Q1. What is charge?
Ans. Charge is a fundamental property of matter that governs all electric and magnetic interactions. It is a key characteristic of atomic particles. Electric charge is responsible for the electric forces that bind atoms together by attracting electrons (negatively charged) to the nucleus (positively charged). 
There are two types of charge:
Positive charge (carried by protons)
Negative charge (carried by electrons)
The nucleus of an atom consists of protons (positively charged) and neutrons (uncharged), while electrons orbit around it.
Key behavior of charges:
Like charges repel each other (e.g., proton-proton or electron-electron).
Opposite charges attract each other (e.g., proton-electron).
The SI unit of charge is the coulomb (C).
The charge of a single electron is approximately −1.6 × 10⁻¹⁹ C.
Q2. What are conductors and insulators?
Ans. ∙ Conductors and Insulators
Materials can be categorized based on their ability to allow the flow of electric charges:
Conductors – Substances that allow electric charges to flow easily. 
Examples: Metals like copper, aluminum, silver, and gold.
Reason: Conductors have free electrons that are loosely bound to their atoms, enabling easy movement of charge.
Insulators – Substances that resist or block the flow of electric charges.
Examples: Glass, rubber, mica, plastic, and dry wood.
Reason: Insulators lack free electrons because their electrons are tightly bound to their atomic nuclei, preventing charge flow.
Key Difference
The presence or absence of mobile electrons determines whether a material is a conductor or insulator:
Conductors → Electrons are loosely held → Charge flows freely.
Insulators → Electrons are tightly bound → Charge cannot flow.
Q3. What is electric potential and potential difference?
Ans. Charge Flow in Conductors
Electric charges (electrons) in a conductor do not move on their own. For charges to flow, there must be a potential difference (also called voltage) across the conductor. This difference in electric pressure is typically provided by a battery or an electric cell.
Role of a Battery (Electric Cell)
- A battery generates a potential difference between its terminals through chemical reactions inside the cell.
- When connected to a circuit, this potential difference drives electrons through the conductor, creating an electric current.
- To sustain the current, the battery continuously expends its stored chemical energy.
- Definition of Potential Difference
The potential difference (V) between two points is defined as:
V=W/Q
where: V = Potential difference (in volts, V)
W = Work done (in joules, J) to move the charge
Q = Charge moved (in coulombs, C)
The Volt (Unit of Potential Difference)
1 volt (V) is the potential difference when 1 joule (J) of work is done to move 1 coulomb (C) of charge.
1 V=1 J /1 C
Measuring Potential Difference
A voltmeter is the instrument used to measure the potential difference between two points in a circuit.
Q4. What is electric current and What are electrical circuits?
Ans. Electric Current and Circuits: Electric current is the rate of flow of electric charge through a conductor (such as a metal wire).
Mathematically, if a net charge Q flows through a cross-section of a conductor in time t, the current I is: I=Q/t
SI Unit: Ampere (A) 
Ampere (A) = 1 Coulomb (C) of charge flowing per 1 second (s).
Measuring Current
An ammeter measures electric current and is always connected in series with the circuit.
Electric Circuits
A closed and continuous path through which current flows is called an electric circuit.
Example: A simple circuit may include:
A cell (providing potential difference),
A bulb (or resistor),
An ammeter,
A switch (plug key).
Direction of Electric Current
Conventional Current: Defined as the flow of positive charges, from the positive terminal to the negative terminal of the cell.
Note: In reality, electrons (negative charges) flow opposite to conventional current.
Circuit Diagrams
Electric circuits can be represented conveniently through a circuit diagram. A diagram which indicates how different components in a circuit have to be connected by using symbols for different electric components is called a circuit diagram.
Q5. What is Ohm’s Law?
Ans. ∙ Ohm’s law is the relation between the potential difference applied to the ends of the conductor and current flowing through the conductor. This law was expressed by George Simon Ohm in 1826.
