Lesson: Electricity

A cell, a resistor, a key and ammeter are arranged as shown in the circuit diagrams figure given below. The current recorded in the ammeter will be:

(a) Maximum in (i)

(b) Maximum in (ii)

(c) Maximum in (iii)

(d) The same in all the cases

b

In the following circuits, heat produced in the resistor or combination of resistors connected to a 12 V battery will be:

(a) Same in all the cases

(b) Minimum in case (i)

(c) Maximum in case (ii)

(d) Maximum in case (iii)

a

Electrical resistivity of a given metallic wire depends upon

(a) Its length

(b) Its thickness

(c) Its shape

(d) Nature of the material

d

A current of 1 A is drawn by a filament of an electric bulb. Number of electrons passing through a cross section of the filament in 16 seconds would be roughly:

(a) ${10}^{20}$

(b) ${10}^{16}$

(c) ${10}^{18}$

(d) ${10}^{23}$

a

In the following circuits, identify the one in which the electrical components have been properly connected.

(a) (i)

(b) (ii)

(c) (iii)

(d) (iv)

b

What is the maximum resistance which can be made using five resistors each of $\frac{1}{5}\Omega$?

(a) $\frac{1}{5}\Omega$

(b) 10 Ω

(c) 5 Ω

(d) 1 Ω

d

What is the minimum resistance which can be made using five resistors each of $\frac{1}{5}\Omega$?

(a) $\frac{1}{5}\Omega$

(b) $\frac{1}{25}\Omega$

(c) $\frac{1}{10}\Omega$

(d) 25 Ω

b

In the given below figure, the proper representation of series combination of cells obtaining maximum potential is:

(a) (i)

(b) (ii)

(c) (iii)

(d) (iv)

a

Which of the following represents voltage?

(a) $\frac{\text{Work}\text{done}}{\text{Current}\text{×}\text{Time}}$ 

(b) $\text{Work done}\text{×}\text{Charge}$

(c)  $\frac{\text{Work}\text{done}\text{×}\text{Time}}{\text{Current}}$

(d) $\text{Work done}\text{×}\text{Charge}\text{×}\text{Time}$

a

A cylindrical conductor of length l and uniform area of cross-section A has resistance R. Another conductor of length 2l and resistance R of the same material has area of cross section:

(a) $\frac{\text{A}}{2}$

(b) $\frac{3\text{A}}{2}$

(c) 2 A

(d) 3 A

c

A student carries out an experiment and plots the V-I graph of three samples of nichrome wire with resistances R₁, R₂ and R₃ respectively. Which of the following is true?

(a) $R_1=R_2=R_3$

(b) $R_1>R_2>R_3$

(c) $R_3>R_2>R_1$

(d) $R_2>R_3>R_1$

c

If the current I through a resistor is increased by100% (assume that temperature remains unchanged), the increase in power dissipated will be:

(a) 100 %

(b) 200 %

(c) 300 %

(d) 400 %

c

The resistivity does not change if:

(a) The material is changed

(b) The temperature is changed

(c) The shape of the resistor is changed

(d) Both material and temperature are changed

c

In an electrical circuit three incandescent bulbs A, B and C of rating 40 W, 60 W and 100 W respectively are connected in parallel to an electric source. Which of the following is likely to happen regarding their brightness?

(a) Brightness of all the bulbs will be the same

(b) Brightness of bulb A will be the maximum

(c) Brightness of bulb B will be more than that of A

(d) Brightness of bulb C will be less than that of B

c

In an electrical circuit two resistors of 2 Ω and 4 Ω respectively are connected in series to a 6 V battery. The heat dissipated by the 4 Ω resistor in 5 s will be:

(a) 5 J

(b) 10 J

(c) 20 J

(d) 30 J

c

An electric kettle consumes 1 kW of electric power when operated at 220 V. A fuse wire of what rating must be used for it?

(a) 1 A

(b) 2 A

(c) 4 A

(d) 5 A

d

Two resistors of resistance 2 Ω and 4 Ω when connected to a battery will have:

(a) Same current flowing through them when connected in parallel

(b) Same current flowing through them when connected in series

(c) Same potential difference across them when connected in series

(d) Different potential difference across them when connected in parallel

b

Unit of electric power may also be expressed as:

(a) Volt ampere

(b) Kilowatt hour

(c) Watt second

(d) Joule second

a

A child has drawn the electric circuit to study Ohm’s law as shown in given below figure. His teacher told that the circuit diagram needs correction. Study the circuit diagram and redraw it after making all corrections.

