Chapter 16: ELECTRICAL ENERGY
Chapter 16: Electrical Energy Class 10 Science and Technology (212) NIOS Textbook Solutions
Chapter 16. ELECTRICAL ENERGY
INTEXT QUESTIONS 16.1
1. Define the units of (i) charge (ii) electric potential.
Answer: (i) Unit of charge is Coulomb.
1C charge is the charge which when placed at a distance of 1 m from an equal like charge repels it with a of force of 9×109 N
.
(ii) Unit of potential is volt.
1 volt is the potential at a point in an electric field such that if 1C positive charge is brought from outside the field to this point against the field 1 J work is done.
2. When a glass rod is rubbed with a piece of silk it acquires +10 micro coulomb of charge. How many electrons have been transferred from glass to silk?
Answer: Here, Q=+10 μC=10×10-6 C 
and e=1.6×10-19 C 
Number of electrons
n=Qe=10×10-61.6×10-19=6.25×1013 
3. How will the force between two small electrified objects vary if the charge on each of the two particles is doubled and separation is halved?
Answer: According to Coulomb’s law ,
F=kq1q2r2 
Here, q'1=2q1 , q'2=2q2 
and r'=r2 
New force
F'=kq'1q'2r'2=k×2q1×2q2r22=4 kq1q2r²4=16×kq1q2r2=16F 
Therefore, the force becomes 16 times the original force.
4. How does the force between two small charged spheres change if their separation is doubled?
Answer: According to Coulomb’s law ,
F=kq1q2r2
Here, r'=2r 
Then the new force,
F'=kq'1q'2r'2=k×q1×q22r2= kq1q24r2=14×kq1q2r2=F4 
5. A particle carrying a charge of 1 micro coulomb (μC) is placed at a distance of 50 cm from a fixed charge where it has a potential energy of 10 J. Calculate
(i) the electric potential at the position of the particle
(ii) the value of the fixed charge.
Answer: (i) Here, U=10 
J , q=1μC=1×10-6 C 
We have,
V=Uq=101×10-6=1×107 volt 
(ii) Here, U=10 
J , q=1μC=1×10-6 C 
,r=50 cm=0.5 m
and k=9×109 Nm²C-2 
We have,
U=KQqr=10×0.59×109×1×10-6=59×103=0.56×10-3C
U=5.6×10-4 C
6. Two metallic spheres A and B mounted on two insulated stands as shown in the Fig. 16.4 are given some positive and negative charges respectively. If both the spheres are connected by a metallic wire, what will happen?
Photo Fig 16.4
Answer: Electrons will flow from sphere B to sphere A through the wire till the potentials of the two spheres become equal.
INTEXT QUESTIONS 16.2
1. Define the SI units of (i) current (ii) resistance.
Answer: 1. (i) Unit of current is ampere. 1A is the current in a wire in which 1C charge flows in 1 second.
(ii) Unit of resistance is ohm. 1 ohm is the resistance of a wire across which when 1V potential difference is applied, 1A current flows through it.
2. Name the instruments used to measure (i) current (ii) potential difference.
Answer: (i) ammeter (ii) voltmeter
3. Why is a conductor different from an insulator?
Answer: A conductor has free electrons, whereas an insulator has no free electrons.
4. How is a volt related to an ohm and an ampere?
Answer: One volt is the potential difference across a conductor when a current of 1 ampere flows through a resistance of 1 ohm.
i.e., 1 volt = 1 ohm × 1 ampere
5. A number of bulbs are connected in a circuit. Decide whether the bulbs are connected in series or in parallel, when (i) the whole circuit goes off when one bulb is fused (ii) only the bulb that get fused goes off.
Answer: (i) If the whole circuit goes off when one bulb is fused, the bulbs are connected in series.
(ii) If any one bulb goes off and the rest of the circuit remains working, the bulbs are connected in parallel.
6. When the potential difference across a wire is doubled, how will the following quantities be affected (i) resistance of the wire (ii) current flowing through the wire?
Answer: 6. (i) Resistance of the wire remains unaffected.
(ii) Current flowing through the wire is doubled.
7. What is the reading of ammeter in the circuit given below?
Fig. 16.10
Answer: Kkkkkkkkkkk
8. How can three resistors of resistance 2Ω, 3Ω and 6Ω be connected to give a total resistance of (i) 11Ω (ii) 4.5Ω and (iii) 4Ω?
Answer: (i) All the three resistors are connected in series.
(ii) Resistors 2Ω and 6Ω are connected in parallel and 3Ω is connected in series to the combination of 2Ω and 6Ω.