Statement of Ohm’s Law
If the physical state of the conductor (Temperature and mechanical strain etc.) remains unchanged, then current flowing through a conductor is always directly proportional to the potential difference across the two ends of the conductor mathematically
V ∝ I or V=IR
where constant of proportionality R is called the electric resistance or simply resistance of the conductor.
Value of resistance depends upon the nature, dimension and physically dimensions of the conductor.
From Ohm’s Law
V = IR … (3)
R =V/I… (4) 
Thus electric resistance is the ratio of potential difference across the two ends of conductor and amount of current flowing through the conductor.
If a graph is drawn between the potential difference readings (V) and the corresponding current value (I), then the graph is found to be a straight line passing through the origin as shown below in the figure 
- From graph we see that these two quantities V and I are directly proportional to one another.
- Also from this graph we see that current (I) increases with the potential difference (V) but their ratio V/I remain constant and this constant quantity as we have defined earlier is called the Resistance of the conductor.
- Electric resistance of a conductor is the obstruction offered by the conductor to the flow of the current through it.
- SI unit of resistance is Ohm (Ω) where 1 Ohm=1 volt/1 Ampere or 1Ω=1VA-1. Bigger units of resistance are Kilo-Ohm and Mega-Ohm
Q6. What is Resistance?
Ans.Resistance is the property of a conductor by which it opposes the flow of electric current through it.
R = V/I
The SI unit of Resistance is ohm (Ω).
1 ohm is the resistance when potential difference across conductor is 1 V and the current through it is 1 ampere. The electrical appliance which is used to oppose the current flow in a circuit is called resistor.
Variable resistance (Rheostat) is the component used to regulate current (electricity) without changing the voltage source.
Q7. What are the factors on which Resistance Depends?
Ans. 1. Length of the Wire 📏 → More Length = More Resistance
Why? A longer wire is like a longer hallway – electrons have to travel farther, facing more “bumps” along the way.
Formula: R ∝ Length (l)
Example: Extension cords get warmer when very long because resistance increases.
2. Thickness of the Wire 🧵 → Thicker Wire = Less Resistance
Why? A thicker wire is like a wider highway – more space for electrons to flow easily.
Formula: R ∝ 1/Area (1/A)
Example: Power lines are thick to reduce resistance and prevent energy loss.
3. Material of the Wire 🔩 → Some Metals Resist More Than Others
Why? Copper has low resistivity (ρ) (lets electrons flow easily). Rubber has very high resistivity (blocks flow).
Formula: R ∝ Resistivity (ρ)
Example: Wires are made of copper, not iron, because copper has lower resistivity.
Combined Formula:
R=ρ× l/A (Resistance = Resistivity × Length / Area)
4. Temperature 🔥 → Hotter Wire = More Resistance
Why? Heating a wire makes atoms vibrate more, blocking electron flow.
Effect: Increase temperature → Increase resistance
Decrease temperature → Decrease resistance
Exception: Some materials (like semiconductors) behave differently.
Real-Life Examples:
✔ Long wires (e.g., extension cords) heat up because resistance increases with length.
✔ Thin wires (e.g., headphone cables) have higher resistance than thick power cables.
✔ Copper vs. Iron: Copper wires are preferred for circuits due to lower resistivity.
Quick Summary Table:
Factor Change Effect on Resistance Example
Length (↑) Longer wire Increases ☝️ Long extension cords
Thickness (↑) Thicker wire Decreases 👇 Power transmission lines
Material Higher ρ (e.g., iron) Increases ☝️ Copper vs. iron wires
Temperature (↑) Hotter wire Increases ☝️ Toaster wires get hot
Q8. What is resistivity or specific resistance?
Ans. It is the property of material. Since, the SI unit of R is Ω, SI unit of Area is m2 and SI unit of length is m,
Unit of resistivity (p) = Ω x m2 / m = Ωm [ p = RA / l ]
Materials having resistivity in the range of 10 – 8 Ω m to 10 – 6 Ω m are considered as very good conductors. Silver has resistivity equal to 1.60 X 10– 8 Ω m and copper has resistivity equal to 1.62 X 10– 8 Ω m.
