Electricity Notes - Electric Current, Potential, Ohm's Law, Resistance, Heating Effect and Power

Electric Current and Circuit

• The flow of electric charges through a conductor (e.g., metallic wire) is called an electric current.
• In a torch, the cells (or a battery, when placed in proper order) provide electric current through the bulb to glow.
• A continuous and closed path of an electric current is called an electric circuit. If the circuit is broken (or the switch is turned off), the current stops.
• Electric current is measured as the amount of charge flowing through a particular area in unit time, i.e., it is the rate of flow of electric charges.
• In a circuit of metallic wire, electrons constitute the flow of charges. Electrons were unknown during the discovery of electricity. So, electric current was considered as the flow of positive charges, and their direction of flow was taken as the direction of electric current.
• Conventionally, in an electric circuit, the direction of electric current is taken as opposite to the direction of the flow of electrons (negative charges).
• Consider a net charge Q flows across a cross-section of a conductor in time t. Then, the current I through the cross-section is:

• The SI unit of electric charge is coulomb (C). It is equivalent to the charge contained in nearly 6 × 10¹⁸ electrons. An electron has a negative charge of 1.6 × 10^–19 C.
• The SI unit of electric current is ampere (A), named after Andre-Marie Ampere (France, 1775–1836).
• One ampere represents the flow of one coulomb of charge per second, i.e., 1 A = 1 C/1 s.
• Small quantities of current are expressed in milliampere (1 mA = 10^–3 A) or in microampere (1 µA = 10^–6 A).
• Ammeter: An instrument to measure electric current in a circuit. It is always connected in series in a circuit.

• In a circuit, electric current flows from the positive terminal of the cell to the negative terminal through the bulb and ammeter.

Problem: A current of 0.5 A is drawn by a filament of an electric bulb for 10 minutes. Find the amount of electric charge that flows through the circuit.

Solution: I = 0.5 A; t = 10 min = 600 s.

Q = I t

= 0.5 A × 600 s = 300 C

‘Flow’ of charges inside a wire: Inside a solid conductor, atoms are packed together. But electrons can easily travel through it, almost as if they are in a vacuum. When a current flows through a conductor, the electrons move with a certain average ‘drift speed’ (1 mm s^-1 for a typical copper wire). However, an electric bulb lights up as soon as the switch is turned on. It is not due to the movement of an electron from one terminal to the other terminal through the bulb.

Electric Potential and Potential Difference

• Electric charges do not flow in a metallic conductor (e.g., copper wire) by themselves. Here, gravity has no role.
• Electrons move only if there is a difference of electric pressure in the conductor. It is called potential difference.
• Potential difference can be produced by a battery. The chemical action within a cell generates the potential difference across the terminals of the cell, even when no current is drawn from it.
• When the cell is connected to a conducting circuit, the potential difference sets the charges in motion and produces an electric current. To maintain this current, the cell has to expend chemical energy stored in it.
• The potential difference between two points in an electric circuit is the work done to move a unit charge from one point to the other.
• Potential difference (V) between two points = Work done (W)/Charge (Q).

V = W/Q       or       W = VQ

• The SI unit of electric potential difference is volt (V), named after Alessandro Volta (Italy, 1745–1827).
• One volt is the potential difference between two points in a current-carrying conductor when 1 joule of work is done to move 1 coulomb charge from one point to another.
• Voltmeter: An instrument to measure potential difference. It is always connected in parallel across points between which the potential difference is to be measured.
• Problem: How much work is done in moving a charge of 2 C across two points having a potential difference of 12 V?

Solution:

Amount of charge Q = 2 C.

Potential difference V = 12 V.

Amount of work W = VQ.

= 12 V × 2 C = 24 J.

Circuit Diagram

Conventional symbols of some commonly used components in circuit diagrams:

Ohm’s Law

• Set up a circuit consisting of a nichrome wire XY (0.5 m length), an ammeter, a voltmeter, and four cells of 1.5 V each. (Nichrome = an alloy of Ni, Cr, Mn, & Fe.)

• Using one cell, note the ammeter reading I for the current and voltmeter reading V for potential difference across the nichrome wire. Repeat this using 2, 3, and 4 cells.

• In each case, V/I value is approximately the same.
• V–I graph is a straight line. Thus, V/I is a constant ratio.

