RESISTANCE Introduction All material offer a little or more opposition to the, flow of electric current through them ~ INFOTECHNO

3.1. INTRODUCTION

All material offer a little or more opposition to the, flow of electric current through them, this property of material is termed as resistance. Its symbol is R and its unit is ohm (Ω). The resistance of a material is not affected by the direction of flow of current, hence it remains the same on D.C. and A.C. course, a difference exists in the electric consumption on D.C. and A.C. of a conductor.

3.2. RESISTANCE LAW

The resistance of a conductor depends on the following three factors :

(i) Length of a conductor R α 1

(ii) Cross-sectional area of a conductor R α 1

(iii) Specific resistance of conductor p (rho).

Hence,

Where, R = resistance of conductor, ohms

I = length of conductor, cm

A = cross-sectional area of conductor, cm²

ρ (rho) = specific resistance of conductor. ohm-centimetre.

3.3. SPECIFIC RESISTANCE

The resistance of a piece of a material having a length of one centimeter and a cross-sectional area of one square centimetre is termed as its specific resistance. It is expressed in ohm-centimetres. It is also known as resistivity.

If I = 1cm; A = 1cm²

Then ρ = R ohm-centimetre.

Table : Specific Resistance

S. No.	Name of Material	Specific Resistance Ohm-cm. at 20⁰C
1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12.	Silver Copper Aluminum Tungsten Iron Platinum Lead Manganin Eureka Carbon Mercury Nichrome	1.64 x 10^-6 1.72 x 10^-6 2.69 x 10^-6 5.51 x 10^-6 10.0 x 10^-6 11.0 x 10^-6 22.6 x 10^-6 40.0 x 10^-6 49.0 x 10^-6 70.0 x 10^-6 95.8 x 10^-6 108.5 x 10^-6

Note. In MKS system, use 10^-6 in place of 10^-6 with the S.R. value of a material.

3.4. TEMPERATURE CO-EFFICIENT

The resistance of a material also depends on its temperature. In most of the materials, resistance increases with an increase in temperature but in certain other materials (e.g.. carbon) the resistance decreases .pith an increase in temperature. Hence. the change in resistance by increasing its temperature by 1⁰C is called the temperature coefficient of the material.

Where, R = resistance at t⁰C, ohms

R₀ = resistance at 0°C. ohms

α = temperature coefficient, per degree centigrade (1°C)

t = temperature. °C.

Example 3.1. Calculate the resistance of a copper wire of 2 km in length and 5 mm² in cross-sectional area if the resistivity of the copper is 1.7 x 10^-6 ohm-cm.

Solution. Given : Length of wire = 2 km = 2 x 10³ m

Cross-sectional area of wire = 5 mm² =5 x 10^-6 m²

Resistivity = 1.7 x 10^-6 ohm-cm

1.7 x 10^-8 ohm-cm

Example 3.2. The resistance of a coil made of copper wire 100 ohms at 0⁰C. Calculate the resistance at 30⁰C. Given α = 0.004/⁰C.

Solution. Given: Resistance at 0⁰C, R₀= 100 ohms

Temperature, t = 30⁰C

Temperature coefficient, α = 0.004/⁰C.

Hence Resistance at 1⁰C, R₁= R₀(1 + αt)

= 100(1 + 0.004 x 30)

= 100(1.12)

= 112 ohms

3.5. OHM’S LAW

In a closed d.c. circuit the potential difference developed across a conductor (or a resistor) is directly proportional to the current flowing through it if the temperature and other physical conditions of the conductor have been kept constant.

Fig. 3.1. Simple Circuit

3.6. CONSUMPTION OF ELECTRICAL POWER

The consumption of electrical power of a circuit is calculated in watts. Its symbol is P.

Where, P = electrical power or consumption, watts

I = current, amps.

V = p.d.. volts,

R = resistance, ohms.

The commercial unit of electric power consumption is kilo watt hour (kWh)

1 kWh is called a unit is also referred as 'Board Of Trade unit’

Example 3.3. Calculate the pid, and wattage of a resistor of 250 ohms if the magnitude of maximum current flowing through it is 250 mA.

Solution, Given : resistance, R = 250 kΩ

Current, I = 250 mA

= 250 x 10^-3 A

Voltage, V = I. R

= 250 x 10^-3x 250

= 62.5 volts

And wattage, P = I²R

= (250 x 10^-3)^-2x 250

= 15.625 watts.

Example 3.4 The power rating of a 2 kilo-ohms resistor is 80 watts. Calculate the maximum permissible current which can flow safely through it.

Soloution. Given: resistance, R = 2 kΩ

= 2 x 10³Ω

wattage, P = 80 watts

:. Using the formula P = I².R

3.7 RESISTORE IN SERIES

A combination of two or more resistors which has one and only one path for the flow of current is called a series circuit. In this circuit each resistor has its individual voltage drop, hence total voltage

Fig. 1.2. Resistors in series

In this way, the total circuit resistance is equal to the sum of all the resistances. If all resistors have the same resistance value (R) and their number is (n) then the total resistance

Total electric consumption:

3.8 RESISTORS IN PARALLEL

A combination of two more resistors in which all the resistors are getting the same voltage (i.e.., they are connected to same source of e.m.f) is called a parallel circuit. The current drainage of each resistor connected in parallel is different, hence total current-

Fig. 3.3. Resistors in parallel

In this way, the total circuit resistance is lesser than the least resistance. If all the resistors have the same resistance value (R) and their number is (n) then the total resistance -

Note. The electric consumption of a resistor depends on its resistance and the current flowing through it (P = I².R), thus the electric consumption formula of resistor connected in series or in parallel is the same (i.e., P_T= P₁+ P₂+ P₃+ …).