Rubber and glass are very bad conductors or very good insulators. They have resistivity in the order of 10 12 Ω
m to 10 17 Ω m.
Resistance, as well as resistivity both, varies with change in temperature.
Q9. Explain the combination of Resistors
Ans. Resistors or conductors can be connected into three ways:
- Series connection or combination
- Parallel connection or combination
- Series-parallel combination
The total resistance of series or parallel combination of resistors is known as equivalent resistance or effective resistance.
Resistors or conductors connected in series:
Two or more resistors or conductors are said to be connected in series if they are connected one after the other such that the same current flows through each conductor or resistor when some potential difference is applied across the combination of resistors.
Assume current I be flowing through each resistance and V be the potential difference across the combination of the resistances.
If V1, V2 and V3 be the potential difference across resistances R1, R2 and R3 respectively, then potential difference across
PQ = Potential difference across R1 + Potential difference across R2 + Potential difference across R3.
| So,
We know that according to Ohm›s law, |
V | = | V1 + V2 + V3 | … (1) |
|---|---|---|---|---|
| V1 V2 V3 | =
= = |
IR1 IR2 IR3 |
By using these values of V1, V2 and V3 in eqn. (1), we get
V = IR1 + IR2 + IR3
V = I (R1 + R2 + R3) … (2)
If Rs = effective resistance of series combination of R1, R2 and R3, Then
Hence, eqn. (2) becomes
V = I x Rs
IRs = I (R1 + R2 + R3)
Rs = R1+ R2 + R3 (3)
For n resistors equivalent resistance of the series combination is given by
Rs = R1 + R2 + R3 +. + Rn
Advantage of series combination of resistors
- The net resistance of the combination is equal to the sum of the resistances of the individual resistors.
- The current flowing through each resistor is the same.
- The voltage applied across the series combination of resistor is equal to the sum of potential differences across
the individual resistors.
Disadvantage of series arrangement of resistors
- Series arrangement of electric appliances is not used in domestic electric circuits. Suppose all electric appliances like bulbs and electric tubes are connected in series in a circuit. If any one of them fuses (breaks), then all the other appliances will also not work. This is because in series arrangement, there is only a single path for the flow of current through all appliances.
- Different electric appliances like bulbs, electric fan, heaters, toaster etc. Have different resistances and hence they require different amount of electric current for their operations. If they all are connected in series, they will not be working properly.
Resistors or conductors connected in parallel
Two or more resistors are said to be connected in parallel if one end of each resistor is connected at one common point and the other end of each resistor is connected at other common point such that the potential difference across each resistor is equal to the applied potential difference across the combination of the resistors.
Assume V be the applied potential difference across P and Q. The current I drawn from the cell divides into three parts
I1, I2 and I3 at point P. Current I1 passing through R1, current I2 through R2 and current I3 through R3.
I = I1 + 12 + 13 … (1)
According to Ohm›s law
Equation (1) becomes
I1 = V/R1, I2 = V/R2, I3 = V/R3 I = V/R1 + V/R2 + V/R3
I = V (1/R1 + 1/R2 + 1/R3) (2)
Rp = equivalent resistance of the parallel combination.
I = V/Rp
Equation (2) becomes For n resistors
Features of Parallel Combination of Resistors
V/Rp = V (1/R1 + 1/R2 + 1/R3) 1/Rp = 1/R1 + 1/R2 + 1/R3
1/Rp = 1/R1 + 1/R2 + 1/R3 + + Rn
The sum of the reciprocal of the resistances of the individual resistors or conductors is equal to the reciprocal of the net or equivalent resistance of the combination.
-
- Different amount of current flows through different resistors.
- The potential difference across each conductor or resistor is equal to the voltage
Applied across the parallel combination of the conductors or resistors.
Advantage of parallel Combination
If any one of the electric appliances or devices connected in parallel does not work (or fuses), then the working of other devices will not be affected. When different electric devices are connected in parallel, the potential difference across each device is equal to the applied potential difference and hence they draw the current as per their requirement and hence they work properly.
Q10. Explain heating effect of electric current. How is it caused?