• In 1827, Georg Simon Ohm (Germany, 1787–1854) found out the relationship between current and the potential difference. Potential difference (V) across the ends of a metallic wire in an electric circuit is directly proportional to the current I flowing through it at constant temperature. It is called Ohm’s law.

V ∝ I

V/I = constant = R

V = IR or I = V/R

• R is a constant for the given metallic wire at a given temperature and is called its resistance. It is the property of a conductor to resist the flow of charges. Its SI unit is ohm (Ω).
• According to Ohm’s law, R = V/I.
• If the potential difference is 1 V and the current is 1 A, then the resistance (R) of the conductor is 1 Ω.
• A conductor having appreciable resistance is called a resistor. It is used to control electric current.
• The current through a resistor is inversely proportional to its resistance. If the resistance is doubled, the current gets halved. It has many practical applications.
• A component used to regulate current without changing the voltage source is called variable resistance. Rheostat is a variable resistor used to change resistance.

Electrical resistance of a conductor:

• Take a nichrome wire, a torch bulb, a 10 W bulb, an ammeter (0–5 A range), a plug key, and some wires.
• Set up a circuit by connecting four dry cells of 1.5 V each in series with the ammeter, leaving a gap XY in the circuit.

• Complete the circuit by connecting the nichrome wire in the gap XY. Plug the key. Note down the ammeter reading.
• Replace the nichrome wire with the torch bulb in the circuit and note down the ammeter reading.
• Repeat this with the 10 W bulb and any material component.
• The current is different for different components. In certain components, there is easy flow of electric current, while others resist the flow.
• Within a conductor, electrons are not completely free to move due to the attraction by atoms. So, resistance is increased, and motion of electrons is retarded.
• A component that conducts electricity and has a low resistance is called a good conductor.
• A component with a high resistance is a poor conductor.
• A component with very high resistance is called an insulator. It does not conduct electricity.

Factors on which the Resistance of a Conductor Depends

Note the ammeter reading in an electric circuit with:

• A nichrome wire of length l. Assume the reading = 1 A.
• Nichrome wire of twice the length (2l). Here, ammeter reading decreases to one-half (0.5 A).
• A thicker (larger cross-sectional area) nichrome wire of the same length l. Here, reading is increased. If the area is doubled, reading is also doubled (2 A).
• A copper wire of same length and cross-sectional area as that of first nichrome wire. Here, reading is changed.

• Thus, resistance of a conductor depends on (i) its length (ii) its area of cross-section and (iii) nature of material.
• Resistance of a uniform metallic conductor is directly proportional to its length (R ∝ l) and inversely proportional to the area of cross-section (R ∝ 1/A).

R = ρl/A

• Where ρ (rho) is a constant of proportionality and is called electrical resistivity of the material of conductor. It is a characteristic property of the material.
• SI unit of resistivity is Ω m.
• Resistivity of metals & alloys is very low (Range: 10^–8 to 10^–6 Ω m). They are good conductors of electricity.
• Resistivity of insulators (rubber, glass etc.) is very high (Range: 10¹² to 10¹⁷ Ω m).
• Resistance and resistivity vary with temperature.
• Resistivity of an alloy is generally higher than that of its constituent metals. Alloys do not oxidise (burn) readily at high temperatures. So, they are used in electrical heating devices, like electric iron, toasters etc.
• Tungsten is used for filaments of electric bulbs. Copper & aluminium are used for electrical transmission lines.

Electrical resistivity of some substances at 20°C

Problem:

• a) How much current will an electric bulb draw from a 220 V source, if the resistance of bulb filament is 1200 Ω?
• b) How much current will an electric heater draw from a 220 V source, if the resistance of heater coil is 100 Ω?

Solution:

• (a) V = 220 V; R = 1200 Ω.
• Current, I = V/R = 220 V / 1200 Ω = 0.18 A.
• (b) V = 220 V; R = 100 Ω.
• Current, I = 220 V / 100 Ω = 2.2 A.
• Thus the current drawn by an electric bulb and electric heater from the same 220 V source is different.

Problem: The potential difference between the terminals of an electric heater is 60 V when it draws a current of 4 A from the source. What current will the heater draw if the potential difference is increased to 120 V?

Solution:

• Potential difference V = 60 V, current I = 4 A.
• Resistance, R = V/I = 60 V / 4 A = 15 Ω.

• When the potential difference is increased to 120 V the current is given by:

I = V/R = 120 V / 15 Ω = 8 A.