Example 3.5. Two resistors of 45 and 67.5 ohms respectively are connected in series across a source of 240 volts. Calculate (i) circuit current, (ii) voltage drop of each resistor.

Solution. Given : Source voltage, V = 240 volts

Values of resistors, R₁= 45 ohms

R₂ = 67.5 ohms

Total resistance, R_T= R₁+ R₂

= 45 + 67.5

= 112.5 ohms

Example 3.6. Calculate the maximum voltage which can be applied across a combination of two resistors connected in parallel.

(i) 50 kilo ohms, 0.5 watts

(ii) 33 kilo ohms, 0.25 watts.

Solution. Given : Resistance R₁= 50 kilo ohms

= 50 x 10³ohms

R₂= 33 kilo ohms

= 33 x 10³ohms

Wattage, P₁= 0.5 watts

P₂= 0.25 watts

Now, using the formula P = I²x R

Note. Resistors may also be connected so as to form a composed circuit. There are following two types of compound circuits.

(i) Parallel Series Circuit. If a few parallel combinations of resistors are connected in

series then they form a parallel-series circuit. Here, calculate equivalent resistance of parallel groups first and then calculate total resistance by using series formula.

(ii) Series-parallel Circuit. If a few series combinations of resistors are connected in

parallel then they form a series-parallel circuit. Here, calculate equivalent resistance of series groups first and then calculate total resistance by using parallel formula.

Example 3.7 Three resistors of 15 ohms each are connected in parallel. Another parallel group of two resistors of 10 ohms each is connected in series with the first group. If the source voltage is 24 volts, calculate total circuit resistance, circuit resistance, circuit current and voltage drop of each group.

Solution. Total resistance of first group,

Example 3.8 Calculate the total circuit resistance of the circuit shown in Fig. 3.4.

Fig. 3.4.

Solution.

Resistance of CDB circuit

= 1 + 1 = 2 ohms

Resistance between C.B.

3.9. KIRCHHOFF’S LAWS

Famous scientist Kirchhoff derived the following two laws for various kinds of d.c. circuits :

(i) Current Law. “The sum of current following towards a junction is equal to the sum of currents following away from the junction.” If the currents following towards the junction are i₁, i₂ and that following away are i₃, i₄, i₅, then

i₁ + i₂ = i₃ + i₄ + i₅

Alternately – The algebraic sum of currents at a junction of a circuit is always zero

Hence current cannot accumulate at any junction of a circuit.

(ii) Voltage Law. “The algebraic sum of the e.m.fs. applied to a circuit is equal to the algebraic sum of products of resistances and currents flowing through them.

Note. Kirchhoff’s Law are used for the calculation of currents in complex circuits.

Example 3.9. Calculate the magnitude of current flowing through 5 ohms resistor of Fig. 3.5

Fig. 3.5 (For example 3.9)

Solution. Let the current through 4 ohms resistor be i₁ and through 6 ohm resistor be i₂. Thus the current through 5 ohm resistor will be i₁+ i₂.

For circuit ABD 6 = 5 (i₁ + i₂) + 4 (i₁)

or 6 = 5 i₁ + 5 i₂+ 4 i₁

or 6 = 9 i₁+ 5 i₂

or 9 i₁+ 5 i₂= 6 …(i)

For circuit ABCD 6-3 = 5 (i₁+ i₂) + 6 (i₂)

or 3 = 5 i₁+ 5 i₂+ 6 i₂

or 3 = 5 i₁+ 11 i₂

or 5 i₁+ 11 i₂ = 3 …(ii)

By solving equation (i) and (ii)

Note. (-) sign indicates that the direction of i₂is opposite to the direction assumed

By substituting i₂ = -0.04 in equation (i)

3.10. WHEATSTONE BRIDGE

Scientist Mr. Wheatstone designed a combination of a resistor for the calculation of ohmic value of an unknown resistor on the basis of Kirchhoff’s laws. The combination is called a Wheatstone Bridge.

Here (in Fig 3.6) P and Q form ration arms. The ratio of P and Q may be selected out of 1 : 1, 1 : 10, 1 : 100 or 10 : 1, 100 : 1, The value of R can be adjusted between 1 to 5000 ohms.

‘Resistor S is an unknown resistor.

Fig. 3.6. Wheatstone Bridge

Principle. For the determination of value of an unknown resistor S select a suitable ration between P and Q and adjust the value of R in such a way that the current flowing through the galvanometer becomes zero.

In this state

This formula is known as Wheatstone Bridge formula.

Use. An apparatus called Post Office Box has been designed on the principle of Wheatstone Bridge for the determination of value of an unknown resistor. Besides it, an Impedance Bridge has also been designed on the same principle for the determination of value of unknown inductors and capacitors.