KPK Grade 10 Chapter No 14 Physics class 10 current electricity important questions, Conceptual Questions, Comprehensive Questions, Short question, long question, and, Numerical Questions.

**Class XYZ Notes Conceptual Questions** **Physics class 10 current electricity Notes Chapter No 14**

**Q.1) As water is made of atoms having protons (charge +e) and electrons (charge -e), does the water flowing through pipe carry an electric current? Explain.**

**Answer:**

No, the water flowing through a pipe does not carry an electric current. An electric current is the flow of electrons from negative to positive terminal due to a potential difference. Water molecule as a whole is neutral and its flow does not create an electric potential difference. So no electric current is present in the pipe.

**Q.2) A car has two headlights, when the filament in one headlight burns out, the other headlight stays on. Are the headlights connected in series or in parallel?**

**Answer:**

The headlights are connected in parallel. In a parallel combination, the circuit has more than one paths to flow. So if one headlight burns out, the resistor (filament) of the second headlight keeps the current flowing.

**Q.3) Qurat-ul-Ain needs a 100-Ω resistor for a circuit, but she only has a box of 300-Ω resistors. What can she do?**

**Answer:**

She can put three resitors of 300 Ω in parallel combination, in this way the equivalent resistance “R_{e}” becomes 100 Ω.

1/R_{e} = 1/300 + 1/300 + 1/300

Taking 1/300 common

1/R_{e} = (1/300) × (1 + 1 + 1)

1/R_{e} = (1/300) × (3)

1/R_{e} = (1/~~3~~00) × (~~3~~)

1/R_{e} = (1/100)

Or**R _{e} = 100 Ω**

**Q.4) A number of light bulbs are connected to a single power outlet. Will they provide more illumination when connected in series or in parallel? Why?**

**Answer:**

If we want to get maximum illumination of bulbs, then we have to connect them in parallel combination.**Reason:**In a parallel combination, the voltage remains constant across each bulb. Therefore the illumination of each bulb will be the same and there will not be a gradual decrease in the illumination of bulbs.

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**Q.5) Explain why light bulbs almost always burn out just as they are turned on and not after they have been on for some time.**

**Answer:**

The function of a light bulb is to provide heat energy at the cost of electrical energy. In other words, they convert electrical energy to heat energy and a small amount will convert to provide light.

That is the reason that they burn out just after they are switched on.

**Q.6) Explain why is it possible for birds to perch safely on high tension wires without being electrocuted?**

**Answer:**

The feet of the bird are insulators. Therefore the potential difference between the wire and feet of the bird is zero. Hence it is possible for a bird to perch on a high voltage wire without being electrocuted.

**Q.7) An electrician working on “live” circuits wears insulated shoes and keeps one hand behind his or her back. Why?**

**Answer:**

An electrician wears insulated shoes to save themselves from getting a shock. This prevents them from getting grounded through their feet.

And the reason that they keep one hand behind their back is to decrease the probability of touching a grounded wire or any conducting object.

**Q.8) Explain why is it dangerous to turn on a lightbulb when you are in a bath tub?**

**Answer:**

The human body is a very good conductor of electricity and so as the water. Therefore, when a person in bath tub will try to switch on a light then he may get an electric shock which could be fatal.

**Q.9) Why circuit breaker, fuses and switches are installed to ‘live wire’?**

**Answer:**

Circuit breaker, fuses and switches are installed to ‘live wire’ because it carries the AC current. If the current increase above the limit it will break the circuit.

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**important questions of current electricity class 10 Chapter No 14**

**Q.1) Define electric current. In what units it is measured? Discuss briefly the direction of current through the conductor.**

**Answer:****Electric Current:***“Rate of flow of charge through a cross-sectional area of a conductor is known as electric current”*

**Mathematical Expression**

Mathematically electric current is given as,

**Unit**

SI unit of charge is ampere. It can be defined as,*“If one coulomb charge will flow through a conductor in a time of one second then the current produced in the conductor will be one ampere”***Direction of Current through a Conductor:**The direction of conventional current is the direction of in which positive charges flow.

