## Chapter 12 Electricity Short Questions And Answers

**Question 1. Redraw the circuit question 1, putting in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the 12Ω resistor. What would be the readings in the ammeter and the voltmeter?**

**Answer:** The required circuit diagram is given below

Reading of ammeter, \(I=\frac{V}{R}\)

⇒ \(=\frac{6}{25}\)

= 0.24A

Reading of voltmeter V=I.R

0.24×12

= 2.88 v

**Question 2. Judge the equivalent resistance when the following are connected:**

- Inland 106£l,
- 1 £l and 103£l, and 106£l.

**Answer:** When the resistors are connected in parallel, the equivalent resistance is smaller than the smallest individual resistance.

Equivalent resistance < 1Ω.

Equivalent resistance <1Ω.

**Read and Learn More CBSE Class 10 Science Short Answer Questions**

**Question 3. What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?**

**Answer:** **The advantages of connecting electrical devices in parallel with the battery are as follows:**

Each device gets the full battery voltage. The parallel circuit divides the current through the electrical devices. Each device gets proper current depending on its resistance. If one device is switched OFF/ON, others are not affected.

**Question 4. Why does the 1 cord of an electric heater not glow while the heating element does?**

**Answer:** Both the cord and the heating element ofan electric heater carry the same current. However, the heating element becomes hot due to its high resistance (H = I2Rf) and begins to glow. The cord remains cold due to its low resistance and does not glow.

**Question 5. Compute the heat generated while transferring 96,000 coulombs of charge in one hour through a potential difference of 50 V.**

**Answer:** Here, Q = 96,000 C, t = 1 hour = 3600 s, V = 50 V

The heat generated, H = VQ

H = 50 V x 96,000 C

H = 48,00,000 J.

**Question 6. An electric iron of resistance 20H takes a current of 5 A. Calculate the heat developed in 30 seconds.**

**Answer:** Here, R = 20 Q, 1 = 5 A, t = 3 s

Heat developed is

H= I2Rt

H=25x20x30

H=15000 J.

**Question 7. An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2h.**

**Answer:**

Here, I = 5 A. V = 220 V, / = 2 h = 7200 s

Power, P = VI

=220×5

=1100W

Energy consumed = px

= 100wx7200s

=7,20,000J.

**Question 8. Compare the power used in the 2Ω resistor in each of the following circuits: 1. a 6 V battery in series with 1Ω and 2 Ω resistors, and a 4 V battery in parallel with 12 Ω and 2Ω resistors.**

**Answer:** The circuit diagram is shown in the figure

Total resistance R=1+2=3

Potential difference v=6v

Current \(\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{6 \mathrm{~V}}{3 \Omega}=2 \mathrm{~A}\)

Power used in 2Ω resistor = I2R= (2)2×2=8w.

**The circuit diagram for this case is shown below:**

Power used in 2Ω1 resistor \(\begin{aligned}

& =\frac{V^2}{R}=\frac{(4)^2}{2} \\

& =8 \mathrm{~W}

\end{aligned}\)

=8w

**Question 9. Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to the electric mains supply. What current is drawn from the line if the supply voltage is 220 V?**

**Answer:** Total power consumed in the circuit = 100 + 60 = 160 W

voltage v=220

power =vI

Current \(I=\frac{\text { Power }}{V}\)

⇒ \(\begin{aligned}

& =\frac{160}{220} \\

& =0.727 \mathrm{~A} .

\end{aligned}\)

=0.727A

**Question 10. Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?**

**Answer:** Energy used by 250 W TV set in 1 hour

= 250 W x 1 h = 250 Wh

Energy used by 1200 W toaster in 10 minutes.

= 1200 W x 12 min.

⇒ \(=1200 \mathrm{~W} \times \frac{12}{60} h\)

= 240 Wh

Thus, the TV set uses more energy than the toaster.

