Lesson: Electricity

What does an electric circuit mean?

An electric circuit is a continuous and closed path through which electric current flows. It consists of electric devices like electric bulbs, fans, etc., a source of electricity, switches and wires that are connected.

Define the unit of current.

The unit of electric current is ampere (A).

1 A is the flow of 1 C of charge in 1 second through a wire.

i.e., $\text{1}\text{A=}\frac{\text{1}\text{C}}{\text{1}\text{s}}$

Calculate the number of electrons constituting one coulomb of charge.

An electron possesses a charge of $1.6\times {10}^{-19}\text{C}$.

Therefore, the number of electrons that contain 1 C of charge

$\begin{array}{l}\begin{array}{l}=\frac{1}{1.6}\times {10}^{-19} \hfill \\ =6.25\times {10}^{18} \hfill \end{array}\\ =6\times {10}^{18}\end{array}$

Name a device that helps to maintain a potential difference across a conductor.

Devices like a battery, cell, power supply, etc., which are sources of electricity, help to maintain a potential difference across a conductor.

What is meant by saying that the potential difference between two points is 1 V?

When 1 joule of work is done to move a charge of 1 coulomb from one of the points to the other, the potential difference between those two points in a current carrying conductor is said to be 1 V.

i.e., $\text{1}\text{V=}\frac{\text{1}\text{}\text{J}}{\text{1}\text{}\text{C}}$

How much energy is given to each coulomb of charge passing through a 6V battery?

The energy given to each coulomb of charge is equal to the amount of work done in moving it through a 6V battery.
Work done = potential difference × charge
Given that the charge = 1 C and the potential difference = 6V

$�$Work done $=6\times 1=6\text{}\text{J}$

On what factors does the resistance of a conductor depend?

Factors on which the resistance of a wire depends:

·         Material used to make the conductor

·         Length of the conductor

·         Cross-sectional area of the conductor

·         Temperature of the conductor

Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?

The resistance to the flow of current in a conductor is inversely proportional to the area of its cross-section. A thick wire will have more area of the cross section. Thus, current will flow more easily through a thick wire than a thin wire of the same material, when connected to the same source.

Let the resistance of an electrical component remains constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it?

According to Ohm’s law,

$\begin{array}{l}V = IR\\ \Rightarrow I=\frac{V}{R}\mathrm{...} \left(1\right)\end{array}$

When the potential difference is decreased to half, the new potential difference will be $\frac{V}{2}$.

As the resistance remains constant,

So, the new current flowing will be $=\frac{\frac{V}{2}}{R}$

$\begin{array}{l}=\frac{1}{2}\times \frac{V}{R}\\ =\frac{1}{2}\times I \\ = \frac{I}{2}\end{array}$

Thus, the amount of current flowing through an electrical component is reduced by half if the potential difference is decreased to half.

Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?

Electrical appliances like electric toasters and electric irons work on the principle of heating effects of current. The coils of electric toasters and electric irons are made of an alloy rather than a pure metal because:

(a) The resistivity of an alloy is higher than the pure metal.

(b) The alloys do not melt readily at high temperature.

Use the data in the table below to answer the questions.

(a) Which among iron and mercury is a better conductor?

(b) Which material is the best conductor?

(a) Resistivity of iron $=10.0\times {10}^{-8} \Omega$
Resistivity of mercury $=94.0\times {10}^{-8} \Omega$
Thus, iron is a better conductor than mercury.

(b) Silver is the best conductor as its resistivity is the lowest among the listed materials.

Draw a schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5 Ω resistor, an 8 Ω resistor, and a 12 Ω resistor, and a plug key, all connected in series.

Redraw the circuit of above question, putting in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the 12 Ω resistor. What would be the readings in the ammeter and the voltmeter?

The ammeter should be connected in the circuit in series with the resistors. To measure the potential difference across the resistor 12 Ω, the voltmeter should be connected in parallel.  

The reading of the ammeter:

Given:
Potential difference, $V =6\text{ V}$
Resistance of the circuit, $R =5+8+12=25\Omega$

According to Ohm’s law,

$\begin{array}{l}V = IR\\ \Rightarrow I = \frac{V}{R} =\frac{6}{25}=0.24\text{ A}\end{array}$

Thus, the reading of the ammeter will be 0.24 A.

The reading of the voltmeter:

Let the potential difference across 12 Ω resistor be $V_1$
Current flowing through the 12 Ω resistor, $I =0.24\text{ A}$
Using Ohm’s law, we get $V_1 = IR =0.24\times 12=2.88 \text{V}$
This the reading of the voltmeter will be 2.88 V

Judge the equivalent resistance when the following are connected in parallel $–$

 (a) 1 Ω and ${10}^6$ Ω,

(b) 1 Ω and ${10}^3$ Ω, and ${10}^6$ Ω.

(a)   When 1Ω and ${10}^6$ Ω are connected in parallel, the equivalent resistance R can be found as:

$\begin{array}{l}\frac{1}{R}=\frac{1}{1}+\frac{1}{{10}^6}\\ \Rightarrow R=\frac{{10}^6}{1+{10}^6}=\frac{{10}^6}{{10}^6}\approx 1\Omega \end{array}$

(b) When 1Ω, ${10}^3$ Ω and ${10}^6$ Ω are connected in parallel, the equivalent resistance R can be found by:

$\begin{array}{l}\Rightarrow \frac{1}{R}=\frac{1}{1}+\frac{1}{{10}^3}+\frac{1}{{10}^6}=\frac{{10}^6+{10}^3+1}{{10}^6}\\ \Rightarrow R=\frac{1000000}{1001001}=0.999\Omega \end{array}$

An electric lamp of 100 Ω, a toaster of resistance 50 Ω, and a water filter of resistance 500 Ω are connected in parallel to a 220 V source.