Problem: Resistance of a metal wire of length 1 m is 26 Ω at 20°C. If the diameter of the wire is 0.3 mm, what will be the resistivity of the metal at that temperature? Predict the material of the wire.

Solution:

• Resistance R of the wire = 26 Ω.
• Diameter d = 0.3 mm = 3 × 10^–4 m.
• Length l of the wire = 1 m.
• Resistivity of the metallic wire, ρ = (RA/l)

= (R π d²/4 l).

= 1.84 × 10^–6 Ω m.

• This is the resistivity of manganese.

Problem: A wire of given material having length l and area of cross-section A has a resistance of 4 Ω. What would be the resistance of another wire of the same material having length l/2 and area of cross-section 2A?

Solution:

• For first wire: R₁ = ρl/A = 4 Ω.
• For second wire: length = l/2, area = 2A.
• R₂ = ρ(l/2)/(2A) = (ρl/A)/4 = 4 Ω / 4 = 1 Ω.

Resistance of a System of Resistors

In electrical gadgets, resistors are used in various combinations based on Ohm’s law.

There are two methods of joining the resistors: Resistors in series and Resistors in parallel.

Resistors in Series

• Join three resistors having resistances R₁, R₂, and R₃ (e.g., 1 Ω, 2 Ω, 3 Ω) in series. Connect them with a 6 V battery, an ammeter, and a plug key. Note the ammeter reading.

• Change the position of the ammeter to anywhere in between the resistors.
• The value of the current in the ammeter is the same, i.e., in a series combination of resistors, the current is the same in every part of the circuit or the same current through each resistor.
• Insert a voltmeter across the ends X and Y of the series combination of three resistors. Note its reading. It gives the potential difference (V) across the series combination of resistors. Now measure the potential difference across the two terminals of the battery. Compare the two values.
• Now insert the voltmeter across the ends X and P of the first resistor.

• Measure the potential differences V₁, V₂, and V₃ across the first, second, and third resistors separately.
• The total potential difference V across a combination of resistors in series is equal to the sum of potential differences across the individual resistors.

V = V₁ + V₂ + V₃

• Let I be the current through this electric circuit. The current through each resistor is also I. The three resistors joined in series can be replaced by an equivalent single resistor of resistance R, such that the potential difference and the current remain the same. Applying Ohm’s law to the entire circuit, we have:

V = I R

• On applying Ohm’s law to the three resistors separately:

V₁ = I R₁,       V₂ = I R₂,       V₃ = I R₃

I R = I R₁ + I R₂ + I R₃

or

R_s = R₁ + R₂ + R₃

• When several resistors are joined in series, the resistance of the combination R_s equals the sum of their individual resistances, R₁, R₂, R₃.

Problem:

An electric lamp, whose resistance is 20 Ω, and a conductor of 4 Ω resistance are connected to a 6 V battery. Calculate (a) the total resistance of the circuit, (b) the current through the circuit, and (c) the potential difference across the electric lamp and conductor.

Solution:

• a) Resistance of electric lamp, R₁ = 20 Ω.

Resistance of the conductor connected in series, R₂ = 4 Ω.

Then the total resistance, R = R₁ + R₂ = 20 Ω + 4 Ω = 24 Ω.

• b) The total potential difference across the two terminals of the battery, V = 6 V.

The current through the circuit is I = V/R_s = 6 V / 24 Ω = 0.25 A.

• c) Potential difference across the electric lamp: V₁ = I R₁ = 20 Ω × 0.25 A = 5 V.

Potential difference across the conductor: V₂ = I R₂ = 4 Ω × 0.25 A = 1 V.

• If the series combination of electric lamp and conductor is replaced by an equivalent resistor, its resistance would be R = V/I = 6 V / 0.25 A = 24 Ω.
• This is the total resistance of the series circuit; it is equal to the sum of the two resistances.

Resistors in Parallel

• Make a parallel combination, XY, of three resistors having resistances R₁, R₂, and R₃ in an electric circuit. Connect a voltmeter in parallel with the resistors.

• Note the ammeter reading (I) and the voltmeter reading.
• Voltmeter shows the potential difference V, across the combination. The potential difference across each resistor is also V. This can be checked by connecting the voltmeter across each individual resistor.
• Insert the ammeter in series with the resistor R₁. Note the ammeter reading, I₁. Similarly, measure the currents I₂ and I₃ through R₂ and R₃ respectively.

