Kerala Syllabus Class 9 Physics Chapter 7 Electric Current Notes Solutions

The comprehensive approach in Kerala Syllabus 9th Standard Physics Textbook Solutions Chapter 7 Electric Current Notes Questions and Answers English Medium ensures conceptual clarity.

Std 9 Physics Chapter 7 Notes Solutions Electric Current

SCERT Class 9 Physics Chapter 7 Notes Solutions Kerala Syllabus Electric Current Questions and Answers

Class 9 Physics Chapter 7 Let Us Assess Answers Electric Current

Question 1.
Which of the following device convert chemical energy into electrical energy?
a) Dry cell
b) Dynamo
c) Solar Cell
Answer:
a) Dry cell

Question 2.
For current to flow from a cell in a closed circuit, the two terminals of the cell must be
a) at high potential
b) having a potential difference between them
c) at different temperatures
d) at different heights
Answer:
b)

Question 3.

In which of the following circuits are the ammeter and the voltmeter connected properly?
Answer:
(c)

Question 4.
Match the terms in column A with those in columns B and C

A	B	C
Potential difference (V)	Q/t	ohm(Ω)
Current (I)	W/Q	volt (V)
Resistance (R)	V/I	ampere (A)

Answer:
Potential difference (V) – W/Q – volt(V)
Current (I) – Q/t – ampere (A)
Resistance (R) – V/I – ohm ( Ω )

Question 5.
50 J of work is done to move an electric charge of 5 C from point M to N in an electric circuit. What is the potential difference between M and N?
Answer:
Work done, W = 50 J
Charge, Q = 5 C
Potential difference = W/Q = 50/5 = 10 V

Question 6.
In which of the following circuits are bulbs connected in series

Answer:
(b) and (c)

Question 7.
Analyse the circuit given below and answer the following questions

a) Calculate the effective resistance in the circuit when the switch is turned on.
b) What is the current through the circuit?
c) What is the potential difference across the 2Ω resistor?
d) What is the current flowing through the 1 Ω resistor?
Answer:
a) R₁ = 1Ω
R₂ = 2Ω
Effective resistance R = R₁ + R₂
= 1 + 2 = 3 Ω

b) Total current through the circuit is I = \(\frac{V}{R}\)
= \(\frac{6}{3}\) 2 A

c) V = IR₂
= 2 × 2 = 4V

d) I = 2 A. The resistors are in series connection.

Question 8.
In an electric circuit, in which way are the fan and its regulator connected? Parallel / series
Answer:
Series

Question 9.
Consider the following circuits. Which voltmeter shows a reading of 10 V?

Answer:
(b), (d)

Question 10.
A wire of length 50 cm has a resistance of 5Ω. If the length is doubled by stretching,
a) What happens to the area of the cross-section (thickness) of the conductor?
b) What will be the resistance of the wire?
Answer:
a) When the length of the conductor is doubled, area decreases to half.
b) If the length of the wire is doubled and its area of the cross-section is decreased to half, then its resistance will increase to four times.

Question 11.
Which of the graph given below represents Ohm’s law?

Answer: (c)
V-I graph is a straight-line graph passes through the origin.

Class 9 Physics Chapter 7 Extended Activities Answers Electric Current

Question 1.
The graph given below represents Ohm’s law. Analyse the graph and find the resistance.

Answer:
Locate two points on the linear portion of the graph. These points will have coordinates (I₂, V₁) and (I₂, V₂).
Using the formula of Ohm’s law, R = \(\frac{V}{I}\)
Slope of V – I graph gives resistance
If you have two points
R = \(\frac{V_2-V_1}{I_2-I_1}\)
This formula gives the slope of the line, which is equal to the resistance.
Take I₂ = 0.5 A, I₂ = 1.0 A
V₁ = 1.5V, V₂ = 3 V
R = \(\frac{V_2-V_1}{I_2-I_1}\)
R = \(\frac{3-1.5}{1-0.5}\)
= \(\frac{1.5}{0.5}\)
R = 3 Ω

Question 2.
Draw a circuit diagram connecting three LEDs of 3 V and a 9 V battery. Construct the circuit and light the LEDs.
Answer:

Components needed
3 LEDs of 3V, Resistor, Battery, connection wire.
Steps:
Connect 3 LEDs. Connect the positive of the 9V battery to one end of the resistor. And connect the other end of the resistor to the positive of one LED.