Three 2 Ω resistors, A, B and C, are connected as shown in the following figure. Each of them dissipates energy and can withstand a maximum power of 18 W without melting.

Find the maximum current that can flow through the three resistors?

Maximum current through resistor A $=\sqrt{\frac{18}{2}}\text{A}=3\text{A}$

Given that the resistance of each of the resistors are 2 Ω each and maximum power the resistors can withstand is 18 W.

The power dissipated is given by, $P=I^{2}R$.

The maximum current that the resistor A can withstand $=\sqrt{\frac{18}{2}}\text{A}=3\text{A}$ 

Since B and C are connected in parallel, the 3A of current that passes through the resistor A splits and passes through the resistor B and C equally. Thus, the maximum current through resistors B and C each,

$=3 \times \frac{1}{2}\text{A}=1.5\text{ A}$.

Should the resistance of an ammeter be low or high? Give reason.

An ammeter measures the current passing through a circuit. Ideally, an ammeter should not have any resistance as resistance of the ammeter reduces the amount of current passing through it and therefore the actual amount of current passing through a circuit cannot be determined. However, in practice, every ammeter will have some amount of resistance at least. Therefore, the best option would be to have an ammeter with minimum possible resistance.

Draw a circuit diagram of an electric circuit containing a cell, a key, an ammeter, a resistor of 2 Ω in series with a combination of two resistors (4 Ω each) in parallel and a voltmeter across the parallel combination. Will the potential difference across the 2 Ω resistor be the same as that across the parallel combination of 4 Ω resistors? Give reason.

When two resistors are connected in series, the potential difference across resistors will be equal only if the resistance of the resistors are equal.

The combined resistance of two resistors connected in parallel is 2 Ω.

 Thus, the potential difference across the 2 Ω resistor will be the same as that across the parallel combination of two 4 Ω resistors.

How does use of a fuse wire protect electrical appliances?

A fuse consists of a piece of wire made up of an alloy or metal of an appropriate melting point. When the current higher than a specified value flows through a wire, the temperature of the fuse wire increases; leading to the melting and breakage of the circuit.

Thus, a fuse stops the flow of any unduly high electric current in a circuit. This helps in protecting electrical appliances from damages.

What is electrical resistivity? In a series electrical circuit comprising a resistor made up of a metallic wire, the ammeter reads 5 A. The reading of the ammeter decreases to half when the length of the wire is doubled. Why?

Resistivity is a measure of resistance offered to an electric current by a conducting substance.

Resistance of a conductor is given by the relation, $R=\rho \frac{l}{A}$

where l is the length, A is the area of cross-section of the wire and $\rho$ is resistivity of the wire.

In simple words, the more is the resistivity of a substance, the more is its resistance.

The SI unit of electrical resistivity is ohm-metre (Ω.m).

When the length of the wire is doubled, the resistance offered by the wire also doubles. This reduces the amount of current by half. This can be understood from the equation $I=\frac{V}{R}$. Here I is the current and R is the resistance in a circuit across a voltage of V.

What is the commercial unit of electrical energy? Represent it in terms of joules.

The commercial unit of electrical energy is kWh.

$1\text{ kWh}=1000\text{ W}\times 60\times 60\text{s}=3.6\times {10}^6\text{J}$

A current of 1 ampere flows in a series circuit containing an electric lamp and a conductor of 5 Ω when connected to a 10 V battery. Calculate the resistance of the electric lamp.

Now if a resistance of 10 Ω is connected in parallel with this series combination, what change (if any) in current flowing through 5 Ω conductors and potential difference across the lamp will take place? Give reason.

Case 1: When the lamp is in a series circuit,

the total resistance of the circuit, $R=\frac{V}{I}$,

$R=\frac{10V}{1A}=10\Omega$

The resistance offered by the lamp

$=\text{total resistance of the circuit and the lamp}-\text{resistance of the circuit}$

$=10\Omega -5\Omega$

$=5\Omega$ 

Case 2: When a 10 Ω resistor is connected in parallel with the series connection, there will be no change in current flowing through 5 Ω conductors, as there will be no change in potential difference across the lamp either.

Why is parallel arrangement used in domestic wiring?

(i)     In a parallel connection, the overall resistance of the circuit becomes less due to which the current supply from the power source is high.

(ii)  Unlike series connection, if one appliance in parallel connection stops working, the other appliances are not affected. Similarly, if we have to use only appliance, we can do so independently.