(iii) Resistors 3Ω and 6Ω are connected in parallel and 2Ω is connected in series to the combination of 3Ω and 6Ω.
9. State two advantages of connecting electrical devices in parallel with the battery instead of connecting them in series.
Answer: Two advantages of connecting electrical devices in parallel with the battery are:
(i) Each device works independently because every appliance takes current according to its requirement. Switching off one device does not affect the others.
(ii) If one component fails, the circuit does not break, so the other electrical devices continue to work properly.
INTEXT QUESTIONS 16.3
1. Which will produce more heat in 1 second – 1 ohm resistance on 10V or a 10 ohm resistance on the same voltage? Give reason for your answer.
Answer: Heat produced can be found using Joule’s law of heating,
H=V² tR 
Where, H = heat produced , V = voltage , R = resistance and t = time
Since both are connected to the same voltage (10 V) and for 1 second:
For 1 Ω resistance:
H=10²×11=100 J 
For 10 Ω resistance:
H=10²×110=10 J 
Therefore, 1 Ω resistance produces more heat.
Heat produced is inversely proportional to resistance when voltage and time remain constant . Hence lower resistance produces more heat.
2. How will the heat produced in a conductor change in each of the following cases?
(i) The current flowing through the conductor is doubled.
(ii) Voltage across the conductor is doubled.
(iii) Time for which current passed is doubled.
Answer: (i) Heat produced becomes four times (ii) heat produced becomes four times(iii) heat produced will be doubled.
3. 1 A current flows though a conductor of resistance 10 ohms for 1/2 minute. How much heat is produced in the conductor?
Answer: Here, i=1 A 
, R=10Ω 
, t=12×60 s=30s 
We have,
Q = i²Rt = 1 × 10 × 30 =
300 J.
4. Two electric bulbs of 40 W and 60 W are given. Which one of the bulbs will glow brighter if they are connected to the mains in (i) series and (ii) parallel?
Answer: (i) When connected in series:
The bulb with higher resistance produces more heat and glows brighter.
Power
Since the 40 W bulb has greater resistance, it produces more heat and light.
Therefore, 40 W bulb glows brighter in series.
(ii) When connected in parallel:
The bulb with lower resistance (higher wattage) consumes more power and glows brighter.
Power 
The 60 W bulb consumes more power, so it glows brighter.
Therefore, 60 W bulb glows brighter in parallel.
5. How is 1 kW h related with SI unit of energy?
Answer: 1 kW h=1000×60×60=3600000= 3.6×106 
j
6. Name two household electric devices based on thermal effect of electric current.
Answer: (i) Electric heater (ii) Electric kettle .
INTEXT QUESTIONS 16.4
1. Which has a higher resistance, a 40W-220 V bulb, or a 1 kW-220V electric heater?
Answer: (i) Here, P=40 W 
, V=220 volt 
For electrical devices operating at the same voltage, resistance is given by
R=V²P=200²40=4840040=1210Ω
(ii) Here, Here, P=1 kW=1000 W 
, V=220 volt
For electrical devices operating at the same voltage, resistance is given by
R=V²P=200²1000=484001000=48.4 Ω 
2. What is the maximum current that a 100W, 220 V lamp can withstand?
Answer: Here, Here, P=100 W 
, V=220 volt
We have,
I=PV=100220=511=4.7 Ω
3. How many units of electricity will be consumed by a 60 W lamp in 30 days if the bulb is lighted 4 hours daily?
Answer: Here, P = 60 W and t = 4h ×30 = 120 h
We have,
Q=Pt=60×120=7200 Wh=7.2 kWh 
4. How many joules of electrical energy will a quarter horse power motor consume in one hour?
Answer: Here,
P=7464W 
, t=1 h=60×60 s=3600 s 
We have,
Q=Pt=7464×3600=746×900=671400 J 
5. An electric heater is used on 220 V supply and draws a current of 5 A. What is its electric power?
Answer: Here, V=220 volt 
and I=5 A 
We have,
P=VI 
P=220×5=1100 W 
6. Which uses more energy, a television of 250 W in 60 minutes or a toaster of 1.2 kW in (1/6) th of an hour?
Answer: Energy used by television = 0.25 kW × 1 h = 0.25 kW h
Energy used by toaster = 1.2 kW × 1/6 h = 0.2 kW h.
TERMINAL EXERCISE
1. Tick mark the most appropriate answer out of four given options at the end of each of the following statements:
(a) A charged conductor ‘A’ having charge Q is touched to an identical uncharged conductor ‘B’ and removed. Charge left on A after separation will be:
(i) Q (ii) Q/2 (iii) Zero (iv) 2Q
Answer: When a charged conductor A touches an identical uncharged conductor B, the total charge distributes equally because both have the same size and capacity. Therefore, each gets = Q/2
So, charge left on A = Q/2 .