Question 3.
Measure the emf across the terminals of a dead torch cell and a new torch cell using a multimeter. Do you get any difference in their value? What is the reason?
Answer:
Yes.
New Torch Cell- A new torch cell shows a voltage close to its rated emf.
Dead Torch Cell – A dead torch cell will show a much lower voltage near zero or significantly below its rated voltage.
A new cell contains reactants that can fully undergo the electrochemical reactions needed to generate voltage. As the cell discharges, these reactants are consumed, leading to a decrease in the ability to produce emf.

Electric Current Class 9 Notes Questions and Answers Kerala Syllabus

Question 1.
What are the electric charges that objects acquire when rubbed against each other?
Answer:
Objects acquire positive and negative charges when rubbed against each other.

Question 2.
What is the charge of electrons?
Answer:
Negative Charge.

Question 3.
What is the unit of electric charge?
Answer:
Coulomb (C).

Question 4.
Connect a charged capacitor (6 V, 500 μF) to an LED as shown in figure.

What is your observation?
Answer:
The LED lights up for a very short time.

Question 5.
What happens to the electric charge in the capacitor when it is connected to an LED?
Answer:
An electric charge flows through the LED.

Question 6.
What could be the reason for the LED not glowing continuously in this arrangement?
Answer:
Because the charge of the capacitor is completely lost.

Question 7.
Connect an LED as shown in figure to a button cell used in calculators and other electronic devices.

What is your observation? Does the LED glow continuously here?
Answer:
LED lights continuously. The movement of charges creates a continuous flow of current through the circuit.
In the first experiment, there is a momentary flow of electric charge, and in the second, there is a continuous flow of electric charge. The flow of charges produces an electric current through a circuit.

Activity

Take equal quantity of water in two identical containers A and B. Add any colour to one of the containers. Fill a plastic tube (siphon) with water and dip it as shown in the figure.

Question 8.
Does water flow from container A to B?
Answer:
No, the water does not flow because there is no difference in the height of the water level in the two containers.

Question 9.
Is the gravitational potential due to the height of the water level in the two containers the same? (Same/different)
Answer:
Same.

Question 10.
Repeat the experiment by keeping the container A slightly elevated. Now, does the water flow from container A to B?
Answer:
Yes. Water flowed from A to B because of the difference in the water level.
It is understood that in the first experiment, water did not flow because the water level in the two containers is the same. But in the second experiment, water flowed from A to B because of the difference in the water level.
Observe two more situations.
Situation 1
An iron rod touches the flame of a candle as shown in figure.

Question 11.
On which part of the rod will the temperature rise first, as soon as it touches the flame? (P/Q)
Answer:
Q

Question 12.
What is the direction of the flow of heat through the rod?
(From P to Q/ From Q to P)
Answer:
From Q to P

Situation 2
A tube filled with air to its maximum is shown in figure.

Question 13.
Where is the air pressure greater? (inside the tube / outside the tube)
Answer:
Inside the tube.

Question 14.
If the valve of the tube is opened, what is the direction of the air flow?
Answer:
From inside the tube to outside.

Question 15.
Tabulate the inference drawn from the above activities.

Situation	Direction of flow
Water flows
Heat flows	From a point of high temperature to a point of lower temperature.
Air flows

Answer:

Situation	Direction of flow
Water flows	From a point of high temperature to a point of lower temperature.
Heat flows	From a point of high temperature to a point of lower temperature.
Air flows	From a point of high pressure to a point of lower pressure.

The flow of water occurred because the two containers are placed at different heights. Similarly, heat flow occurred because there were two points with different temperatures. Air flow occurred because there were two regions having a difference in pressure.

Observe the figure.

While using the siphon, the flow of water from container A to B is possible only until the water levels in both containers become equal.
A pump is used to pump back the water from container B to A. The pump works to keep the rate of flow of water from B to A the same as that from A to B.

Question 16.
Is there a continuous flow of water while the pump is working? What may be the reason? Answer: Yes. Because here, a gravitational potential difference is maintained. For the continuous flow of water, a difference in water levels in containers A and B has to be maintained. This difference is maintained by using a pump which acts as an external energy source.

Question 17.
Observe the circuit in figure.

Does the bulb glow when the switch is turned on? What may be the reason?
Answer:
If the switch is turned on here, the bulb will not light up. Because there is no potential difference between terminals C and D to allow charge to flow. For the bulb in the circuit to glow continuously, there must be a flow of electric charge through the bulb.

