Electricity

The electric circuit is the continuous path which allows electric current or electrons to flow through it. An electric circuit consists of connecting wires, other elements (like electric bulb, resistances, key etc.) and a battery.

The SI unit of electric current is Ampere, commonly denoted by letter A. 1 Ampere is defined as - When an amount of charge equivalent to 1 coulomb flows through a cross-section of a conductor in 1 second.

We know that, the charge of an electron, e = 1.6 X 10^-19 Coulomb

Therefore, 1.6 X 10^-19 C of charge is equivalent to = 1 electron

By unitary method,
If the charge is 1C then no of electrons

An electric cell or a battery helps to maintain a potential difference across a conductor.

The potential difference between two points is said to be 1 volt (1V) if the amount of work done in moving a charge of 1 Coulomb (1C) between the two points is equivalent to 1 Joule (1J)

The charge passing through the battery is 1 coulomb. The potential difference is 6V. We have to find out the energy which is equal to the work done in moving the charge through 6V. Now,

Since the work done on each coulomb of charge is 6 joules. Therefore, the energy given to each coulomb of charge is also 6 joules.

The electrical resistance of a conductor depends on the factors listed below:

1) Nature of the material of the conductor – eg. A wire made of copper would be a better conductor than a wire made of iron

2) Length of the conductor – eg. A shorter wire would be a better conductor than a longer wire

3) Area of cross-section of the conductor – eg. A thick copper wire is a better conductor than a thin copper wire

4) Temperature of the conductor – resistance goes up with increase in temperature

The current will flow more easily through a thick wire than a thin wire of the same material when connected to the same source. This is mainly because the resistance of a wire is inversely proportional to the square of its diameter. And, a thick wire has greater diameter than a thin wire. Therefore, thick wire will offer less resistance to the current flow and thin wire will offer more resistance to the current flow. Thus, current flows easily through thick wire.

We know, From Ohm’s Law:

Since the resistance is constant and the potential difference is decreased to half of the former value. Thus, the current through the component will also decrease to half of its former value.

The coils (or heating elements) of toasters and electric irons are made of an alloy rather than a pure metal due to the following two reasons:

1) The resistivity of an alloy is much higher than that of a pure metal, and

2) An alloy does not undergo oxidation easily even at high temperature

(a) The electrical resistivity of iron is . And, the electrical resistivity of mercury is . Since the resistivity of mercury is greater than that of iron, it will offer more hindrance to current flow as compared to iron. Therefore, iron is a better conductor than mercury.

(b) Silver metal has the lowest resistivity nearly about 1.60 x 10^-8m. Therefore, silver metal is the best conductor of electricity.

In this problem, we have 3 cells of voltage2 V each. Thus, the total voltage of the three cells connected together to form a battery will be 3 X 2V = 6V. The circuit also consists 3 resistors of 5 , 8 , 12 respectively and a plug key, all connected in series as shown in the figure given below:

We redraw the circuit by including an ammeter circuit and a voltmeter across the 12 resistor, as shown in figure. Please note that the ammeter is always connected n series with the circuit but the voltmeter has been put in parallel with the 12 resistor

We will now calculate the current reading in the ammeter and potential difference reading in the voltmeter:

(i) Calculation of current flowing in the circuit. The three resistors of 5 , 8 and 12 are connected in series. So, net resistance in series is the sum of all the resistances.

Ans: Total resistance,

Potential difference, V = 6 V

And, Current, I = ?

We know,

Since the current in the circuit is 0.24 A, therefore, the ammeter will show reading of 0.24 A.

(ii) Calculation of potential difference across the 12 resistor. From (i) we have calculated that a current of 0.24 A flows in the circuit. Since, in series same current flows through the whole circuit. The same current of 0.24 A also flows through the 12 .

Now, for the 12 resistor:

Current, I = 0.24 A

Resistance, R = 12

And, Potential difference, V =?

We know that,

Thus, the potential difference across the 12 resistor is 2.88V. So, the voltmeter will show a reading of 2.88 V.

(a) The equivalent resistance of two resistances 1and 10⁶ connected in parallel will be less than 1.This is because when a number of resistances are connected in parallel, then their equivalent resistance is less than the smallest individual resistance. Here, since the smallest resistance is 1. Therefore, the net resistance will be less than 1

(b) The equivalent resistance of three resistances1,10³and 10⁶ connected in parallel will be less than 1 because the smallest resistance is 1.