(iii)             All the appliances connected in parallel connection can get the same voltage as supplied by power supply.

${\text{B}}_{\text{1}}{\text{, B}}_{\text{2}}{\text{and B}}_{\text{3}}$ are three identical bulbs connected as shown in given below figure. When all the three bulbs glow, a current of 3A is recorded by the ammeter A.

(i) What happens to the glow of the other two bulbs when the bulb B₁ gets fused?

(ii) What happens to the reading of ${\text{A}}_{\text{1}}{\text{, A}}_{\text{2}}{\text{, A}}_{\text{3}}$ and A when the bulb B₂ gets fused?

(iv) How much power is dissipated in the circuit when all the three bulbs glow together?

(i) The glow of the bulbs B₂ and B₃ will remain the same because glow of bulbs depends on power. Power is given as $P=\frac{V^2}{R}$
The potential difference (V) and resistance (R) of B₂ and B₃ remain the same. Therefore, there will be no change in the glow of b.

(ii) The amount of current flowing through each of the bulbs remain the same as the current, $I=\frac{V}{R}$. Both V and R in this case remains the same.

(iii) The net resistance (R') of the three resistors connected in parallel can be found from the following equations.

$\frac{1}{R\text{'}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}=\frac{3}{R}$

as $R_1=R_2=R_3=R$

Therefore,

$\begin{array}{l}R\text{'}=\frac{R}{3}\\ \Rightarrow V=IR\\ \Rightarrow 4.5=3\times \left(\frac{R\text{'}}{3}\right)\\ \Rightarrow R\text{'}=4.5 \Omega \\ P=V\times I=4.5\times 3=13.5W\end{array}$

Three incandescent bulbs of 100 W each are connected in series in an electric circuit. In another circuit another set of three bulbs of the same wattage are connected in parallel to the same source.

(a) Will the bulb in the two circuits glow with the same brightness?

Justify your answer.

(b) Now let one bulb in both the circuits get fused. Will the rest of the bulbs continue to glow in each circuit? Give reason.

(a) No. The combined resistance of the bulbs in series will be three times the resistance of the bulbs connected in parallel. Therefore, the current in the series combination will be one-third of the current in each bulb in the parallel combination. Therefore, the bulbs in the parallel connections will glow brighter.

(b) When one of the bulbs in both the connections gets fused:

·         The remaining bulbs connected in the series stop glowing as the circuit is broken.

·         There will be no impact on the two bulbs as the circuits for these bulbs remain intact without any change in the voltage and the amount of current flowing through the circuits.

State Ohm’s law? How can it be verified experimentally? Does it hold good under all conditions? Comment.

Ohm’s law states that the potential difference across the ends of a resistor is directly proportional to the current passing through the resistor provided its temperature remains the same. If I is the current flowing through a conductor, and V is the potential difference at its ends, then $V\propto I\Rightarrow V=IR$

where R is the constant of proportionality and commonly called the resistance of a conductor.

Experiment to verify Ohm’s Law:

Steps:

a)      We can set up a circuit as shown in the figure. The circuit consists of a nichrome wire of length, say 1 m, an ammeter, a voltmeter and four cells of 2.5 V each.

b)      We will use only one cell as the source in the circuit. We will note down the reading in the ammeter and the voltmeter for the potential difference across the nichrome wire in the circuit and tabulate these values.

c)      Next, we will connect two cells in the circuit and note the respective readings of the ammeter and voltmeter.

d)      We will repeat the above steps using three cells and then four cells in the circuit, separately.

e)      From the table, we can calculate the value of the V and I for each observation.

S.No	Number of cells used in the circuit	Current through the nichrome wire, I (ampere)	Potential difference across the nichrome wire, V (volt)	$\frac{V}{I}$ ( $\frac{\text{Volt}}{\text{Ampere}}$ )
1 2 3 4	1 2 3 4

 

f)       We will then plot a graph between V and I by taking V along X-axis.

Observation: The V$–$I graph is a straight line that passes through the origin of the graph.

Conclusion- $\frac{V}{I}$ is a constant ratio. This verifies Ohm’s Law.

This law does not hold good for all situations. For example:

·         This law is valid only for conductors, provided the temperature and other physical conditions remain constant.

·         It is not followed in case of insulators.

What is electrical resistivity of a material? What is its unit?

Describe an experiment to study the factors on which the resistance of conducting wire depends.

Resistivity is numerically equal to the resistance of a wire which is 1 metre long having a cross section of 1 square metre.

