10th Science Chapter 4: Electricity Book Back Questions and Answers

Book Back Questions with Answers - Electricity | Science

I. Choose the best answer

1. Which of the following is correct?

• a) Rate of change of charge is electrical power.
• b) Rate of change of charge is current.
• c) Rate of change of energy is current.
• d) Rate of change of current is charge.

2. SI unit of resistance is

• a) mho
• b) joule
• c) ohm
• d) ohm meter

3. In a simple circuit, why does the bulb glow when you close the switch?

• a) The switch produces electricity.
• b) Closing the switch completes the circuit.
• c) Closing the switch breaks the circuit.
• d) The bulb is getting charged.

4. Kilowatt hour is the unit of

• a) resistivity
• b) conductivity
• c) electrical energy
• d) electrical power

II. Fill in the blanks

1. When a circuit is open, current cannot pass through it.
2. The ratio of the potential difference to the current is known as resistance.
3. The wiring in a house consists of parallel circuits.
4. The power of an electric device is a product of voltage and current.
5. LED stands for Light Emitting Diode.

III. State whether the following statements are true or false: If false correct the statement.

1. Ohm’s law states the relationship between power and voltage. - False

   Ohm's law states the relationship between current and voltage.

2. MCB is used to protect house hold electrical appliances. - True

3. The SI unit for electric current is the coulomb. - False

   The SI unit for electric current is the ampere.

4. One unit of electrical energy consumed is equal to 1000 kilowatt hour. - False

   One unit of electrical energy consumed is equal to 1 kilowatt hour.

5. The effective resistance of three resistors connected in series is lesser than the lowest of the individual resistances. - False

   The effective resistance of three resistors connected in series is greater than the highest of the individual resistances.

IV. Match the items in column-I to the items in column-II:

Column - I	Column - II
(i) electric current	(a) volt
(ii) potential difference	(b) ohm meter
(iii) specific resistance	(c) watt
(iv) electrical power	(d) joule
(v) electrical energy	(e) ampere

Answer:

1. electric current - ampere
2. potential difference - volt
3. specific resistance - ohm meter
4. electrical power - watt
5. electrical energy - joule

V. Assertion and reason type questions:

Mark the correct choice as

• a) if both the assertion and the reason are true and the reason is the correct explanation of the assertion.
• b) if both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.
• c) if the assertion is true, but the reason is false.
• d) if the assertion is false, but the reason is true.

1. Assertion: Electric appliances with a metallic body have three wire connections.

Reason: Three pin connections reduce heating of the connecting wires.

Answer: c) if the assertion is true, but the reason is false.

2. Assertion: In a simple battery circuit the point of highest potential is the positive terminal of the battery.

Reason: The current flows towards the point of the highest potential.

Answer: c) if the assertion is true, but the reason is false.

3. Assertion: LED bulbs are far better than incandescent bulbs.

Reason: LED bulbs consume less power than incandescent bulbs.

Answer: a) if both the assertion and the reason are true and the reason is the correct explanation of the assertion.

VI. Very short answer questions.

1. Define the unit of current.

The current flowing through a conductor is said to be one ampere, when a charge of one coulomb flows across any cross section of a conductor, in one second. Hence,

$$1 \text{ ampere} = \frac{1 \text{ coulomb}}{1 \text{ second}}$$

2. What happens to the resistance, as the conductor is made thicker?

(i) Decreases: The resistance decreases as the conductor is made thicker.

(ii) Reason: Resistance is inversely proportional to area of cross section A.

i. e., $R \propto \frac{1}{A}$

here, $A = \pi r^2$

Where, r is the radius which determines the thickness.

3. Why is tungsten metal used in bulbs, but not in fuse wires?

Tungsten has high melting point, it can bear high heat for glowing. But in fuse wire, the wire used in it should melt. So a metal (wire) which has low melting point should be used in a fuse wire, but not tungsten wire.

4. Name any two devices, which are working on the heating effect of the electric current.

Electric iron, and electric toaster, or electric oven, and electric heater.

VII. Short answer questions

1. Define electric potential and potential difference.

Electric potential : The electric potential at a point is defined as the amount of work done in moving a unit positive charge from infinity to that point against the electric force.