In conductors (e.g metals), the actual flow of current is due to electrons which flows from negative terminal to the positive terminal of the battery. However, before the discovery of electrons, the positive to the negative terminal of the battery even for conductors.

**Q.3) State and explain Ohm’s Law. What are its limitations?**

**Answer:****Ohm’s Law:**

In 1926, a German scientist George Simon Ohm performed a number of experiments and observed that if we held temperature and other physical condition to be constant then the current passing through an electrical conductor is directly proportional to the applied voltages.**Explanation**

According to Ohm’s law, if we connect a wire to a 3 V battery and another same wire to a 1.5 V battery then the current in first wire will be of twice magnitude as that will flow through second wire. There is another factor which affects the current of the wire known as resistance. Current and resistance are inversely proportional to each other i.e. if we double the resistance then current will reduce to half.

Experimental set up for the verification of Ohm’s law is shown in figure (a). If we draw a graph between applied voltage and current then it will give us a straight line as shown in figure (b).

**Mathematical Expression**

According to the statement of Ohm’s law we have,

Here, ‘R’ is the constant of proportionality and known as the resistance of a conductor.**Limitation**

Ohm’s law is applicable only for those materials which are metallic conductors. And all those materials for which temperature cannot be held constant, Ohm’s law is not applicable.

**Q.4) What is resistance and in what units we measure resistance?**

**Answer:***” The opposition offered to the flow of charges is called electrical resistance “.*

By Ohm’s law the resistance is the ratio of the voltage ‘V’ across the conductors to the current ‘I’ it carries, mathematically

R = V/I**Unit:**The SI unit of resistance is ohm an is represented by Greek letter (omega) Ω.

The resistance of a wire is one ohm if the potential difference of one volt applied across its ends and causes a current of one ampere to flow through it.

1 Ω = 1 V/1A

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**Q.5) What is series combination of resistors? How we can determine equivalent resistance for different resistors connected in series?**

**Answer:****Series Combination of Resistors:**

In a series combination of resistors, current will pass through each resistance one by one i.e. one after other.

Consider three resistances R_{1}, R_{2} and R_{3} are connected in a series combination as shown in the figure. Now let a current ‘I’ is flowing through the circuit such that it will pass through each resistance one by one i.e. same current I pass through resistance R_{1}, R_{2, }and R_{3}. If a potential difference ‘V’ is applied to the circuit the V_{1}, V_{2} and V_{3} be the potential differences across resistances R_{1}, R_{2} and R_{3}. Then according to Ohm’s law, we have,

V_{1} = IR_{1} ………………………….. (1)

V_{2} = IR_{2} ………………………….. (2)

V_{3} = IR_{3 } ………………………….. (3)

And if R_{e} be the equivalent resistance of the combination then,

V = IR_{e} ………………………….. (4)

And we know that for a series combination, net voltage is given by,

V = V_{1} + V_{2} + V_{3}

Putting values from equation (1), (2), (3) and (4) we get

IR_{e} = IR_{1 }+ IR_{2} + IR_{3}

Or,

R_{e} = R_{1} + R_{2} + R_{3}**Characteristics for a series combination is given below,**

1. The voltage across each resistance in series combination is different.

2. Current flowing through the whole circuit is constant as it has only one path to

flow.

3. Sum of individual resistances is equal to the equivalent resistance of the circuit.

**Q.6) What is parallel combination of resistors? How we can determine equivalent resistance for different resistors connected in parallel?**

**Answer:****Parallel Combination of Resistors:**

If we connect a number of resistors such that one end of each resistor is connected to one point and other ends to other points then such a combination is called a parallel combination of resistors. In a parallel combination of resistors, current will have different ways to flow.