• It is observed that the total current I, is equal to the sum of the separate currents through each branch.

I = I₁ + I₂ + I₃

• Let R_p be the equivalent resistance of the parallel combination of resistors.

I = V/R_p

• On applying Ohm’s law to each resistor, we have:

I₁ = V/R₁,       I₂ = V/R₂,       I₃ = V/R₃

V/R_p = V/R₁ + V/R₂ + V/R₃

1/R_p = 1/R₁ + 1/R₂ + 1/R₃

• Thus, the reciprocal of the equivalent resistance of a group of resistances joined in parallel is equal to the sum of the reciprocals of the individual resistances.

Problem:

In the circuit diagram given, suppose the resistors R₁, R₂, and R₃ have the values 5 Ω, 10 Ω, 30 Ω, respectively, which have been connected to a battery of 12 V. Calculate (a) the current through each resistor, (b) the total current in the circuit, and (c) the total circuit resistance.

Solution:

• R₁ = 5 Ω, R₂ = 10 Ω, and R₃ = 30 Ω.
• Potential difference across the battery, V = 12 V.
• This is also the potential difference across each of the individual resistors.
• The current I₁ through R₁ = V/R₁ = 12 V / 5 Ω = 2.4 A.
• The current I₂ through R₂ = V/R₂ = 12 V / 10 Ω = 1.2 A.
• The current I₃ through R₃ = V/R₃ = 12 V / 30 Ω = 0.4 A.
• The total current in the circuit, I = I₁ + I₂ + I₃ = (2.4 + 1.2 + 0.4) A = 4 A.
• The total resistance R_p is:

Thus, R_p = 3 Ω.

Problem:

If in the figure given below, R₁ = 10 Ω, R₂ = 40 Ω, R₃ = 30 Ω, R₄ = 20 Ω, R₅ = 60 Ω, and a 12 V battery is connected to the arrangement. Calculate (a) the total resistance in the circuit, and (b) the total current flowing in the circuit.

Solution:

• Suppose we replace the parallel resistors R₁ and R₂ by an equivalent resistor of resistance, R'.
• Similarly, we replace the parallel resistors R₃, R₄, and R₅ by an equivalent single resistor of resistance R''.
• 1/R' = 1/R₁ + 1/R₂ = 1/10 + 1/40 = 5/40, i.e., R' = 8 Ω.
• 1/R'' = 1/R₃ + 1/R₄ + 1/R₅ = 1/30 + 1/20 + 1/60 = 6/60, i.e., R'' = 10 Ω.
• Thus, the total resistance, R = R' + R'' = 8 Ω + 10 Ω = 18 Ω.
• The current in the circuit, I = V/R = 12 V / 18 Ω = 0.67 A.

Disadvantages of Series Circuit:

• In a series circuit, the current is constant throughout the electric circuit. So, it is impracticable to connect an electric bulb and an electric heater in series, because they need currents of different values.
• When one component fails, the circuit is broken, and none of the components works. For example, it is very difficult to locate the dead bulb in fairy lights.

Advantages of Parallel Circuit:

• A parallel circuit divides the current through the electrical gadgets.
• The total resistance in a parallel circuit is decreased. This is helpful when each gadget has different resistance and requires different current to operate properly.

Heating Effect of Electric Current

• A battery or a cell is a source of electrical energy. It generates potential difference that sets the electrons in motion to flow the current through a resistor or a system of resistors.
• A part of the source energy to maintain the current may be consumed into useful work (e.g., rotation of an electric fan). The rest of the energy is lost as heat. For example, an electric fan becomes warm if used for a longer time.
• If an electric circuit is purely resistive (i.e., a configuration of resistors only connected to a battery), the source energy is dissipated entirely as heat. This is called the heating effect of electric current.
• Consider a current I flowing through a resistor of resistance R. Let the potential difference across it be V and t be the time during which a charge Q flows across.

• The work done in moving the charge Q through a potential difference V is VQ. Therefore, the source must supply energy equal to VQ in time t. Hence, the power input to the circuit by the source is:

• Or the energy supplied to the circuit by the source in time t is P × t, i.e., VIt. This energy is dissipated in the resistor as heat (H).