The combined resistance R of three resistors (or electrical devices) R₁ R₂ and R₃, connected in parallel is given by the formula:

Here, Resistance of electric lamp, R₁ = 100

Resistance of toaster, R₂ = 50

Resistance of water filter, R 3 = 500

Putting these values of R₁ R₂ and R₃ in the above formula, we get:

Thus, the resistance of electric iron will also be 31.25.

Now, Potential difference, V= 220 V

Current, I =?

Thus, the current passing through the electric iron is 7.04 amperes.

The advantages of connecting electrical devices in parallel rather than series are as follows:

a) The voltage is same across all the elements in case of parallel circuits. Even if any one of the electrical devices stop working, it will not affect other electrical appliances and they will work normally. On the other hand, in series circuit, the current through all devices is same. If one electrical appliance stops, Then the whole circuit is broken and no current flows. As a result, all other appliances also stop working.

b) In case of parallel circuits, each electrical appliance works independently, i.e. it has its own switch. Each device can be turned on or turned off independently, without affecting other appliances. In series circuit, all the electrical appliances have only one switch due to which they cannot be turned on or turned off independently.

c) In parallel circuits, each electrical appliance gets the same voltage as that of the battery due to which all the appliances work properly. In series circuit, the appliances do not get the same voltage but get the same current. Each appliance has a different voltage.

d) Parallel circuits divide the current among electrical devices hence each device get necessary current to operate properly.

(a) In order to obtain a total resistance of 4 from three resistors of 2 , 3 and 6 .

Firstly, connect the two resistors of 3 and 6 in parallel to get a total resistance of 2 which is less than the lowest individual resistance.

R = R₁+ R₂

R = 2 + 2

R = 4

The arrangement of three resistors of 2,3 and 6 which gives a total resistance of 4 can now be represented as follows :

Thus, we can obtain a total resistance of 4by connecting parallel combination of 3 and 6 resistors in series with 2 resistor.

(b) In order to obtain a total resistance of 1 from three resistors of 2, 3 and 6Q, all the three resistors should be connected in parallel. This is because in parallel combination:

(a) The highest resistance can be secured (or obtained) by connecting all the four coils in series. In this case:

R = R₁ + R₂ + R₃ + R₄

R = 4 + 8 + 12 + 24

R = 48

Thus, the highest resistance which can be secured is 48 ohms.

(b) The lowest resistance can be secured by connecting all the four coils in parallel. In this case:

Thus, the lowest resistance which can be secured is 2

Electric power of the appliance the rate at which energy is delivered by a current.

The heating element of an electric heater is made of an alloy example nichrome, which has high resistance whereas the cord is made of copper metal which has very low resistance as compared to the heating element. Now, the heating element of an electric heater made of nichrome glows because it becomes red-hot due to the large amount of heat produced on passing current. On the other hand, the connecting cord of the electric heater made of copper does not glow because negligible heat is produced in it by passing the same current (due to its extremely low resistance).

First of all we will calculate the current (I) by using the given values of charge (Q) and time (t). We know that:

Current , I =

We will now calculate the resistance by using Ohm's law:

Resistance, = 1.87

And, Time, t= 1h = 60x60s = 3600 s

Heat generated, H=

=

= 4788397J

= or

Thus, the heat generated is 4.8 x 10⁶ joules.

Here,

Voltage, V = 220 V

Current, I = 5 A

Power, P = V x I = 220V X 5 A = 1100 Watts

Thus, the power of the motor is 1100 watts.

Now,

Power, P = 1100 W (Calculated above)

Time, t = 2 hour = 2 x 60 min x 60 sec

=

= 7200 s

So, Energy, E = P x t

E = 1100W X 7200 sec

= 7.92xJ

Thus, the energy consumed is 7.92 x 10⁶ joules.

Here, Current, I = 5 A

Resistance, R = 20

Time, t = 30 s

Now, we know, Heat produced is given as,

Thus, the heat developed is 15000 joules.