(b) JC-1
is the unit of
(i) Current (ii) Charge (iii) Resistance (iv) Potential
Answer: (iv) Potential
[ 1 volt=1 joule per coulomb=1 JC-1 
Potential difference is defined as the work done per unit charge, so =1 JC-1 is the unit of electric potential (volt).]
(c) Which of the following materials is an electrical insulator?
(i) Mica (ii) Copper (iii) Tungsten (iv) Iron
Answer: (i) Mica
[ Mica does not allow electric current to pass through easily, so it is an electrical insulator. Copper, tungsten and iron are conductors.]
(d) The device which converts chemical energy into electrical energy is called
(i) Electric fan (ii) Electric generator
(iii) Electric cell (iv) Electric heater
Answer: (iii) Electric cell .
[ An electric cell converts chemical energy into electrical energy through chemical reactions and supplies electric current to a circuit.]
(e) The resistance of a conductor does not depend on its
(i) Temperature (ii) Length (iii) Thickness (iv) Shape
Answer: (iv) Shape .
[ The resistance of a conductor depends on its length, thickness (area of cross-section), and temperature, but it does not depend on its shape.]
(f) There are four resistors of 12 Ω each. Which of the following values is possible by their combination (series and/or parallel)?
(i) 9 Ω (ii) 16 Ω (iii) 12 Ω (iv) 30 Ω
Answer: kkkkkkkkk
(g) In case of the circuit shown below in Fig. 16.12, which of the following statements is/are true:
(i) R1 , R2 and R3
are in series
(ii) R2 and R3 
are in series
(iii) R2 and R3 
are in parallel
(iv) The equivalent resistance of the circuit is [R1+ (R2 R3/R2 + R3)]
R1 R3R2+R3 
Fig. 16.12 Photo
(h) The equivalent resistance of two resistors of equal resistances connected in parallel is………….the value of each resistor.
(i) Half (ii) Twice (iii) Same (iv) One fourth
Answer: (i) Half .
If two resistors of equal resistance RRR are connected in parallel:
1Rp=1R+1R =2R ⇒Rp=R2 
So, the equivalent resistance is half the value of each resistor.
2. Fill in the blanks.
(a) When current is passed through a conductor, its temperature .................
(b) The amount of ................ flowing past a point per unit ................ is defined as electric current.
(c) A current carrying conductor carries an ................ field around it.
(d) One ampere equals one ................ per .................
(e) Unit of electric power is .................
(f) Of the two wires made of the same material and having same thickness, the longer one has ................ resistance.
Answer: (a) When current is passed through a conductor, its temperature increases.
(b) The amount of charge flowing past a point per unit time is defined as electric current.
(c) A current carrying conductor carries an magnetic field around it.
(d) One ampere equals one coulomb per second.
(e) Unit of electric power is watt.
(f) Of the two wires made of the same material and having same thickness, the longer one has more (higher) resistance.
3. How many types of electric charge exist?
Answer: here are two types of electric charge:
(i) Positive charge (+)
(ii) Negative charge (–)
4. In a nucleus there are several protons, all of which have positive charge. Why does the electrostatic repulsion fail to push the nucleus apart?
Answer: Although protons in a nucleus repel each other because they all carry positive charge, the nucleus does not break apart because of the strong nuclear force. This force acts between protons and neutrons at very short distances and is much stronger than electrostatic repulsion inside the nucleus, holding the nucleus together.
5. What does it mean to say that charge is conserved?
Answer: Charge conservation means the total electric charge in an isolated system remains constant over time. Charge cannot be created or destroyed, only transferred from one object to another.
6. A point charge of +3.0 μC is 10 cm apart from a second point charge of –1.5 μC. Find the magnitude and direction of force on each charge.
Answer: Here, k=9×109Nm2C-2
, q1=+30.μC=3.0×10-6 C 
, q2=-1.5μC=-1.5×10-6 C 
r=10 cm=10100=0.1 m 
We use Coulomb's Law,
F=k q1q2r2 
F=9×109×3×10-6×1.5×10-6 0.12 
F=40.5×109-120.01 =40.5×10-310-2 
F=40.5×10-1=4.05 N 
7. Name the quantity measured by the unit (a) VC (b) Cs-1 
Answer: (a) VC→ Electrical energy (Work done)
(b) Cs-1 → Electric current
8. Give a one word name for the unit (a) JC-1
(b) Cs-1
Answer: (a) JC-1 → Volt (b) Cs-1 → Ampere
9. What is the potential difference between the terminals of a battery if 250 J of work is required to transfer 20 C of charge from one terminal of the battery to the other?