Potential difference : The electric potential difference between two points is defined as the amount of work done in moving a unit positive charge from one point to another point against the electric force.

2. What is the role of the earth wire in domestic circuits?

The earth wire provides a low resistance path to the electric current. The earth wire sends the current from the body of the appliance to the Earth, whenever a live wire accidentally touches the body of metallic electric appliance. Thus, the earth wire serves as a protective conductor, which saves us from electric shocks.

3. State Ohm’s law.

Ohm’s law states that, at a constant temperature, the steady current 'I' flowing through a conductor is directly proportional to the potential difference 'V', between two ends of the conductor.

$I \propto V$

$I = (\frac{1}{R}) V$

$V = IR$

4. Distinguish between the resistivity and conductivity of a conductor.

Resistivity

1. Electrical resistivity of a material is defined as the resistance of a conductor of unit length and unit area of cross section.
2. Its unit is ohm metre.
3. Electrical resistivity of a conductor is a measure of the resisting power of a specified material to the passage of an electric current. It is a constant for a given material.

Conductivity

1. The reciprocal of electrical resistivity of a material is called its electrical conductivity.
2. It's unit is mho metre^-1.
3. Electrical conductivity of a conductor is a measure of its ability to pass the current through it.

5. What connection is used in domestic appliances and why?

1. All the circuits in a house are connected in parallel, so that the disconnection of one circuit does not affect the other circuit.
2. One more advantage of the parallel connection of circuits is that each electric appliance gets an equal voltage.

VIII. Long answer questions.

1. With the help of a circuit diagram derive the formula for the resultant resistance of three resistances connected: a) in series and b) in parallel

Resistors in series

1. $R_1, R_2$ and $R_3$ are connected in series. Here current through them is same, but voltage is different.
2. i.e., $V = V_1 + V_2 + V_3$
3. Here, $V = IR$, $V_1 = IR_1$, $V_2 = IR_2$, $V_3 = IR_3$
4. $\therefore IR = IR_1 + IR_2 + IR_3$
5. $IR = I (R_1 + R_2 + R_3)$
6. i.e., $\mathbf{R_S = R_1 + R_2 + R_3}$
7. When a number of resistors are connected in series, their effective resistance is equal to the sum of the individual resistances. i.e., $\mathbf{R_S = nR}$
8. The effective resistance in a series combination is greater than the highest of the individual resistances.

Resistors in parallel

1. $R_1, R_2$ and $R_3$ are connected in parallel. Here current through them is different, but voltage is same.
2. i.e., $I = I_1 + I_2 + I_3$
3. Here, $I = V/R$, $I_1 = V/R_1$, $I_2 = V/R_2$, $I_3 = V/R_3$
4. $\therefore V/R = V/R_1 + V/R_2 + V/R_3$
5. $V(1/R) = V(1/R_1 + 1/R_2 + 1/R_3)$
6. i.e., $\mathbf{\frac{1}{R_P} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}}$
7. When a number of resistors are connected in parallel, the reciprocal of the effective resistance is equal to the sum of the reciprocals of the individual resistances. i.e., $\mathbf{\frac{1}{R_P} = \frac{n}{R}}$ or $\mathbf{R_P = \frac{R}{n}}$
8. The equivalent (or) effective resistance in a parallel combination is less than the lowest of the individual resistances.

2. a) What is meant by electric current? b) Name and define its unit. c) Which instrument is used to measure the electric current? How should it be connected in a circuit?

a) Electric current is defined as the rate of flow of charges in a conductor. If a net charge 'Q' passes through any cross section of a conductor in time 't', then the current flowing through the conductor is $I = Q / t$.

b) The unit of current is Ampere. The current flowing through a conductor is said to be one ampere, when a charge of one coulomb flows across any cross section of a conductor, in one second. Hence, $1 \text{ ampere} = 1 \text{ coulomb} / 1 \text{ second}$.

c) Ammeter is used to measure electric current. It should be connected in series in a circuit.