In simple words, the more is the resistivity of a substance, the more is its resistance per unit length and per unit area.

The SI unit of electrical resistivity is ohm-metre (Ω$�$m).

Factors that affect the resistance of a wire:

a) Material (resistivity)

b) Cross- sectional area

c) Length

d) Temperature 

Experiments:

a) Let us take wires of different materials like copper, aluminium, iron of same

cross-sectional area and of the same length.

b) Let us connect to a circuit as shown in the figure. Here AA’ represents a wire.

(i) Let us connect each wire one by one between A and A’. After that, let us insert

the key into the plug and note down the reading of the ammeter.

Observation: The value of the current is different for different wires.

Conclusion: Since the same cell is used every time, the potential difference (V) across the wires is same. This means different wires used in this experiment draw different currents when the same potential difference is applied across them.

Hence, R α ρ

(ii) Let’s perform the same activity as done earlier by taking the same wire of different cross sections.

Observation: The value of the current is different for different cross sections of the same wire. The more is the cross section, the more is the current flowing through the wire.

Conclusion: Resistance is inversely proportional to the area of cross-section (A)

$R\alpha \frac{1}{A}$ 

(iii) Let’s perform the same activity as done before by taking the same wire of different lengths.

Observation: As the length increases, the current in the circuit decreases.

Conclusion: The resistance is proportional to length. 

(iv) Finally, let’s perform the same experiment by increasing the temperature of the lab where the experiment is being performed.

Observation: The higher the temperature, the lesser is the current.

Conclusion: The higher the temperature, the higher is the resistance.

How will you infer with the help of an experiment that the same current flows through every part of the circuit containing three resistances in series connected to a battery?

We can use three bulbs as resistors to perform this activity. We can make the connections as shown.

We will take readings of the ammeter by positioning it before the resistor $R_1$, after the resistor $R_3$, between $R_1\text{and}{R}_2\text{and between}{R}_2\text{and}{R}_3$.

Observation: The ammeter reading for the given setup remains the same for every position.

Conclusion: The same current flows through every part of the circuit containing three resistances in series.

How will you conclude that the same potential difference (voltage) exists across three resistors connected in a parallel arrangement to a battery?

This can be proved by a very simple logic. Let us take a parallel connection of three resistors are shown. The potential difference across the resistor R₁ is same as the difference in the electric potential between the point X and Y.

Similarly, the potential difference across the resistors $R_2\text{and}{R}_3$ are same as the difference in the electric potential between the points X and Y.

Thus, we can conclude that the same potential difference (voltage) exists across the three resistors connected in a parallel arrangement to a battery.

What is Joule’s heating effect? List its four applications in daily life.

The heating of resistor because of dissipation of electrical energy is commonly known as “Heating Effect of Electric Current”. The heat lost in the process is explained by the Joule’s Law of Heating. The law states that the heat produced in a resistor is:

(i) Directly proportional to the square of current for a given resistance

(ii) Directly proportional to resistance for a given current, and

(iii) Directly proportional to the time for which the current flows through the resistor.

Four applications of Joule’s law of heating are:

a)   Electrical appliances such as electric iron, electric toaster, etc.

b)   Filaments of the electric bulbs,

c)   Utilised in electric fuse for protection of household wiring and electric appliances,

d)   Electric heater, room heater, geyser, etc.

Find out the following in the electric circuit given in following figure.

(a) Effective resistance of two 8 Ω resistors in the combination

(b) Current flowing through 4 Ω resistor

(c) Potential difference across 4 Ω resistance

(d) Power dissipated in 4 Ω resistor

(e) Difference in ammeter readings, if any

(a) The two given resistors of 8 Ω each are connected in parallel.

For 2 resistors connected in parallel, the effective resistance R is given by:

$\begin{array}{l}\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}\\ R=\frac{R_1{R}_2}{R_1+R_2}\\ \Rightarrow R=\frac{8\times 8}{8+8}\\ \Rightarrow R=\frac{64}{16}\\ \Rightarrow R=4\Omega \end{array}$

(b) The current (I) flowing through the 4Ω resistor $=\frac{V}{R}=\frac{8}{4+4} =1\text{A}$.

(c) The potential difference across the 4 Ω resistor $=IR=1\times 4=4\text{V}$.

(d) The power dissipated in 4 Ω resistor $=I^{2}R=1^2 \times 4=4\text{W}$.

(e) Both the ammeters will show the same current reading as the same current

passes through each of these.