Answer: Here, W=250 J 
and Q=20 C 
We have,
V=WQ=25020=12.5 Volt 
10. Give the symbols of (a) cell (b) battery (c) resistor (d) voltmeter.
Answer: (a) Cell: Two parallel lines: one longer (positive) and one shorter (negative).
(b) Battery: Two or more cell symbols in series.
(c) Resistor: A zigzag line (USA) or a rectangle (IEC standard).
(d) Voltmeter: A circle with the letter V inside.
11. What is the conventional direction of flow of electric current? Do the charge carriers in the conductor flow in the same direction? Explain.
Answer: The conventional direction of electric current is taken as the direction in which positive charges would flow, that is, from the positive terminal to the negative terminal of a source.
No, the charge carriers in a conductor do not flow in the same direction. In metallic conductors, current is carried by electrons, which are negatively charged. Electrons move from the negative terminal to the positive terminal, opposite to the direction of conventional current.
12. Out of ammeter and voltmeter which is connected in series and which is connected in parallel in an electric circuit?
Answer: An ammeter is connected in series in an electric circuit because it measures the current flowing through the circuit, so the entire current must pass through it.
A voltmeter is connected in parallel across the component because it measures the potential difference (voltage) between two points without interrupting the current flow.
13. You are given two resistors of 3 Ω and 6 Ω, respectively. Combining these two resistors what other resistances can you obtain?
Answer: Here, R1=3 Ω 
, R2=6 Ω 
(i) Series combination:
Rs=R1+R2=3+6=9Ω 
(ii) Parallel combination:
1Rp=13+16=2+16=36=12 
Rp =2 Ω 
So, the other resistances that can be obtained are 9 Ω and 2 Ω.
14. What is the current in SI unit if+100 coulombs of charge flows past a point every five seconds?
Answer: Here, Q=100 C 
and t=5 s 
We have,
I=Qt=1005=20 A 
So, the current is 20 ampere (A) in SI unit.
15. Deduce an expression for the electrical energy spent in flow of current through a conductor.
Answer: Electrical energy spent in the flow of current through a conductor is equal to the work done in moving charges.
We know that,
V=WQ 
⇒W=VQ= V×It=V×VR×t 
⇒W=V²Rt 
16. Find the value of resistor X as shown in Fig. 16.13.
Photo
17. 17. In the circuit shown in Fig. 16.14, find (i) Total resistance of the circuit. (ii) Ammeter reading and (iii) Current flowing through 3 Ω resistor.
Photo
18. For the circuit shown in Fig. 16.15, find the value of:
(i) Current through 12Ω resistor.
(ii) Potential difference across 6Ω and 18Ω resistor.
Photo
19. You are given three resistors of 1 Ω, 2 Ω and 3 Ω. Show by diagrams, how will you connect these resistors to get (a) 6/11 Ω (b) 6 Ω (c) 1.5 Ω?
Answer: Here, R1=1Ω 
, R2=2Ω 
, R3=3Ω
(a) Connect all three resistors in parallel.
1RP=1R1+1R2+1R3 
⇒1RP=11+12+13=6+3+26=116 
⇒RP=611 Ω 
(ii) Connect all three resistors in series.
Rs=R1+R2+R3=1+2+3=6Ω 
(iii) Connect 1 Ω and 2 Ω in series, then connect that combination in parallel with 3 Ω.
Rs=R1+R2=1+2=3Ω 
1RP=1Rs+1R3=13+13=23
⇒RP=32 Ω=1.5 Ω 
20. A resistor of 8 Ω is connected in parallel with another resistor of X Ω. The resultant resistance of the combination is 4.8 ohm. What is the value of resistor X?
Answer: Here, R=4.8 Ω 
, R1=8Ω 
, R2=XΩ 
We have,
1R=1R1+1R2 
⇒14.8=18+1X 
⇒1X=14.8-18 
⇒1X=1048-18=10-648=448=112 
⇒X=12Ω 
21. In the circuit Fig. 16.16, find
(i) Total resistance of the circuit.
(ii) Total current flowing through the circuit.
(iii) The potential difference across 4 Ω resistor.
Photo
22. How many 132 Ω resistors should be connected in parallel to carry 5 A current in 220 V line?
Answer: Here, R=132 Ω 
, V=220 volt , Itotal=5A
Let n be the number of resistors .
We have,
I1=VR=220132 
A/Q, Itoal=n×I1 
⇒5=n×220132 
⇒n=5×132220=5×0.6 
⇒n=3 
Therefore, 3 resistors should be connected in parallel.
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