**Question 11. An electric heater of resistance 8 Q draws 15 A from the service mains for 2 hours. Calculate the rate at which heat is developed in the heater.**

**Answer:** Here, R = 8 £2,I = 15A, t = 2h

The rate at which heat is developed in the heater is equal to the power

Therefore \(\begin{aligned}

\mathrm{P} & =\mathrm{I}^2 \mathrm{R}=(15)^2 \times 8 \\

& =\mathbf{1 8 0 0} \mathbf{J s}^{-1}

\end{aligned}\)

**Question 12. Explain why a conductor offers resistance to the flow of current. Differentiate between conductor, resistor, and resistance.**

**Answer:** We know that the motion of electrons in an electric circuit constitutes an electric current. However, the electrons are not completely free to move within the conductor. They are restrained by the attraction of the atoms among which they move.

**Conductor:** Any substance that is capable of readily transmitting heat, electricity, etc., is called a conductor.

**Resistor:** A conductor having some appreciable resistance is called a resistor.

**Resistance:** It is the property of a material by which it opposes the flow of current through it.

**2. A piece of wire of resistance 6Q is connected to a battery of12 V. Find the amount of current flowing through it. Now, the same wire is redrawn by stretching it to double its length. Find the resistance of the new (redrawn) wire.**

**Answer:** Here, resistance (R) = 6 Q, potential difference (V) = 12 V

V=IR

Or \(I=\frac{V}{R}=\frac{12 \mathrm{~V}}{6 \Omega}=2 \mathrm{~A}\)

**Original Resistance**

⇒ \(\mathrm{R}=\rho \frac{l}{\mathrm{~A}}\)

If the wire is doubled, then its new lengthl’ = 21 and new area of cross-section \(\mathrm{A}^{\prime}=\frac{\mathrm{A}}{2} \text {. }\)

**The New Resistance**

⇒ \(\mathrm{R}^{\prime}=\rho \frac{l^{\prime}}{\mathrm{A}^{\prime}}=\rho \frac{2 l}{\frac{\mathrm{A}}{2}}=\frac{4 \rho l}{\mathrm{~A}}=4 \mathrm{R}\)

or R’ = 4R, i.e., the resistance increases to four times

**Question 13. Mention the condition under which charges can move in a conductor. Name the device which is used to maintain this condition in an electric circuit.**

**Answer:** Charges can move, if there is a potential difference across the two ends ofthe conductor. An electric cell or a battery consisting of one or more cells can maintain an electric current in an electric circuit.

**Question 14. Define the term ‘volt’. State the relation between work, charge, and potential difference for an electric circuit. Calculate the potential difference between the two terminals of a battery, if100joules ofwork is required to transfer 20 coulombs of charge from one terminal ofthe battery to the other.**

**Answer:** One-volt potential difference between two points in a current-carrying conductor when 1 joule of work is done to move a charge of coulomb from one point to the other.

The relation is V = W/Q

Given, W = 100 J, Q = 20, V = ?

Using the relation, V = W/Q, we have

V = 100/20 = 5 V

**Question 15. (a) How is the direction of electric current related to the direction of flow of electrons in a wire? Calculate the current in a circuit, if 500 C of charge passes through it in 10 minutes.
**

**Answer:**The direction of electric current is opposite to the direction of flow of electrons in a wire.

Given Q = 500 C,

t = 10 minute

= 10 x 60 = 600 s,I = ?

Using the relation, \(I=\frac{Q}{T}\)

= 500/600=083A

**Question 16. Define electric current and state its SI unit. With the help of Ohm’s law explain the ****meaning of 1-ohm resistance.**

**Answer:** It is defined as the rate of flow of electric charge through an electric circuit. Its SI unit is ampere (A). Ohm’s law is V = IR Let V = 1 ohm and1=1 ampere, then R = 1 volt/1 ampere =1 ohm Thus, 1 ohm is the resistance of a conductor, under the effect of a potential difference of volt across it a current of ampere flows through it.