Question 18.
Doesn’t this require an external source sufficient to maintain a potential difference between points C and D?
Answer:
Yes.
A source of electricity is a source of energy used to maintain a potential difference between two points in an electric circuit.

Question 19.
The commonly used torch cell is an example of a source of emf. Write down other sources of emf known to you.
Answer:

• Button cell
• Dynamo
• Storage battery

Question 20.
Write down the energy change that occurs in each of these in the science diary.
Answer:

Source of emf	Energy change
Torch cell (Dry cell)	Chemical energy into electrical energy
Generator	Mechanical energy into electrical energy
Button cell	Chemical energy into electrical energy
Fuel cell	Chemical energy into electrical energy

Question 21.
Alessandro Volta first designed a device that could be used as a source of emf. This is volta cell. Find more information about him and write down in the science diary.
Answer:

Let’s make a simple volta cell. Place a zinc rod and a copper rod in a beaker containing dilute sulphuric acid as the electrolyte as shown in figure. Connect a voltmeter to the copper rod and the zinc rod.

Question 22.
What do you observe?
Answer:
The electrons are transferred due to the reaction of zinc with dilute sulphuric acid and flow through the external circuit. Hence the needle of the voltmeter deflects. Copper rod is at a high energy level of electric charges (positive potential) and zinc rod is at a low energy level of electric charges (negative potential). Therefore, current flows from the copper rod to the zinc rod in the external circuit. Connect an LED instead of the voltmeter, and Mg instead of Zn in the above arrangement.

Question 23.
What do you observe?
Answer:
When Mg is used instead of Zn and LED is used instead of voltmeter, LED lights up.

Question 24.
What happens to the intensity of light if this cell is operated for a short interval of time?
Answer:
Intensity decreases.

Question 25.
What may be the reason?
Answer:
As the rate of chemical reaction decreases, the electric energy obtained also decreases. Primary cells are those which cannot be reused after using it for a certain period.

Question 26.
Write examples for primary cells.
Answer:

• Dry cell
• Volta cell
• Button Cell

Secondary cells (also known as storage cells) are energy sources that can be recharged and reused. A battery is a system in which multiple cells are arranged and used as a single source of electricity.

Question 27.
Write down more examples for secondary cells/ Batteries.
Answer:

• Mobile battery
• Lithium-ion battery
• Car battery
• Emergency lamp battery

Question 28.
How is the potential difference between the terminals of a cell/battery measured?
Answer:
Voltmeter is a device for measuring potential difference.
Observe the diagram, in which the potential difference is measured using the voltmeter.
We know that a cell has two terminals, positive (+) and negative (-). Voltmeter also has two terminals.

Question 29.
Which terminal of the cell is connected to the positive terminal of the voltmeter?
Answer:
The positive terminal of the voltmeter is connected to the positive of the cell.

Question 30.
What about the negative terminal?
Answer:
The negative terminal of the voltmeter is connected to the negative of the cell.

Question 31.
Measure and tabulate the potential difference across the terminals of the following sources of electricity using a voltmeter.
Answer:

Sources	Potential Difference (V)
Torch Cell (Dry Cell)	1.5
Button Cell	3
Volta Cell	1.1
Mobile Battery	3.6 to 3.8

Question 32.
There are various types of electric sources from the early volta cell to today’s solar panel. Gather information about them, prepare and present a seminar paper about their characteristics and where they are used.
Answer:
(Hints)
Introduction
Electricity powers our daily lives, and many sources have been developed to generate it. Voltaic Cell (1800)

• First battery, using zinc, copper, and a salt solution.
• Used for early experiments.

Daniell Cell (1836)

• Improved battery, reducing corrosion with separate solutions.
• Used in telegraphs.

Lead-Acid Battery (1859)

• First rechargeable battery, using lead and acid.
• Still used in car batteries.

Dry Cell (1886)

• Portable, non-rechargeable battery (like AA batteries).
• Powers devices like torches and radios.

Lithium-Ion Battery (1970s)

• Rechargeable, lightweight battery.
• Used in phones, laptops, and electric vehicles.

Solar Panel (Modern)

• Converts sunlight to electricity.
• Powers homes and industries with renewable energy.

Conclusion
Electric sources have advanced from simple chemical batteries to modern solar panels, meeting various needs over time.

Question 33.
Arrange two torch cells in three different ways, as shown in the figures (circuits). Arrange the potential differences across each of them using a voltmeter and tabulate them.