Ans. The production of heat in a conductor due to the flow of electric current through it is called heating effect of electric current.
When electric current passes through a high resistance wire, the wire becomes and produces heat. This is called heating effect of current.
This phenomenon occurs because electrical energy is gets transformed into heat energy when current flows through a wire of some resistance say R Ω.
Role of resistance in electrical circuits is similar to the role of friction in mechanics.
Potential difference = work / charge
V = W/q or W = V x q
If I current flowing through the wire or conductor for time t then charge q is given by
Q = I/t
So, W = V x It
This work done is equal to the energy supplied to the conductor.
H = VIt … (1)
We know that V = IR
| Eqn 1 can also be written as | ||
|---|---|---|
| H = I²Rt | … (2) | |
| But I = V/R | ||
| So, | H = V²/R x t | … (3) |
| Equation 2 and 3 known as Joule’s law of heating. | ||
| Cause of heating effect | ||
We know that conductor consists of large number of free electrons. When the conductor is connected to the electric energy like a dry cell or a battery then free electrons start to move from negative terminal to positive terminal but the motion of the electrons in the conductor is not that simple. Electrons collide with the atoms or ions of the conductor. Due to these collisions the kinetic energy of the free electrons is transferred to the atoms or ions of the conductor. That’s why atoms or ions start vibrating with large amplitudes about their main positions. These vibrations increase the total energy of the ions or atoms. Due to this reason temperature of the conductor also increases. That’s why heat is produced when electric run flows through the metallic conductor or wire.
Q11. Explain Joule’s heating effect of current.
Ans. When current flows, the electrons move and when they move, they collide with each other. When they collide, heat is produced. Heat produced depends upon the square of current, resistance and temperature.
According to Joule’s Law of Heating , Heat produced in a resistor is
- Directly proportional to the square of current for a given resistor.
- Directly proportional to resistance of a given resistor.
- Directly proportional to time for which current flows through the resistor.
i.e.,: H = I2RT
Also called as joules heating effect.
Q12. List some practical Applications of the Heating Effect of Electric Current.
Ans. 1. Heating Appliances: Devices that convert electrical energy into heat using high-resistance wires:
Electric iron – Heating coil inside gets hot to press clothes.
Toaster – Nichrome wires glow red-hot to toast bread.
Electric kettle – Heats water quickly using a high-current coil.
Room heater – Nichrome coils heat up and warm the air.
Why? These devices have high-resistance wires that heat up when current passes through them.
2. Electric Bulb (Light Production) 💡
The filament (usually tungsten) heats up to ~2500°C and glows white-hot.
Why tungsten? Very high melting point (won’t melt easily).
Does not oxidize quickly at high temperatures.
Downside: Only ~5% of energy becomes light; 95% is wasted as heat (that’s why LEDs are better).
3. Electric Fuse (Safety Device) ⚡
A thin wire (e.g., copper, aluminum) placed in series with a circuit.
How it works: If too much current flows (short circuit/overload), the fuse wire heats up (Joule’s heating). When the temperature exceeds its melting point, the wire breaks, stopping current flow.
Purpose: Protects expensive appliances from damage.
Example: A 5A fuse melts if current exceeds 5A, cutting off power.
Q13. What is electric power?
Ans. Electric power is defined as the amount of electric energy consumed in a circuit per unit time. If W be the amount of electric energy consumed in a circuit in t seconds, then electric power is
| P | = | W/t | |||
|---|---|---|---|---|---|
| But we know that W = vit | P | = | VIt/t or P = VI | … (1) | |
| According to Ohm’s law | V | = | IR | ||
| From eqn (1) | P | = | I²R | …(2) | |
| Also, I = V/R | |||||
| From eqn (2) | P | = | V²/R | ||
| SI unit of power is Watt (W) | |||||
Q14. Define one Watt?
Ans. Electric power is said to be one what if one ampere current flows through a circuit having one-volt potential difference.