What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it?

Given:

·         Resistance of the electric lamp, $R_1 =100\Omega$

·         Resistance of the toaster, $R_2 =50\Omega$

·         Resistance of the water filter, $R_3 =500\Omega$

·         Potential difference of the source, $V =220 \text{V}$
As these are connected in parallel, the corresponding circuit diagram is given as:

Let R be the equivalent resistance of the circuit.

$\begin{array}{l}\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}=\frac{1}{100}+\frac{1}{50}+\frac{1}{500}\\ =\frac{5}{500}+\frac{10}{500}+\frac{1}{500}\\ =\frac{16}{500}\end{array}$

As per Ohm’s law,

$\begin{array}{l}V=IR\\ \Rightarrow I=\frac{V}{R}\\ \Rightarrow I=\frac{220}{\frac{500}{16}}=\frac{220\times 16}{500}=7.04\text{A}\end{array}$

So, 7.04 A of current is drawn by all the three given appliances.

$\Rightarrow$ The current drawn by the electric iron connected to the same source of potential $220\text{ V}=7.04\text{A}$

If $R\text{'}$ is the resistance of electric iron then $V=IR\text{'}$

$R\text{'}$$=\frac{V}{I}=\frac{220}{7.04}=31.25\Omega$

What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?

The advantages of connecting electrical devices in parallel are:

a)      The potential difference across all appliances connected in parallel is equal to the supplied voltage. Hence, there is no division of voltage among the appliances.

b)      The total effective resistance of the circuit when connected in parallel is reduced. This results in more flow of current for a given voltage.

How can three resistors of resistances 2 Ω, 3 Ω, and 6 Ω be connected to give a total resistance of (a) 4 Ω, (b) 1 Ω?

(a)   To get 4 Ω, we can make the connections as shown in the circuit diagram.

The equivalent resistance of 6 Ω and 3 Ω when connected in parallel

$=\frac{1}{\frac{1}{6}+\frac{1}{3}}=\frac{6\times 3}{6+3}=2\Omega$

Therefore, the equivalent resistance of the circuit 2 Ω connected with 2 Ω in series is 4 Ω.
(b) To get 1 Ω, we can make the connections as shown in the circuit diagram.

Here, the resistors are connected in parallel.

The equivalent resistance $=\frac{1}{\frac{1}{2}+\frac{1}{3}+\frac{1}{6}}=\frac{1}{\frac{3+2+1}{6}}=\frac{6}{6}=1\Omega$

What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistance 4 Ω, 8 Ω, 12 Ω, 24 Ω?

There are four coils of resistances 4 Ω, 8 Ω, 12 Ω and 24 Ω respectively.

(a) The highest resistance can be secured by connecting these coils in series.

The resultant resistance when connected in series is $4+8+12+24=48\Omega$.
(b) The lowest resistance can be secured by connecting these coils in parallel. The resultant resistance is

$\frac{1}{\frac{1}{4}+\frac{1}{8}+\frac{1}{12}+\frac{1}{24}}=\frac{1}{\frac{6+3+2+1}{24}}=\frac{24}{12}=2\Omega$

Why does the cord of an electric heater not glow while the heating element does?

A cord of an electric heater is usually made up of aluminium or copper. These are good conductors of electricity. Thus, they offer very little resistance to the flow of electric current.

The heating element of the heater is made up of an alloy that has a very high resistance. Thus, it offers very high resistance when current flows through it. Due to the high resistance, the alloy becomes very hot and glows.

Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.

Given:

Charge, $Q=96000\mathrm{C}$
Time, $t=1\text{hr}=60\times 60=3600\text{s}$

Potential difference, $V=50\text{ volts}$
As $I=\frac{Q}{t}$

$\Rightarrow I=\frac{96000}{3600}=\frac{80}{3}\text{A}$

As $H=VIt$

$\Rightarrow H=50\times \frac{80}{3}\times 60\times 60=4.8\times {10}^6\text{J}$

Therefore, the heat generated is 4.8 x 10⁶ J.

An electric iron of resistance 20 Ω takes a current of 5 A. Calculate the heat developed in 30 s.

The amount of heat (H) produced is given by the equation, $H=Vlt$
Here,
The current $I =5\text{ A}$

Time, $t =30\text{ s}$

Voltage, $V =I\times R=5\times 20=100 \text{V}$

$\Rightarrow H=100\times 5\times 30=1.5\times {10}^4 \text{J}$

Therefore, the amount of heat developed in the electric iron is $1.5\times {10}^4 \text{J}$.

What determines the rate at which energy is delivered by a current?

The rate at which energy is delivered by a current is called power. So, the power (of an appliance) determines the rate at which energy is delivered by a current.

An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.

Power (P) of the motor, $P = VI$,

where the voltage, $V =220\text{ V}$ and the current, $I =5\text{ A}$

Therefore, $P=220\times 5=1100\text{ W}$

Energy consumed by the motor $=Pt$

Here, the time, $t =2\text{ hr}=2\times 60\times 60=7200\text{ s}$

Therefore, the energy $=1100\times 7200=7.92\times {10}^6\text{ J}$