Answer:

Circuit	Number of cells	Voltmeter Reading/Potential Difference
1	2	3
2	2	1.5
3	2	0

Question 34.
In which arrangement was the maximum potential difference obtained?
Answer:
Circuit 1

Question 35.
Write down in your science diary the instances where cells are connected in series.
Answer:

• Remote control of TV
• Remote control of electrical equipment
• Toys
• Torch
• Radio
• 9V battery
  

Question 36.
What is the maximum potential difference obtained using these? Illustrate the arrangement.
Answer:
Potential difference E = n × e = 4 × 1.5 = 6 V

Question 37.
If six identical cells are connected in series to form a battery of 9 V, what is the emf of one cell?
Answer:
e = \(\frac{E}{n}\)
= \(\frac{9}{6}\) = 1.5 V

Question 38.
If a charge Q flows through a circuit in a time t calculate the charge that flows in one second?
Answer:

Charge flowing per second = \(\frac{Q}{t}\)
Let’s find the unit of current from this equation.
unit of current = \(\frac{\text { unit of charge }}{\text { unit of time }}\)
The unit of current is given as C/s.
It is known as ampere (A).
The unit ampere is given to intensity of electric current in honour of the scientist, Andre Marie Ampere. Smaller units milli ampere (mA) and micro ampere (μA) are also used to measure current.
1 A = 1000 mA
1 mA = 1000μ A

Question 39.
If 10 C charge flows through a conductor in 5 s, what is the current?
Answer:
I = \(\frac{Q}{t}\)
= \(\frac{10 c}{5 s}\)
I = 2 A

Question 40.
If a current of 1.5 A flows through a conductor for 3s, calculate the quantity of electric charge that passes through the conductor.
Answer:
Q = I × t = 1.5 × 3 = 4.5 C

Question 41.
If an ammeter connected in an electric circuit shows a reading of 1A, what is the charge flowing through the ammeter in one second?
Answer:
Q = I x t
= 1 × 1 = 1C

Question 42.
Draw a circuit using the following table. Write down the names of any three components and their use.

Answer:

Battery – Acts as a source of electricity
Switch – Used to turn the circuit on and off
Bulb – Converts electrical energy into light.

Question 43.
Construct a small torch using two torch cells, a suitable LED, wires, switch and PVC pipe.

Answer:
(Hints)
Prepare the Pipe: Cut PVC pipe to fit two batteries and the LED.
Insert Batteries: Place the two cells in series (positive to negative).
Connect LED: Attach the positive terminal of the top battery to the longer leg of the LED.
Add Switch: Connect the shorter leg of the LED to one terminal of the switch, and the other switch terminal to the negative terminal of the bottom battery.
Test Circuit: Flip the switch to turn the LED on/off.
Assemble: Secure the LED and switch on the PVC pipe.

Question 44.
Observe the circuit in figure. Tabulate the symbols of the ammeter and voltmeter, their uses and how they are connected in the circuit.

Answer:

Question 45.
Measure the potential difference and current through the circuit and tabulate. Repeat the activity by increasing the number of cells in series.

Answer:

Question 46.
What change in voltmeter reading (potential difference) is observed when the number of cells increases?
Answer:
Potential difference increases when the number of cells increases.

Question 47.
What change in potential difference is observed when the current (ammeter reading) increases?
Answer:
Potential difference increases.

Question 48.
Isn’t the value of V/I approximately equal?
Answer:
Yes.
It is understood that as the current increases the potential difference also increases and is a constant. That is, 1 ∝ V or V ∝l
V = a constant × I
∴ \(\frac{V}{I}\) = a constant
This constant will be equal to the resistance in the circuit. It is denoted by the letter R.
R = \(\frac{V}{I}\)

Question 49.
What will be the unit of resistance?
Answer:
R = \(\frac{V}{I}\)
Unit of resistance = \(\frac{\text { unit of potential difference }}{\text { unit of current }}\) = \(\frac{\text { Volt }}{\text { Ampere }}\)
The unit of resistance volt/ampere is known as ohm (Ω).
Larger units such as kilo ohm (kΩ) and mega ohm (MΩ) are also used.

Question 50.
What happens to the current through the circuit when the resistance increases?
increases/decreases
Answer:
decreases

Question 51.
What is the SI unit of resistance?
Answer:
SI unit of resistance is Ω

Question 52.
Complete table based on Ohm’s law and draw a graph. Plot a graph by taking current on the X-axis and voltage on the Y-axis. What is the peculiarity of the graph obtained?