**Question 17. Find the current drawn from the battery by the network of four resistors**

**Answer:** Equivalent resistance of the given network is

⇒ \(\frac{1}{\mathrm{R}}=\frac{1}{\mathrm{R}_4}+\frac{1}{\mathrm{R}_1+\mathrm{R}_2+\mathrm{R}_3}=\frac{1}{10}+\frac{1}{10+10+10}\)

⇒ \(=\frac{1}{10}+\frac{1}{30}=\frac{3+1}{30}=\frac{4}{30}\)

⇒ \(\mathrm{R}=\frac{30}{4}=7.5 \Omega\)

⇒ \(\begin{aligned}

& \mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{3}{7.5}=\frac{30}{75}=\frac{2}{5} \\

& \mathrm{I}=0.4 \mathrm{~A}

\end{aligned}\)

**Question 18. State the formula co-relating the electric current flowing in a conductor and the voltage applied across it. Also, show this relationship by drawing a graph. What would be the resistance of a conductor, if the current flowing through it is 0.35 ampere when the potential difference across it is 1.4 volts?**

**Answer:** It states that Physical conditions remain the same the current flowing through the conductor is directly proportional to the potential applied across its two ends.”

**The graph is shown below**

Given V = 1.4 V,I = 0.35 A

Now, resistance is given by the expression

⇒ \(\mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}}=\frac{1.4}{0.35}=4 \Omega\)

**Question 19. The resistivity of two elements A and B are 1.62 x 10~8 Q m and 520 x respectively. Out ofthese two, name the element that can be used to make:**

- The filament of the electric bulb.
- Wires for electrical transmission lines. Justify your answer in each case.

**Answer: Element B:** It has more resistivity (520 x 10-8 £2 m).

**Element A:** It has less resistivity and hence less heating effect/dissipation of energy during transmission of power.

**Question 20. Draw the nature of the V-Igraph for a nichrome wire (V – Potential difference, I – Current) A metallic wire of625 mm in length offers a 4 12 resistance. If the resistivity ofthe metal is 4.8 x IQ-7 ohm-meter, then calculate the area of the cross-section of The wire. **

**Answer: The graph is as shown:**

Given p = 4.8 x 10-7n m,

L = 625 mm = 0.625 m, R = 4 Q, A = ?

Using the expression \(\mathrm{R}=\frac{\rho \mathrm{L}}{\mathrm{A}}\) we have

⇒ \(A=\frac{\rho L}{R}\)

⇒ \(\begin{aligned}

& =\frac{4.8 \times 10^{-7} \times 0.625}{4} \\

& =0.75 \times 10^{-7} \mathrm{~m}^2

\end{aligned}\)

**Question 21. Derive the relation R=R1+R2 + R3 when three resistors Rp R2 andR3 are connected in series in an electric circuit.**

**Answer:** Consider three resistors of resistances, R1, R2, and R3 connected in series to a cell of potential V as shown in the figure. Since the three resistors are connected in series, therefore, the current through each of them is the same. By Ohm’s law

V1=iR1 V2=IR 2 V3=IR3

IfRs are the equivalent resistance ofthe series combination, then on applying a potential difference V across it, the same must flow through it.

Therefore,

V = IRS

v = v1 + v2 + v3

IRS = IR1 + IR2 + IR3

Rs = Rx + R2 + R3

**Question 22. (a) Nichrome wire of length L and radius ‘R’ has a resistance of 10 Q. How would the resistance ofthe wire change when:**

- only the length ofthe wire is doubled.
- only diameter ofthe wire is doubled1? Justify your answer.
- Why are elements of electrical heating devices made made-up of alloys?

**Answer:**

R⇒ L

So resistance becomes two times,

i.e., R = 2 x 10 = 20 Q.

⇒ \(\mathrm{R} \propto \frac{1}{\mathrm{~A}} \propto \frac{1}{\mathrm{D}^2}\)

So when the diameter is doubled, resistance becomes \(\frac{1}{4}[latex] of its original

i.ec., R=10/4=2.5

This is because alloys do not oxidize readily at high temperatures.