Consider three resistances R_{1}, R_{2} and R_{3} are connected in parallel combination as shown in figure (a) Now let a current ‘I’ is flowing through the circuit such that it is divided into three parts i.e. I1 passes through resistance R_{1}, I_{2} through R_{2} and similarly I_{3} is flowing through R_{3}. If a potential difference ‘V’ is applied between points A and B then according to Ohm’s law we have,

And if R_{e} be the equivalent resistance of the combination then,

And we know that for a parallel combination, net current is given by,

I = I_{1} + I_{2} + I_{3}

Putting values from equation (1), (2), (3) and (4) we get

**Characteristics**

Characteristics for a parallel combination is given below,

1. The voltage across each resistance in parallel combination is constant.

2. Current flowing in each resistance is different.

3. Sum of reciprocals of individual resistances is equal to the reciprocal of equivalent resistance of the circuit.

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**Q.7) Explain the factors on which the resistance of metallic conductor depends.**

**Answer:**

The factors on which the resistance of a metallic conductor depends are:**A. Length**

Resistance increases with increase in length

R ∝ L**B. Cross-sectional Area**

Resistance increases by decreasing cross-sectional area

R ∝ 1/A**C. Temperature**

Resistance increases as temperature increases**D. Material**

Keeping Length, Cross-sectional Area and Temperature constant, resistance also varies by using different materials.

**Q.8) What are ohmic and non-ohmic devices? Sketch and interpret the VI characteristic graph to justify metallic conductor, filament lamp and thermistor as ohmic or non-ohmic materials.**

**Answer:***” Materials that obey Ohm’s law, and hence have a constant resistance over a wide range of voltages, are said to be Ohmic. And materials having a resistance that changes with voltage or current are non-Ohmic “.*Ohmic conductors have a linear voltage-current relationship over a large range of applied voltages. Only metals show Ohmic behavior.

The graph of filament bulb and thermistor are curved, therefore, they are termed as non-Ohmic conductors.

**A. Metallic Conductors:**

For metallic conductors and some alloys, the graph of ‘I’ versus ‘V’ is a straight line as shown in the graph below. For example, when the potential difference is doubled, the current through metallic conductors also doubles.

**B. Filament Bulb:**The graph of filament bulb shows that current saturates as it is increased and at large value and even a large change in voltage ‘V’ will show a small change in the current ‘I’ as shown in the graph below.

This is because the tungsten wire in the filament of the bulb heats up with the increase in applied potential difference and free electrons collide more with lattice atoms causing the resistance to increase and consequently the rate and the rate of change in current decreases.

**C. Thermistor:**Thermistor is a device whose resistance significantly (highly) with temperature. Usually, thermistor’s resistance decreases with increase in temperature. The IV graph of the thermistor shows that resistance decreases sharply. Thus, at a large value even for a small change in applied voltage ΔV, it will show a large change in current ‘I’ as shown in the graph below.

**Q.9) Explain the concept of electric energy and electric power. What is the commercial unit for the consumption of electric energy?**

**Answer:****Electric Energy:**

We know that if we want to move a charge from one point to another point in an electric field then we need to supply some energy to that charge. Similarly, if some amount of charge is flowing from one point to another point then they will liberate some amount of energy. This energy is known as electrical energy.

Consider the potential difference of ‘V’ volt is applied between two points. If ‘Q’ coulomb charge passes between these two points then the energy supplied by this charge will be ‘W’. This amount of energy is given by,

Related: Biology Class 10 Notes Chapter 18 Pharmacology Short Questions**W = Q V**

We know that,**V = IR and Q = It**

Putting these values, electrical energy will be given by

W = IR × It = I^{2 }Rt**Unit**

SI unit of electrical energy is joule and is defined as,

“One joule is the amount of energy liberated when one coulomb of charge will move from one point to another due to the potential difference of one volt.”