Answer:

Question 53.
In which case is resistance higher?
Answer:
The resistance will be higher as soon as the heater is turned off.

Question 54.
Haven’t you understood that resistance decreases with decrease in temperature?
Answer:
Yes. Resistance decreases as the temperature decreases.

Question 55.
Do any other factors influence resistance?
Answer:
Yes.

Activity
Arrange a circuit as shown in figure. CD is a nichrome wire of 40 cm long.

EF is a nichrome wire having the same length and twice the thickness of CD. GH is a copper wire of the same length and diameter as CD. They are screwed on a wooden board. M is the midpoint of the nichrome wire CD. J is the free end of the conductor whose other end is connected to the negative terminal of the battery as shown in figure above. Touch the free end J to M, D, F and H one after the other.

Question 56.
Observe the changes in the intensity of light of the bulb and the ammeter reading. Tabulate the observations. Write answers to the questions based on observations and analysis of the table.
Answer:

In all cases, the potential difference applied across the circuit is the same. According to Ohm’s law, V = IR.

Question 57.
Isn’t the change in the ammeter reading attributed to the change in the resistance of the conductor?
Answer:
Yes.

Question 58.
In which situation is the ammeter reading (current) the least?
Answer:
The current is smallest when point D touches J.

Question 59.
What is the reason? (more resistance / less resistance)
Answer:
More resistance.

Question 60.
What happens to the resistance as the length of the same conductor increases?
Answer:
Resistance Increases.

Question 61.
What happens to the resistance as the thickness of the same conductor increases?
Answer:
Resistance decreases.

Question 62.
Which has more resistance, nichrome or copper, having the same length and thickness?
Answer:
Nichrome.

Question 63.
Based on these activities, write down the factors that influence the resistance of a conductor.
Answer:

• The nature of the material of the conductor.
• The thickness (area of cross-section) of the conductor.
• Length of the conductor
• Temperature

The resistance of a conductor increases as length increases and decreases as the area of cross-section increases. Different substances also have different resistances. Consider wires of nichrome, tungsten, copper, aluminium and silver of the same length and thickness. Among these, nichrome and tungsten have relatively high resistance while aluminium and copper have very low resistance. Silver has the least resistance.

Question 64.
What is the working principle of a rheostat?
Answer:
As the length of a conductor increases without any change in its diameter, its resistance also increases. A rheostat is a device that works on this principle.

Question 65.
A rheostat has 100 turns of resistance wire. The resistance of one turn is 0.15 Ω. Then which of the following values of resistance cannot be included in a circuit using this rheostat? Justify your answer.
a) 3 Ω
b) 7.5 Ω
c) 4 Ω
d) 8.25 Ω
Answer:
A rheostat cannot include any value of resistance between 0 and maximum. A rheostat includes one coil, two coils, three coils, etc., when the sliding contact moves. Only resistances that are multiples of the round resistance can be included (here, multiples of 0.15). So, the answer is 4 Ω.

Question 66.
Turn on S₁ and S₂ in both circuits. In which circuit does the bulb glow more brightly?
Answer:
The bulb in figure (d) is brighter.

Question 67.
Turn on only the switch S1 in both circuits. What is your observation?
Answer:
If only S1 is turned on, the bulbs in the circuit of (c) will not light. In the circuit of Fig. (d) only the bulb adjacent to S1 will light up.

Question 68.
Why do the bulbs in figure(d) glow relatively brighter if all the switches in both circuits are turned on?
Answer:
The bulbs in figure (d) glow brighter because the bulbs received the same voltage and current. The difference in voltage and current to the bulbs is the reason for the increase and decrease in light intensity.
The difference in potential difference and current are the reasons for the increase or decrease in the intensity of light. The arrangement of the bulbs as shown in figure (c) is the connection in series. The arrangement of the bulbs as shown in figure (d) is the connection in parallel. More characteristics of series and parallel connections can be understood by using circuits with resistors instead of bulbs.

Question 69.
When resistors are connected in series, what will be the potential difference between the ends of the resistor with higher resistance? more/less.
Answer:
More.

Question 70.
If resistors of the same value are connected in series, what will be the potential difference across the ends of the resistors? equal/ different
Answer:
Equal.