The resistance of wire is R. This wire is cut into five equal pieces as a result resistance of each piece of wire will be R⁄5.Now, when these five pieces are connected in parallel, the equivalent resistance for the combination is R’. We can write,
(Assume that each piece of wire is named R₁, R₂, R₃, R_4, R₅₎

Thus, the correct answer is: Option (d)

the electrical power is not represented as IR²

The bulb parameters are as given below:

Power, P = 100W

Potential difference, V = 220V

Then Resistance, R = ?

Now,

This resistance of the bulb will remain constant i.e. 484

Now, when we operate this bulb on 110 V, then,

Power, P =?

Potential difference, V = 110V

And, resistance, R = 484

P =

P = = = 25 W

Thus, the correct answer is: (d) 25 W.

Suppose the resistance of each one of the two wires is R.

Case (i) When both wires are connected in series. Then, net resistance will be R+
R = 2R.

Then, heat produced will be ( As H = I² X R X T)

Case (ii) When both wires are connected in parallel. Then, net resistance will be

Then, heat produced will be

Now, The ratio after simplifying will be as follow:-

Thus, option (c) is correct.

The volt meter is always connected in parallel connection across the two points where the potential difference is to be measured in the circuit.

Given:

The radius of wire,
r =

Area of the cross-section can be calculated by,

The Resistivity of the resistor,

Resistance, R= 10

Formula used:
Resistance,
Where,
ρ is the resistivity of the copper wire
L is the length of the copper wire
A is the area of the cross-section of the wire
Solving the above equation for the length

Putting the values in the above equation, we get

Thus, the length of copper wire required to make 10 resistance will be 122.5 meters.

(b) If the diameter is twice, D'=2D
Where,
D' is the new diameter
D is the initial diameter
Then, according to the question, the new diameter is twice the initial diameter.
D' = 2(0.5mm) = 1mm

Now, the new radius R' =D (Initial diameter of the cross-section)= D'/2 = 0.5 mm
As we know,

Putting the values in the above formula, we get

Thus, Resistance, R=2.5 Ω
Thus, if we twice the diameter of the copper wire the resistance become the quarter part of the initial resistance.

The graph between V and I plotted by using the given data is shown below:

We know that Resistance = Slope of the Graph=

In this problem,

The current, I =

Potential difference, V = 12 V

And, Resistance, R=?

And, From Ohm's law:

The value of the resistor is 4800

All the resistors are connected in series.

Thus, equivalent resistance, R = 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4

Potential difference, V = 9 V

And, Current, I=?

Since in series connection, the current throughout the circuit is same. Therefore, 0.67 A will flow through the 12 resistor.

Here, Potential difference = is 220 V

And the current, I = 5 A.

Thus, the resistance of the circuit is 44 ohms.

Now, we need to find the number of resistors should be connected in parallel to obtain a resultant resistance of 44. Suppose the number of 176 resistors required is ‘n’.

Then,

Thus, 4 resistors of 176 each should be connected in parallel.

(i) In order to get a resistance of 9 from three resistors of 6 each, we connect two 6 resistors in parallel and then this parallel combination is connected in series with the third 6 resistor .

For parallel connection:

Or R= 3

And for series connection: R= 3 + 6= 9

(ii) In order to get a resistance of 4 , their are multiple cases
(a) 1st Case
From three resistors of 6 each, we connect all the three 6 resistors in parallel:

Then, Net Resistance R= 2+ 2= 4.
(b)2nd Case
We connect two resistors of 6 Ω in series and another resistor of 6 Ω in parallel to them
R= 6 Ω + 6Ω= 12Ω
Then Net Resistance
Then R= 4 Ω

First we will calculate the resistance R of one bulb.

We know,

Therefore, resistance of one bulb is 4840

We will now calculate the total resistance of the circuit by using Ohm’s Law:

Thus, the total resistance of the circuit is 44.

Now, we need to find out the number of 4840 bulbs should be connected in parallel to obtain a total resistance of 44. Suppose the number of 4840 bulbs required is x. Then:

=110

Thus, 110 bulbs can be connected in parallel in this circuit.