Some other units of electrical energy are kilojoule.**Electric Power:**

“Rate of doing work in an electrical circuit is known as electric power.”**Mathematical Expression**

Mathematically power is given as,

**Q.10) State the functions of live, neutral and earth wires in domestic main supply.**

**Answer:**

The electric power enters our house through three wire. One is called **earth wire **or ground wire (E). The earth wires carries no electricity it is connected to a large metal plate buried deep in the ground near the house.

The other wire is maintained at zero potential by connecting it to the Earth wire at the power station itself and is called **neutral wire (N)**. This wire provide the return path for the current.

The third wire is at a higher potential and is called **live wire (L)**. The potential difference between the live wire and the neutral wire is 220 V.

All electrical appliances are connected across the neutral and the live wires. The same potential difference is therefore applied to all of them and hence these are connected in parallel to the power source.

**Q.11) Explain why the domestic appliances are connected in parallel?**

**Answer:**

Every circuit is connected in parallel with the supply, i.e. across the live and neutral, and receives the p.d. of 220 V from supply mains. The advantages of having appliances connected in parallel rather than in series, can be seen by studying the lightening circuit in figure shown below.

i) The p.d across both lamps is fixed (same as the supply potential difference) so the lamp shines with the brightness irrespective of how many other lamps are switched on.

ii) Each lamp can be turned on and off independently; if one lamp fails, the others can still operated.

**Q.13) What are the hazards of electricity? What safety measures are taken in household electricity to safeguard for these hazards?**

**Answer:****Hazards of electricity:****1. Electric Shock**

Electric Shock occurs if current flows from an electric circuit through a person’s body to earth. This can happen if there is a damaged insulation or faulty wiring. The typical resistance of dry skin is about 10,000 Ω, so if a person touches a wire carrying electricity at 240 V, an estimate of the current flowing through them to earth would be I = V/R = 240/10000 = 0.024 A = 24 m A.

For wet skin, the resistance is lowered to about 1000 Ω (since water is good conductor of electricity) so the current would increase to around 240 m A.

It is the size of the current (not the voltage) and the length of time for which it acts which determines the strength of an electric shock.**Precautions:**i. Switch off the electrical supply to an appliance before starting repairs.

ii. Use plugs that have an earth pin and a cord grip; an insulating casing (a rubber or plastic case) is preferred.

iii. Do not allow appliances or cables to come into contact with water.

iv. Do not have long cables trailing across a room, because the insulation can become damaged.

**2. Fire risks**

If the electrical wiring in the walls of a house becomes overheated, a fire may start. Wires become hot when they carry electrical currents, the larger the current carried, the hotter a particular wire will become.

**Precautions:**

To reduce the risk of fire through over heated cables, the maximum current in a circuit should be limited by taking these precautions:i. Use plugs that have a correct fuse.

ii. Do not attach too many appliances to a circuit (for example an extension box).

iii. Don’t over load circuits by using too many adapters.

iv. Thick wires have lower resistance, therefore appliances such as heaters requiring large amount of power must not be operated with thin wires.

v. Damaged insulation or faulty wiring which leads to a large current flowing to earth through flammable material can also start a fire.

**Numerical Questions Physics Class 10 Notes Chapter No 14 current electricity PDF**

**Q.1) A small electric heater has a resistance of 15 ohms when the current in it is 2 amperes. What voltage is required to produce this current?**

**Answer:****Given Data:**

resistance = R = 15 Ω

current = I = 2 A**To Find:**

Voltage = V = ?**Solution:**

According to Ohm’s Law

V= IR

Putting values in above equation we have,

V = 2 A × 15 Ω**V = 30 V**

**Q.2) If a potential difference of 10 V is maintained across a 1 m length of the Nichrome wire having a resistance of 3.1 Ω, what is the current in the wire?**

**Answer:****Given Data:**resistance = R = 3.1 ΩVoltage = V = 10 V

**To Find:**

current = I = ?