3. a) State Joule’s law of heating. b) An alloy of nickel and chromium is used as the heating element. Why? c) How does a fuse wire protect electrical appliances?

a) Joule's law of heating states that the heat produced in any resistor is:

1. Directly proportional to the square of the current passing through the resistor.
2. Directly proportional to the resistance of the resistor.
3. Directly proportional to the time for which the current is passing through the resistor.

b) An alloy of nickel and chromium (Nichrome) is used as the heating element because it has the following properties:

1. It has high resistivity.
2. It has a high melting point.
3. It is not easily oxidized.

c) The fuse wire is connected in series in an electric circuit. When a large current passes through the circuit, the fuse wire melts due to Joule's law of heating and hence the circuit gets disconnected. Therefore, the circuit and the electric appliances are saved from any damage.

4. Explain about domestic electric circuits. (circuit diagram not required)

1. The first stage of domestic circuit is to bring the power supply to the main-box from a distribution panel, such as a transformer. The important components of the main-box are (i) a fuse box and (ii) a meter.
2. The meter is used to record the consumption of electrical energy. The fuse box contains either a fuse wire or a miniature circuit breaker (MCB).
3. The function of the fuse wire or a MCB is to protect the house hold electrical appliances from overloading due to excess current.
4. The electricity is brought to houses by two insulated wires, out of these two wires, one wire has a red insulation and is called the 'live wire'. The other wire has a black insulation and is called the 'neutral wire'.
5. The electricity supplied to house is actually an alternating current having an electric potential of 220 V. Both, the live wire and the neutral wire enter into a box where the main fuse is connected with the live wire.
6. After the electricity meter, these wires enter into the main switch, which is used to discontinue the electricity supply whenever required.
7. After the main switch, these wires are connected to live wires of two separate circuits. Out of these two circuits, one circuit is of a 5A rating, which is used to run the electric appliances with a lower power rating, such as tube lights, bulbs and fans. The other circuit is of a 15A rating, which is used to run electric appliances with a high power rating. Such as A/C. refrigerators, electric iron and heaters.

5. a) What are the advantages of LED TV over the normal TV? b. List the merits of LED bulb.

a) Advantages of LED TV:

1. It has brighter picture quality.
2. It is thinner in size.
3. It uses less power and consumes very less energy.
4. Its life span is more.
5. It is more reliable.

b) Merits of LED bulb:

1. As there is no filament, there is no loss of energy in the form of heat. It is cooler than the incandescent bulbs.
2. In comparison with the fluorescent light, the LED bulbs have significantly low power requirement.
3. It is not harmful to the environment.

IX. Numerical problems:

1. An electric iron consumes energy at the rate of 420 W when heating is at the maximum rate and 180 W when heating is at the minimum rate. The applied voltage is 220 V. What is the current in each case?

Given (Maximum Rate):

Power, P = 420 W

Voltage, V = 220 V

Solution (Maximum Rate):

We know, $P = V \times I$. Therefore, $I = P/V$.

$I = 420 / 220 = 21 / 11 \approx 1.909$ A

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Given (Minimum Rate):

Power, P = 180 W

Voltage, V = 220 V

Solution (Minimum Rate):

$I = P/V = 180 / 220 = 9 / 11 \approx 0.818$ A

2. A 100 watt electric bulb is used for 5 hours daily and four 60 watt bulbs are used for 5 hours daily. Calculate the energy consumed (in kWh) in the month of January.

Energy consumed by 100W bulb:

Energy = Power × time × days = 100 W × 5 hr/day × 31 days = 15500 Wh = 15.5 kWh

Energy consumed by four 60W bulbs:

Energy = (4 bulbs × 60 W/bulb) × 5 hr/day × 31 days = 240 W × 5 hr/day × 31 days = 37200 Wh = 37.2 kWh

Total Energy Consumed:

Total = 15.5 kWh + 37.2 kWh = 52.7 kWh

Total energy consumed = 52.7 kWh

3. A torch bulb is rated at 3 V and 600 mA. Calculate it’s a) power b) resistance c) energy consumed if it is used for 4 hour.

Given:

Voltage, V = 3 V

Current, I = 600 mA = $600 \times 10^{-3}$ A = 0.6 A

a) Power:

P = V × I = 3 V × 0.6 A = 1.8 W

b) Resistance:

R = V / I = 3 V / 0.6 A = 5 Ω

c) Energy consumed in 4 hours:

Energy = Power × time = 1.8 W × 4 hr = 7.2 Wh

4. A piece of wire having a resistance R is cut into five equal parts.

a) How will the resistance of each part of the wire change compared with the original resistance?