Case (i): When each coil is used separately

Here, Potential difference, V = 220 V

Resistance of a coil, R = 24

Using Ohm’s Law, the current passing through one coil is given as,

Case (ii) Calculation of current when the two coils are connected in series

When the two coils of 24 each are connected in series, then their combined resistance will be R= 24 + 24 = 48

Potential difference, V = 220 V

And, Resistance, R = 48 Q

Again, Current passing through both coils connected in series, is given as:

Case (iii) Calculation of current when the two coils are connected in parallel

When the two coils of 24 each are connected in parallel, then their combined resistance will be given as:

Here, Potential difference, V = 220 V

Resistance of two coils in parallel connection, R = 12 (Two coils in parallel)

Now, Current, I =

(i) In the First Case:

Potential difference, V = 6 V

And, Net Resistance, R= 1 + 2 = 3 (Resistors are connected in series)

Now, by Ohm’s law:

Thus, the current flowing in the circuit containing 1 and 2 resistors in series is 2 A. In a series circuit, the same current flows throughout the circuit. So, the current flowing through the 2 resistor is also 2A. Now, we know that the current (in 2 resistor) is 2 A and its resistance is 2 .

Now, Power used in 2 resistor is given as:

(ii) In the Second Case:

Here, 4 V battery is attached across the parallel combination of 12 and 2 resistors, so the potential difference across the 2 resistor will also be 4 V. Because in parallel connection

Voltage is same.

Now, Potential difference, V= 4 V

And Resistance, R = 2

So, Power used in 2 resistor, P= =

Thus, the power used in the 2 resistor in the second case is also 8W.

From both the cases we can say that the 2 resistor uses equal power in both the circuits.

Given the power of the two lamps, we will calculate the resistances R₁ and R₂ of two lamps separately.

Now, The two lamps of resistances 484 and 806.7 are connected in parallel.

So, the combined resistance of two lamps can be calculated as follows:

Total resistance, R= 302.5

Current drawn from the line can now be calculated by using Ohm's law as follows:

Thus, the current drawn from the line is 0.72 A

(i) For TV set:

Power, P= 250W=

Time, t = 1 h

So, Electric energy, E= P x t = 0.25Kw x 1 h = 0.25 kWh

(ii) For toaster:

Power, P= 250W=

Time, t= 10 min=

So, Electric energy, E= P x t= 1.2 kW x = 0.20kWh

From the above calculations, it is clear that the TV set uses more energy (0.25 kWh) whereas toaster uses less energy (0.20 kWh).

Given
Resistance, R =8Ω
Current, I=15 A
Time period,T = 2 hours
Power, P is the rate at which the heat is developed by the heater
Formula used:
Power, P= I² R
Where
I is the current
R is the resistance
Putting the values in the above equation, we get
H = 15 Χ 15 Χ 8 = 1800 W
Thus, the rate at which heat is developed in the heater is 1800 joules per second.

(a) Tungsten is used almost exclusively for making the filaments of electric lamps and electric bulbs because it has very high melting point nearly 3380°C due to which the tungsten filament can be kept white-hot without melting away.

(b) The coils (or heating elements) of toasters and electric irons are made of an alloy rather than a pure metal because the resistivity of an alloy is much higher than that a pure metal, and an alloy does not undergo oxidation easily even at high temperature, when it is red hot but metals can get easily oxidized at higher temperatures.

(c) The series arrangement is not used for domestic circuits because of the following disadvantages:

(i) In series arrangement, the current throughout the circuit is same. If one electrical appliance stops working due to some defect, then all other appliances also stops working (because the whole circuit is broken).

(ii) In series arrangement, all the electrical appliances can have only one switch due to which they cannot be turned 'on' or 'off' independently.

(iii) In series arrangement, all the appliances do not get the same voltage (220 V) as that of the power supply line (because the line voltage is shared by all the appliances). Due to this, the appliances do not work properly.

(iv) In series arrangement of electrical appliances, the overall resistance of the circuit increases too much because in series the net resistance is the addition of all the resistances. Due to this the current from power supply is low. Because of this, all the appliances of different power ratings cannot draw sufficient current for their proper working.

(d) The resistance of a wire is inversely proportional to its area of cross-section i.e. Thus, when the area of cross-section of wire increases (or thickness of wire increases), then its resistance decreases. Similarly, when the area of cross-section of wire decreases (or thickness of wire decreases), then its resistance increases.

(e) Copper and aluminium wires are usually employed for transmission of electricity because copper and aluminium are good conductors of electric current due to their low electrical resistivity.