 

 

 

 

 

Fig. 5.41: Internal resistance of a cell

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Using the equation E = V + Ir and hence V = E – Ir, the gradient of the graph gives the internal
resistance r of the cell.

If the graph is extrapolated so as to cut the voltage axis, the point at which it does so
gives the e.m.f. of the cell.

Method 2

Apparatus

Ammeter, voltmeter, variable resistor, cells, connecting wires.

 

Fig. 5.44: Determination of internal resistance of a cell

Procedure

Table 5.5

 

The graph is a straight line with a positive gradient.

 

 

 

 

 

 

 

 

Fig. 5.45: Graph of fagainst R

The gradient of the graph gives ~. Internal resistance can be obtained in two ways:
(i) Extrapolating the graph to cut R axis gives r, see figure 5.45.

(ii) If the intercept on t axis is A, then, A = ~

 

 

 

 

Example 21

A battery consisting of four cells in series, each of e.m.f. 2.0 V and internal resistance 0.6 Q, is
used to pass a current through a 1.6 Q resistor. Calculate the current through the battery.

 

 

 

The e.m.f. of the battery is the sum of the e.m.f. of all the cells while the internal resistance of
the battery is the sum of all the internal resistances of the cells.

:. Current through battery =

Example 22

A cell drives a current of 2.0 A through a 0.6 Q resistor. When the same cell is connected to a
0.9 Q resistor, the current that flows is 1.5 A. Find the internal resistance and the e.m.f of the cell.

Solution

 

 

Taking E as e.m.f. ofthe cell and R the internal resistance;
E = 1R + Ir

From figure 5.46 (a);

 

 

 

 

E == (2.0 x 0.6) + 2.0r

=: 1.2 + 2r ……………………………………….. (1)

From figure 5.46 (b);

E = (1.5 x 0.9) + 1.5r

E =: 1.35 + 1.5r ……………………………….. (2)

Since the e.m.f. is the same in both circuits;
1.2 + 2r =: 1.35 + 1.5r

2r – 1.5r =: 1.35 – 1.2

0.5r = 0.15

r = 0.3 Q

Substituting for r in equation (1);
E =: 1.2 + 2r

E =: 1.2 + 2 x 0.3
E = 1.2 + O~6

= 1.8V

Exampie23

A battery consists of two identical cells, each of e.m.f. 1.5 V and internal resistance 0.6 Q,
connected in the parallel. Calculate the current the battery drives through a 0.7 Q resistor.

Solution

When identical cells are connected in parallel, see figure 5.47, the equivalent e.m.f. is equal to that of only one cell. The equivalent internal resistance is equal to that of two such resistors connected in parallel.

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