**According to Ohm’s Law**

Solution:

Solution:

V= IR

I = V/R

Putting values in above equation we have,

I = 10 V / 3.1 Ω

**I = 3.2 A**

**Q.3) What resistor would have a 15 mA current if connected across the terminals of a 9.0 V battery?**

**Answer:****Given Data:**current = I = 15 mA = 0.015 AVoltage = V = 9.0 V

**To Find:**

resistance = R = ?

**According to Ohm’s Law**

Solution:

Solution:

V= IR

R = V/I

I = 9.0 V / 0.015 A

Putting values in above equation we have,

**R = 600 Ω**

**Q.4) Consider a circuit with three resistors R**_{1} = 250.0 Ω, R_{2} = 150.0 Ω, R_{3} = 350.0 Ω, connected in parallel with a 24.0-V battery. Find the total current supplied by the battery.

_{1}= 250.0 Ω, R

_{2}= 150.0 Ω, R

_{3}= 350.0 Ω, connected in parallel with a 24.0-V battery. Find the total current supplied by the battery.

**Answer:****Given Data:**R

_{1}= 250.0 Ω

R

_{2}= 150.0 Ω

R

_{3}= 350.0 Ω

Voltage = V = 24.0 V

**To Find:**

Total current = I = ?

**For Parallel combination the equivalent resistance is**

Solution:

Solution:

1/R

_{e}= 1/R

_{1}+ 1/R

_{2}+ 1/R

_{3}

1/R

_{eq}= 1/250.0 + 1/150.0 + 1/350.0

1/R

_{eq}= 0.0134

R

_{eq}= 74.6 Ω

According to Ohm’s law we have,

V = IR

_{eq}

I = V/R

_{eq}

I = 24.0 V / 74.6 Ω

**I = 0.32 A**

**Q.5) An electric hairdryer is rated at 1,875 watts when operating on 120 volts. What is the current flowing through it? If the hairdryer is used for 3 minutes, how much energy does it consume?**

**Answer:****Given Data:**Power = P = 1875 wattsPotential = V = 120 V

t = 3 minutes = 180 seconds

**To Find:**

Current = I = ?

Energy consumed = E = ?

Solution:

Solution:

As we know,

P = VI

I = P/V

I = 1875 watts / 120 V

**I = 15.6 A**

Now for energy consumed,

E = P × t = 1875 watts × 180 s

**E = 337500 J**

**Q.6) A battery with an emf of 12 V is connected to a 545 Ω resistor. How much energy is dissipated in the resistor in 65 s?**

**Answer:****Given Data:**resistance = R = 545 Ωemf = V = 12.0 V

t = 65 s

**To Find:**

The energy dissipated on the resistor = E = ?

Solution:

Solution:

As we know,

P = VI = V

^{2}/ R

P = (12 V)

^{2}/ (545 Ω) = 0.26 W

Now the energy dissipated is

E = P × t = 0.26 W × 65 s=

**17 J**

**Q.7) If the unit of electricity cost 8.11 Rs/kWh, what is the cost of running two 160W fans and four 100 W light bulbs for 6 hours in school?**

**Answer:****Given Data:**Cost of one unit of electricity = Rs 8.11

Power of fan = P

_{f}= 160 W

Power of light blubs = P

_{b}= 100 W

Time = t = 6 hours

**To find:**

a. Cost of electricity due to two fans =?

b. Cost of electricity due to four light blubs =?

**Calculation:**

We know that

Cost of electricity = (Watt × Time of use in hours × cost of one unit)/ 1000

Now,

**a.**

Cost of electricity due to two fans = 2 × (160 × 6 × 8.11)/1000

Cost of electricity due to two fans =

**Rs 15.75**

**b.**

Cost of electricity due to four light blubs = 4 × (100 × 6 × 8.11)/1000

Cost of electricity due to four light blubs =

**Rs 19.46**

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