Resistance is proportional to length ($R \propto l$). If the wire is cut into five equal parts, the length of each part is $l/5$. Therefore, the resistance of each part will be $R/5$.

b) If the five parts of the wire are placed in parallel, how will the resistance of the combination change?

The resistance of each part is $R' = R/5$. For parallel connection:

$\frac{1}{R_p} = \frac{1}{R'} + \frac{1}{R'} + \frac{1}{R'} + \frac{1}{R'} + \frac{1}{R'} = \frac{5}{R'}$

$\frac{1}{R_p} = \frac{5}{R/5} = \frac{25}{R}$

$R_p = \frac{R}{25}$

c) What will be ratio of the effective resistance in series connection to that of the parallel connection?

Effective resistance in series ($R_s$):

$R_s = R' + R' + R' + R' + R' = 5R' = 5(\frac{R}{5}) = R$

Ratio:

$\frac{R_s}{R_p} = \frac{R}{R/25} = 25$

Ratio $R_s : R_p$ is 25:1

XI. HOTS:

1. Two resistors when connected in parallel give the resultant resistance of 2 ohm; but when connected in series the effective resistance becomes 9 ohm. Calculate the value of each resistance.

Given:

Series Resistance: $R_s = R_1 + R_2 = 9 \, \Omega$ (Equation 1)

Parallel Resistance: $R_p = \frac{R_1 R_2}{R_1 + R_2} = 2 \, \Omega$ (Equation 2)

Solution:

Substitute Eq. 1 into Eq. 2: $\frac{R_1 R_2}{9} = 2 \implies R_1 R_2 = 18$

From Eq. 1, $R_2 = 9 - R_1$. Substitute this into the product equation:

$R_1 (9 - R_1) = 18$

$9R_1 - R_1^2 = 18$

$R_1^2 - 9R_1 + 18 = 0$

Factoring the quadratic equation: $(R_1 - 3)(R_1 - 6) = 0$

So, $R_1 = 3 \, \Omega$ or $R_1 = 6 \, \Omega$.

If $R_1 = 3 \, \Omega$, then $R_2 = 9 - 3 = 6 \, \Omega$.

If $R_1 = 6 \, \Omega$, then $R_2 = 9 - 6 = 3 \, \Omega$.

The resistances are 3 Ω and 6 Ω.

2. How many electrons are passing per second in a circuit in which there is a current of 5 A?

Given:

Current, I = 5 A

Time, t = 1 s

Charge of one electron, $e = 1.6 \times 10^{-19}$ C

Solution:

We know that $I = Q/t$ and total charge $Q = n \times e$, where 'n' is the number of electrons.

So, $I = \frac{ne}{t}$

Rearranging for n: $n = \frac{I \times t}{e}$

$n = \frac{5 \, A \times 1 \, s}{1.6 \times 10^{-19} \, C} = \frac{5}{1.6} \times 10^{19}$

$n = 3.125 \times 10^{19}$ electrons

Number of Electrons, n = $3.125 \times 10^{19}$ electrons

3. A piece of wire of resistance 10 ohm is drawn out so that its length is increased to three times its original length. Calculate the new resistance.

Given:

Original Resistance, $R_{old} = 10 \, \Omega$

New Length, $L_{new} = 3 \times L_{old}$

Solution:

The formula for resistance is $R = \rho \frac{L}{A}$. When the wire is drawn out, its volume (V = L × A) remains constant.

$L_{old} \times A_{old} = L_{new} \times A_{new}$

$L_{old} \times A_{old} = (3 L_{old}) \times A_{new} \implies A_{new} = \frac{A_{old}}{3}$

Now, calculate the new resistance:

$R_{new} = \rho \frac{L_{new}}{A_{new}} = \rho \frac{3 L_{old}}{A_{old}/3} = 9 \left(\rho \frac{L_{old}}{A_{old}}\right)$

Since $R_{old} = \rho \frac{L_{old}}{A_{old}}$, we have:

$R_{new} = 9 \times R_{old} = 9 \times 10 \, \Omega = 90 \, \Omega$

New Resistance = 90 Ω

Concept Map