H = VIt

• Applying Ohm’s law (V = IR):

H = I²Rt

• This is known as Joule’s law of heating. It implies that heat produced in a resistor is directly proportional to:
• The square of current for a given resistance.
• Resistance for a given current.
• The time for which current flows through the resistor.
• In practical situations, when an electric appliance is connected to a voltage source, the equation H = I²Rt is used after calculating the current using the relation I = V/R.

Problem:

An electric iron consumes energy at a rate of 840 W when heating is at the maximum rate and 360 W when the heating is at the minimum. The voltage is 220 V. What are the current and the resistance in each case?

Solution:

• Power input, P = VI
• Thus, the current I = P/V
• (a) When heating is at the maximum rate:

I = 840 W / 220 V = 3.82 A

Resistance of the electric iron is R = V/I = 220 V / 3.82 A = 57.60 Ω

• (b) When heating is at the minimum rate:

I = 360 W / 220 V = 1.64 A

Resistance of the electric iron is R = V/I = 220 V / 1.64 A = 134.15 Ω

Problem:

100 J of heat is produced each second in a 4 Ω resistance. Find the potential difference across the resistor.

Solution:

• H = 100 J,       R = 4 Ω,       t = 1 s,       V = ?
• H = I²Rt

So, the current through the resistor is:

I = √(H / Rt) = √[100 J / (4 Ω × 1 s)] = 5 A

• Thus, the potential difference across the resistor is:

V = IR = 5 A × 4 Ω = 20 V

Practical Applications of Heating Effect of Electric Current

Due to the heating effect of electric current, electrical energy is lost as heat. Also, it may alter the properties of components in electric circuits. But heating effect (Joule’s heating) has many useful applications:

• To make devices such as electric laundry iron, electric toaster, electric oven, electric kettle, and electric heater.
• To produce light in an electric bulb. Here, the filament made of metals with a high melting point can retain much heat. So, it gets very hot and emits light.

  For example, tungsten (melting point 3380°C) is used to make filaments. The filament should be thermally isolated, using insulating support. The bulbs are filled with chemically inactive nitrogen and argon gases to prolong the life of the filament. Most of the power consumed by the filament appears as heat, but a small part is radiated as light.

• To make a fuse used in electric circuits. It protects circuits and appliances by stopping the overflow of electric current. The fuse is placed in series with the device. It consists of a piece of wire made of a metal or an alloy of suitable melting point (aluminium, copper, iron, lead, etc.). During the overflow of the current, the temperature of the fuse wire increases. It melts the fuse wire and breaks the circuit. The fuse wire is encased in a cartridge of porcelain or similar material with metal ends.

  The fuses used for domestic purposes are rated as 1 A, 2 A, 3 A, 5 A, 10 A, etc. For example, when an electric iron which consumes 1 kW electric power is operated at 220 V, 4.54 A current (1000/220) flows in the circuit. In this case, a 5 A fuse must be used.

Electric Power

• Power is the rate of doing work or rate of consumption of energy.
• The equation H = I²Rt gives the rate at which electric energy is dissipated or consumed in an electric circuit. This is also termed as electric power. The power P is given by:

P = VI

P = I²R = V²/R

• The SI unit of electric power is watt (W). It is the power consumed by a device that carries 1 A of current when operated at a potential difference of 1 V.

1 W = 1 volt × 1 ampere = 1 V A

• Watt is a very small unit. So, practically, a much larger unit called kilowatt (1000 watts) is used.
• Electrical energy is the product of power and time. Its unit is watt hour (W h). One watt hour is the energy consumed when 1 watt of power is used for 1 hour.
• The commercial unit of electric energy is kilowatt hour (kW h), commonly known as ‘unit’.

1 kW h = 1000 watt × 3600 second = 3.6 × 10⁶ watt second = 3.6 × 10⁶ joule (J)

• In an electric circuit, electrons are not consumed. We pay for energy to move electrons through electric gadgets.

Problem:

An electric bulb is connected to a 220 V generator. The current is 0.50 A. What is the power of the bulb?

Solution:

• P = VI
• P = 220 V × 0.50 A = 110 W

Problem:

An electric refrigerator rated 400 W operates 8 hours/day. What is the cost of the energy to operate it for 30 days at Rs 3.00 per kW h?

Solution:

• Total energy consumed by the refrigerator in 30 days:

400 W × 8.0 hr/day × 30 days = 96000 W h = 96 kW h

• Cost of energy to operate the refrigerator for 30 days:

96 kW h × Rs 3.00 per kW h = Rs 288.00