Question 71.
8 Ω, 4 Ω resistors and a 6 V battery are provided.
a) Draw a circuit diagram in which these resistances are connected in series.
b) Find the effective resistance in the circuit.
c) Calculate the current in the circuit.
Answer:

b) R = R₁ + R₂ = 8Ω + 4Ω = 12Ω
c) I = \(\frac{V}{R}\) = \(\frac{6}{12}\) = 0.5 A

Question 72.
When ten resistors of 2 Ω each are connected in series, what is the effective resistance?
Answer:
R = n × r 10 × 2 Ω = 20 Ω

Question 73.
How many resistors of resistance 6 N each should be connected in series to get an effective resistance of 42 Ω ?
Answer:
n = \(\frac{R}{r}\) = \(\frac{42}{6}\) = 7

Question 74.
Resistors 6 Ω, 3 Ω and a 6 V battery are given.
a) Draw the circuit diagram connecting them in parallel.
b) Find the effective resistance of the circuit.
c) Calculate the current through each resistor.
Answer:

b) R = \(\frac{R_1 R_2}{R_1+R_2}\)
= \(\frac{6 \times 3}{6+3}\) = \(\frac{18}{9}\) = 2Ω

c) Current through 6Ω = I₁ = \(\frac{\mathrm{V}}{R_1}\) = \(\frac{6}{6}\) = 1A
Current through 3 Ω = I₂ = \(\frac{\mathrm{V}}{R_2}\) = \(\frac{6}{3}\) = 2A

Question 75.
If resistors of the same value are connected in parallel, what will be the current through each resistor?
Answer:
It will be equal.

Question 76.
What is the effective resistance when five 10 N resistors are connected in parallel?
Answer:
\(\frac{R}{n}\) = \(\frac{10}{5}\) = 2 Ω

Question 77.
Find the effective resistance of the arrangement given in the figure.

Answer:
R = R₁ + \(\frac{R_2 R_3}{R_2+R_3}\) = 2 + 1 = 3 Ω

Question 78.
Write down in the science diary the differences between series combinations and parallel combinations of resist ors.
Answer:

Resistors in series	Resistors in parallel
Effective resistance increases	Effective resistance decreases
Current through each resistor will be equal	The current flowing through each resistor is also divided based on the value of the resistor.
The potential difference across each resistor is divided based on the value of the resistor.	The potential difference across each resistor will be the same.
It cannot be controlled using a separate switch.	It can be controlled using a separate switch.

Resistors reduce the current through a circuit. will be the same.
Usually combination of resistors of suitable values are connected in circuits to reduce the potential difference or current in accordance with our requirement.
Electric devices such as bulbs, heaters, and electric iron used in a circuit have resistors.

Question 79.
How are electric appliances such as bulbs, fans etc., connected in household circuits?
Answer:
Parallel connection.

Question 80.
What are the advantages of connecting electric appliances in parallel?
Answer:

• For each appliance,
• Gets the same voltage
• Gets the required current
• It can operate at the rated power
• It can be controlled by separate switches.

A circuit consisting of a 3 V bulb, a 12 Ω resistor and a 9 V battery is given in Figure (e). Another circuit without the 12 Ω resistor is given in Figure (f). The resistance of the bulb is 6 Ω.
Analyse the circuit and find answers to the following questions.

Question 81.
Which circuit has more resistance?
Answer:
Figure (e)

Question 82.
In which circuit, is a potential difference of 3 V obtained between the ends of the bulb?
Answer:
In Fig (e), Current I = \(\frac{V}{R}\) = \(\frac{9}{6+12}\) = \(\frac{9}{18}\) = \(\frac{1}{2}\)
Potential difference between the ends, V = I × R = (1/2) × 6 = 3 V
In Fig (f), Current I = \(\frac{V}{R}\) = \(\frac{9}{6}\) = \(\frac{3}{2}\)
Potential difference between the ends, V = I × R= (3/2) × 6 = 9V
The potential difference between the ends of the 6 2 resistor (bulb) in the circuit of Fig. (e) is 3 V. A potential difference of 3 V across the ends of the bulb causes the bulb to light up.

Question 83.
What happens to the bulb in the circuit (f) when it is switched on?
Answer:
As the bulb receives more potential difference and current than required, the bulb gets damaged.

Question 84.
Is there any difference between the electricity we obtain from battery / cell and the electricity we get in our houses?
Answer:
The electricity from the battery flows only in one direction (Direct Current). But the electricity in our houses changes direction at regular intervals of time (Alternating Current) and is of high voltage (230 V). Hence, do not connect the circuits to household electricity while doing the experiments that you have done in the class. Discuss with elders the precautions to be taken while using electricity.