**Question 23. Draw schematic diagrams ofan electric circuit comprising of3 cells and an electric bulb, ammeter, and plug-key in the ON mode and another with the same components but with two bulbs in parallel and a voltmeter across the combination.**

**Answer:**

**Question 24. The wattage of a bulb is 24 W when it is connected to a 12 V battery. Calculate its ****effective wattage if it operates on a 6 V battery (Neglect the change in resistance due to ****unequal heating ofthe filament in the two cases).**

**Answer: Given:** = 24 W, Vx = 12 V, P2 = ?, V2 = 6 V

Using [latex]\begin{aligned}

\mathrm{P} & =\frac{\mathrm{V}^2}{\mathrm{R}} \\

\frac{\mathrm{P}_1}{\mathrm{P}_2} & =\frac{\mathrm{V}_1^2}{\mathrm{~V}_2^2}

\end{aligned}\)

⇒ \(\begin{aligned}

\mathrm{P}_2 & =\left(\frac{\mathrm{V}_2}{\mathrm{~V}_1}\right) \times \mathrm{P}_1 \\

& =\left(\frac{6}{12}\right)^2 \times 24 \\

& =\frac{1}{4} \times 24 \\

& =6 \mathrm{~W}

\end{aligned}\)

**Question 25. How much current will an electric bulb draw from a 220 V source, if the resistance of the bulb is 1200 Q? If in place of the bulb, a heater of resistance 100 Q is connected to the sources, calculate the current drawn by it.**

**Answer: Given:** V:220 V, Rx = 1200 £2,1, =?

R2 = 100 Q, I2 =?

Using OHm’s Law

⇒ \(\begin{aligned}

\mathrm{V} & =\mathrm{I}_1 \mathrm{R}_1 \\

\mathrm{I}_1 & =\frac{\mathrm{V}}{\mathrm{R}_1} \\

& =\frac{220}{1200} \\

& =0.18 \mathrm{~A}

\end{aligned}\)

And, \(\begin{aligned}

\mathrm{I}_2 & =\frac{\mathrm{V}}{\mathrm{R}_2} \\

& =\frac{220}{100} \\

& =2.2 \mathrm{~A}

\end{aligned}\)

**Question 26. Out of the two wires X and Y shown below, which one has greater resistance? Justify your answer.
**

**Answer:**Wire ‘Has greater resistance as it has more length than wire It is because the resistance of wire is directly proportional to the length of wire for the same area of cross-section for the same material.

**Question 27. An electric iron has a rating of 750 W, 220 V. Calculate the current flowing through it, and its resistance when in use.**

**Answer: Given:** P = 750 W, V = 220 V

p=VI

750=220xI

⇒\(I=\frac{750}{220}=3.40 \mathrm{~A}\)

⇒\(\begin{aligned}

\mathrm{P}=\frac{\mathrm{V}^2}{\mathrm{R}} \Rightarrow \mathrm{R}=\frac{\mathrm{V}^2}{\mathrm{P}} & =\frac{220^2}{750} \\

& =64.53 \Omega

\end{aligned}\)

**Question 28. Why is the parallel arrangement used in domestic wiring?**

**Answer:** Parallel arrangement is used in domestic wiring because of the following:

All the appliances work at the same voltage as that of the electric supply.

If one ofthe appliances is out of order, e.g., if a bulb gets fused, all other appliances keep on working as the circuit is not broken in the parallel arrangement of devices.

**Question 29. Let the resistance of an electrical component remain 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?**

**Answer:** When the potential difference is halved, the current through the component also decreases to half of its initial value. This is per Ohm’s law, i.e., V °c I.

**Question 30. On what factors does the resistance of a conductor depend?**

**Answer:** The resistance of the conductor depends

- On its length,
- On its area of cross-section and the nature of its material.

**Question 31. Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?**

**Answer:** The current will flow more easily through a thick wire than a thin wire of the same material. The larger the area of the cross-section of a conductor, the greater the ease with which the electrons can move through the conductor. Hence the lower the resistance of the conductor.

**Question 32. Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?**

**Answer:** **The coils of electric toasters and electric irons are made of alloys instead of pure metal due to the following reasons:**

Alloys have higher resistivity than that of their constituent metals.

Alloys do not oxidize (or burn) readily at high temperatures.