Units of Power
1 kilowatt (kw) = 10³ W
1 Megawatt (MW) = 10⁶ W 1 gigawatt (GW) = 10⁹ W horsepower =746 watts Power in terms of I and R
We know that P = VI
Now from Ohm’s law V = IR
Putting above equation in equation 15 we get
Power in terms of V and R
P = I×R×I
Power, P = I2×R
We know that P=VI Now from Ohm›s law V = IR Or we have I = V 
Putting this value of I in equation 15 we get
P = V×VR P = V2R
Q15. What is commercial unit of electrical energy?
Ans. The commercial unit of electrical energy is a kilowatt-hour (kWh).
1kWh = 3,600,000J = 3.6×106J
One kilowatt-hour is defined as the amount of energy consumed when 1kW of power is used for 1 hour.
Q16. What is electric fuse? How does it work?
Ans. It is a safety device that protects us from fire. It is shown in figure. It has two terminals T1 and T2 to which fuse wire is attached. The property of fuse wire is that it has very very low melting point. It is made up of an alloy consisting of lead and tin.
It works under two conditions:
Overloading: When many appliances are running together at the same time, they draw a large amount of current which produces a lot of heat that can even lead to fire.
Short circuiting: When live wire and neutral wire touch each other, the current exceeds the normal value and it can also cause fire.
In both the conditions, when current exceeds the normal value, excess heat is produced which melts the fuse wire and hence, fire is caused. 
Text Book Questions
Q1. What does an electric circuit mean?
Ans. A continuous closed path made of electric components through which an electric current flows is known as an electric circuit. A simple circuit consists of the following components:
-
- Conductors
- Cell
- Switch
- Load
Q2. Define the unit of current.
Ans. The unit of current is ampere. Ampere is defined by the flow of one coulomb of charge per second.
Q3. Name a device that helps to maintain a potential difference across a conductor.
Ans. Battery consisting of one or more electric cells is one of the devices that help to maintain a potential difference across a conductor.
Q4. What is meant by saying that the potential difference between two points is 1 V?
Ans. When 1 J of work is done to move a charge of 1 C from one point to another, it is said that the potential difference between two points is 1 V.
Q5. How much energy is given to each coulomb of charge passing through a 6 V battery? Ans. We know that the potential difference between two points is given by the equation,
V = W/Q,
where, W is the work done in moving the charge from one point to another Q is the charge
From the above equation, we can find the energy given to each coulomb as follows:
Substituting the values in the equation, we get
Hence, 6 J of energy is given to each coulomb of charge passing through a 6 V of battery.
Q6. On what factors does the resistance of a conductor depend? Ans. The resistance of the conductor depends on the following factors:
- Temperature of the conductor
- Cross-sectional area of the conductor
- Length of the conductor
- Nature of the material of the conductor
Q7. Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?
Ans. Resistance is given by the equation,
R = ρ l/A
where, ρ is the resistivity of the material of the wire, l is the length of the wire A is the area of the cross-section of the wire.
From the equation, it is evident that the area of the cross-section of wire is inversely proportional to the resistance. Therefore, thinner the wire, more the resistance and vice versa. Hence, current flows more easily through a thick wire than a thin wire.
Q11. An electric lamp of 100 Ω, a toaster of resistance 50 Ω, and a water filter of resistance 500 Ω are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it?
Ans. The electric lamp, the toaster and the water filter connected in parallel to a 220 V source can be shown as using a circuit diagram as follows:
The equivalent resistance of the resistors can be calculated as using Ohm’s law, the current flowing across the circuit can be calculated as follows:
I = 
I = 
= 7.04A
As the appliances are connected in parallel, the current across all of them is 7.04A
Hence, the current drawn by electric iron connected in parallel to the same source is 7.04A
We find the resistance of iron box using Ohm’s law as follows:
R =31.25 Ω
The resistance of the electric iron box is 31.25 Ω.
Q12. What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?
Ans. When the electrical devices are connected in parallel there is no division of voltage among the appliances. The potential difference across the devices is equal to supply voltage. Parallel connection of devices also reduces the effective resistance of the circuit.
Q13. How can three resistors of resistances 2 Ω, 3 Ω, and 6 Ω be connected to give a total resistance of (a) 4 Ω, (b) 1 Ω?
