Electric Cell

The device which converts chemical energy into electrical energy is known as electric cell. Cell is a source of constant emf but not constant current.

The direction of flow of current inside the cell is from negative to positive electrode while outside the cell is form positive to negative electrode.

Internal resistance: When the current flows through the cell, its electrolyte offers resistance to the flow of current. It is because of the fact that ions have to move against the background of other ions and the neutral atoms. It is denoted by r. For ideal cell, it is zero.

EMF

Emf (E): e.m.f. of a cell is the energy supplied by the cell to drive a unit charge round the complete circuit. So, emf of cell .

The potential difference between the two poles of the cell in an open circuit (when no current is drawn from the cell) is also called the electromotive force (e.m.f.) of the cell.

The potential difference between the two poles of a cell in a closed circuit (when current is drawn from the cell) is called the terminal potential difference of the cell. It is denoted by V.

Potential difference across the terminal of a cell is discharging,
Potential difference across the terminal of a cell is charging ,

Grouping of Batteries

Series grouping

Equivalent emf of the combination =

As cells are identical so,

Internal resistance

Main current = Current from each cell

Note:

(a) Equivalent emf E_eq = E₁ E₂

(b) Current

(c)

in the above circuit cell 1 is discharging so it's equation is

and cell 2 is charging so it's equation

Parallel grouping

Equivalent emf

Equivalent internal resistance

Main current

Current from each cell

Condition for max power and the maximum power is

Note

(i)

(a) Equivalent emf

(b) Main current

(ii)

In this combination if cell's are connected with reversed polarity as shown in figure then:

Equivalent emf

Mixed Grouping

If n identical cell's are connected in a row and such m row's are connected in parallel as shown.

Equivalent emf of the combination

Equivalent internal resistance of the combination

Main current flowing through the load

Current from each cell

Condition for maximum power

(= internal resistance of the battery)

Total number of cell = mn Important Points

In parallel grouping of two identical cell having no internal resistance

When two cell's of different emf and no internal resistance are connected in parallel then equivalent emf is indeterminate, note that connecting a wire with a cell but with no resistance is equivalent to short circuiting. Therefore the total current that will be flowing will be infinity.

It is a common misconception that "current in the circuit will be maximum when power consumed by the load is maximum."

Actually current is maximum (= E/r) when R=min = 0 with while power consumed by the load E²R/(R + r)² is maximum (= E²/4r) when R= r and

Illustrations

1. Six lead-acid type of secondary cells, each of e.m.f. 2.0 V and internal resistance 0.015 W are joined in series to provide a supply to a resistance of 8.5 W. What is the current drawn from the supply and its terminal voltage?

Solution: When the cells are connected in series, the total e.m.f. is equal to sum of their e.m.fs. Therefore, total e.m.f. of the cells,

E = 2.0 × 6 =12.0 V

Also, total internal resistance,

r = 0.015 × 6 = 0.09 W

External resistance, R = 8.5 W

Therefore, the current drawn from the supply,

Terminal voltage, V = I R = 1.397 × 8.5 = 11.875V.

2. Three identical cells each of e.m.f. 4 V and internal resistance r are connected in series to a 6 W resistor. If the current flowing in the circuit is 1.5 a, calculate (a) internal resistance of each cell and (b) the terminal voltage across the cells.

Solution: The three cells are connected in series to the external resistance R as shown in figure.

(a) Here E = 4 V; I = 1.5 A and R = 6 W

The total e.m.f. in the circuit,

E' = 3 E = 3 ´ 4 = 12 V

Total resistance of the circuit,

R' = R + 3 r = 6 + 3r

Therefore, the current in the circuit,

or or W

(b) The terminal voltage across each cell,

= 3 V.

3. Four identical cells, each of e.m.f. 2 V, are joined in parallel providing supply of current to external circuit consisting of two 15 W resistors joined in parallel. The terminal voltage of the cells as read by an ideal voltmeter is 1.6 V. Calculate the internal resistance of each cell.

Solution: Let r be the internal resistance of each cell and 1, and current in the circuit. Since the cells are connected in parallel, total e.m.f. in the circuit = e.m.f. of one cell = 2 V

Further, total internal resistance of the cells is given by or

Let R be the resistance of the parallel combination of two

15 W resistors. Then, the total external resistance, W

Also, the internal resistance of the parallel combination of the cells is given by

or or W.