**Question 33. Use the data to answer the following:**

- Which among iron and mercury is a better conductor?
- Which material is the best conductor?

**Answer:** Resistivity of iron = 10.0 x 10-8 ft m Resistivity of mercury = 94.0 x 10-8 ft m Thus, iron is a better conductor because it has lower resistivity than mercury.

As silver has the lowest resistivity (= 1.60 x 10-8 ft m), silvers are the best conductor.

**Question 34. Draw a schematic diagram of a circuit consisting of a battery of three cells of Veach, a 5 ft resistor, an S ft resistor a 12 ft resistor, and a plug key, all connected in series.**

**Answer:** The Required circuit diagram is shown

**Question 35. What determines the rate at which energy is delivered by a current **

**Answer:** The resistance of the circuit determines the rate at which energy is delivered by a current.

**Question 36. How is a voltmeter connected in the circuit to measure the potential difference between two points?**

**Answer:** A voltmeter is connected in parallel to measure the potential difference between two points in a circuit

## Chapter 12 Electricity Multiple Choice Questions

**Question 1. A piece of wire of resistance is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance ofthis combination is’, then the ratio’ is:**

- 1/25
- 1/5
- 5
- 25

**Answer:** Resistance of each part= r/5

When the five parts are connected in parallel, the equivalent resistance R’ is given by

⇒ \(\begin{aligned}

& \frac{1}{\mathrm{R}^{\prime}}=\frac{5}{\mathrm{R}}+\frac{5}{\mathrm{R}}+\frac{5}{\mathrm{R}}+\frac{5}{\mathrm{R}}+\frac{5}{\mathrm{R}}=\frac{25}{\mathrm{R}} \\

& \frac{\mathrm{R}}{\mathrm{R}^{\prime}}=25

\end{aligned}\)

**Question 2. Which of the following terms does not represent electrical power in a circuit?**

- IR2
- \P/R
- PR
- VI

**Answer:** IR2 does not represent electrical power.

**Question 3. An electric bulb is rated 220 V and 100 W. When it is operated at 110 V, the power consumed will be:**

- 75 W
- 25 W
- 100 W
- 50 W

**Answer:** 4. Resistance \(\mathrm{R}=\frac{\mathrm{V}^2}{\mathrm{P}}\)

⇒ \(\begin{aligned}

& =\frac{(220)^2}{100} \\

& =484 \Omega

\end{aligned}\)

When operated at 110 V, the power consumed will be

⇒ \(\begin{aligned}

P^{\prime} & =\frac{V^2}{R} \\

& =\frac{(110)^2}{484} \\

& =25 \mathrm{~W}

\end{aligned}\)

**Question 4. Two conducting wires ofthe same material and of equal lengths and equal diameters are first connected in series and then in parallel in a circuit across the same potential difference. The ratio of produced in series and parallel combinations would be**

- 1:2
- 2:1
- 1:4
- 4:1

**Answer:** Let R be the resistance of each wire. In series combination, the total resistance will be 2R. Heat produced.

**Question 2. Why do V birds sitting on live wire get an electric shock?**

**Answer:** Birds when sitting on the live wire just add to the circuit, it does not get shocked because the current is not flowing out of their body to any other material.

**Question 3. There is a frill of 20 bulbs connected in series in a room. One bulb gets fused. The remaining 19 are again joined in series and connected to the same supply. Will the light increase or decrease in the room?**

**Answer:** The light will increase.If the voltage remains the same, then the power available to the frill

\(\mathrm{P}=\frac{\mathrm{V}^2}{\mathrm{R}}\) where R is resistance ofthe frill. When one bulb is removed, the resistance of the frill decreases and hence the power output will increase.

**Question 4. Two perform the experiments on series and andparallel combinations of two given ****resistors R1 and R2 and plot the following V-Igraphs**

**Which of the graphs is (are) correctly labeled in terms ofthe words ‘series and ‘parallel’1? Justify your answer.**

**Answer:** In a series combination for a given voltage, the current is less as compared to that in a parallel combination. Therefore, both graphs are labeled correctly.