Energy Transfer in an Electric Circuit

Thermal Energy Produced in the Resistor

Thus, a current i for a time t through a resistance R increases the thermal energy by . The power developed is

Using Ohm's law, this can also be written as

Illustrations

1. A resistor develops 400 J of thermal energy in 10 s when a current of 2 A is passed through it. (a) Find its resistance. (b) If the current is increased to 4 A, what will be the energy developed in 10 s.

Solution: (a) Using ,

or 10 W

(b) The thermal energy developed, when the current is 4 A, is

= .

2. How much current will a heater rated 2.5 kW draw, when connected to a 250 V electric supply?

Solution: Here, P = 2.5 kW = 2500 W; V = 250 volt

Now, A

Efficiency of a Cell

In general, the efficiency of any system is given by;

Efficiency, h =

Consider a cell of e.m.f. E and internal resistance r delivering power to a resistance R as shown in Fig. 31. If I is the circuit current, then useful power developed is I²R and power wasted in the internal resistance of the cell is I² r. Total power developed (input power) by the cell is I²R + I²r (= E I).

or,

It is clear that greater the value of external resistance (R), the higher is the efficiency of the cell.

Potentiometer

It is primarily used for measuring emf and the internal resistance of an unknown battery. In its simplest form it is straight piece of resistance wire, usually a metre long, fixed between two points A and B with a battery of output voltage connected between the two ends, as shown in figure. The voltage drop along the wire AB is assumed to be uniform and that in the connecting leads zero.

Consider a battery of emf E with a galvanometer G, as shown in figure whose one end is connected with the terminal A and the other to a sliding contact P that moves over the wire.

The aim is to locate a point on the wire at which the galvanometer shows no deflection. At this position, the voltage drop across the length AP of the wire is equal to the emf of the battery. That is,

Note : that E must be less than .

Uses of Potentiometer

Figure (a)

Comparison of emf's. . see fig. (a)

Figure (b)

Measurement of internal resistance

. see fig. (b)

Illustrations

1. A battery is connected to a potentiometer and a balance point is obtained at 84 cm along the wire. When its terminals are connected by a 5 W resistor, the balance point changes to 70 cm.

Calculate the internal resistance of the cell.
Find the new position of the balance point when the 5 W resistor is changed by 4 W, resistor.

Solution: (i) Here, = 84 cm; l = 70 cm; R = 5 W

We know W

(ii) If R = 4 W, then cm.

2. The resistance of a potentiometer wire of length 10 m is 20 ohm. A resistance box and a 2 volt accumulator are connected in series with it. What resistance should be introduced in the box to have a potential drop of one microvolt per millimetre of the potentiometer wire?

Solution: Here, resistance of the potentiometer wire, R= 20W

Required potential drop along the potentiometer wire = 1 mV mm = ( V) ´ ( m)

= V m.

The length of the potentiometer wire, l = 10 m

Therefore, potential drop along the potentiometer wire, V

If I is current through the potentiometer wire, then A

If is the resistance that should be introduced from the resistance box, then

or or 3980 W

3. With a certain cell, the balance point is obtained at 65 cm from the end of a potentiometer wire. With another cell, whose e.m.f. differs from that of first by 0.1 V, the balance point is obtained at 60 cm. Find the e.m.f. of each cell.

Solution: Let and be the e.m.fs. of two cells.

Here, the balancing length for the first cell, =

65 cm and the balancing length for the second cell,

= 60 cm

Since balancing length for the first cell is less than that for the second cell, the e.m.f. of second cell is less than that of the first i.e. . As the e.m.f. of two cells differ by 0.1 V it follows that

Now, or or = 1.3 V

Also, = 1.3 _ 0.1 = 1.2 V.

4. (a) If the galvanometer in the circuit shown in figure reads zero, calculate the value of the resistor R, assuming that the 12 V source has negligible internal resistance.

(b) If cool air is blown across the wire-wound resistor, what effect will be noticed and why?

Solution: (a) If I is current through the main circuit i.e. through the series combination of the resistors R and connected to 12 V source, then

…(i)

Also, …(ii)

Substituting for I in equation (i), we have

or W = 2 kW

(b) When the current flows through a resistor, its temperature increases due to heat produced across it. If cool air is blown across the wire-wound resistor, its temperature will get lowered. Due to this, the resistance of the wire-would resistor will decrease and the current in the circuit will increase. As a consequence of it, the potential difference across the unknown resistor R will not remain 2 V. Since the e.m.f. of cell connected across R will no longer balance the potential drop across R, the galvanometer will show deflection.