Ans. (a) The circuit diagram below shows the connection of three resistors
From the circuit above, it is understood that 3 Ω and 6 Ω are connected in parallel. Hence, their equivalent resistance is given by
R = 2 Ω
The equivalent resistor 2 Ω is in series with the 2 Ω resistor. Now the equivalent resistance can be calculated as follows:
Req = 2 Ω +2 Ω = 4 Ω
Hence, the total resistance of the circuit is 4 Ω.
(b) The circuit diagram below, shows the connection of three resistors.
From the circuit, it is understood that all the resistors are connected in parallel. Therefore, their equivalent resistance can be calculated as follows:
R = 1Ω
The total resistance of the circuit is 1 Ω.
Q14. What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistance 4 Ω, 8 Ω, 12 Ω, 24 Ω?
Ans. (a) If the four resistors are connected in series, their total resistance will be the sum of their individual resistances and it will be the highest. The total equivalent resistance of the resistors connected in series will be 4 Ω + 8 Ω
+ 12 Ω + 24 Ω = 48 Ω.
(b) If the resistors are connected in parallel, then their equivalent resistances will be the lowest. Their equivalent resistance connected in parallel is 
R = = 2 Ω
Hence, the lowest total resistance is 2 Ω.
Q15. Why does the cord of an electric heater not glow while the heating element does?
Ans. The heating element of an electric heater is made of an alloy which has a high resistance. When the current flows through the heating element, the heating element becomes too hot and glows red. The cord is usually made of copper or aluminum which has low resistance. Hence the cord doesn’t glow.
Q16. Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.
Ans. The heat generated can be computed by Joule’s law as follows:
H = VIt
where,
V is the voltage, V = 50 V I is the current
t is the time in seconds, 1 hour = 3600 seconds
The amount of current can be calculated as follows:
Substituting the value we get
I = 26.66 A
H = 50 x 26.66 x 3600 = 4.8 x 106 J
Q17. An electric iron of resistance 20 Ω takes a current of 5 A. Calculate the heat developed in 30 s.
Ans. The amount of heat generated can be calculated using the Joule’s law of heating, which is given by the equation
H = VIt
Substituting the values in the above equation, we get,
H = 100 × 5 × 30 = 1.5 × 104 J
The amount of heat developed by the electric iron in 30 s is 1.5 × 104 J.
Q18. What determines the rate at which energy is delivered by a current?
Ans. Electric power is the rate of consumption of electrical energy by electric appliances. Hence, the rate at which energy is delivered by a current is the power of the appliance.
Q19. An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.
Ans. The power of the motor can be calculated by the equation,
P = VI
Substituting the values in the above equation, we get
P = 220 V × 5 A = 1100 W
The energy consumed by the motor can be calculated using the equation,
E = P × T
Substituting the values in the above equation, we get
P = 1100 W × 7200 = 7.92 × 106 J
The power of the motor is 1100 W and the energy consumed by the motor in 2 hours is 7.92 × 106 J.
Q20. How is a voltmeter connected in the circuit to measure the potential difference between two points?
Ans. To measure the voltage between any two points, the voltmeter should be connected in parallel between the two points.
Q21. When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor
Ans. The value of the resistor can be calculated using Ohm’s Law as follows: 
R = V/I
Substituting the values in the equation
R = = 4.8 x 103 kΩ
Q22. A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω, respectively. How much current would flow through the 12 Ω resistor?
Ans. In series connection, there is no division of current. The current flowing across all the resistors is the same.
To calculate the amount of current flowing across the resistors, we use Ohm’s law. But first, let us find out the equivalent resistance as follows:
R = 0.2 Ω + 0.3 Ω + 0.4 Ω + 0.5 Ω + 12 Ω = 13.4 Ω
Now, using Ohm’s law,
I = V/R = = 0.671 A
The current flowing across the 12 Ω is 0.671 A.
Q23. How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line?
Ans. Let a resistor of 176 Ω are joined in parallel. Then their combined resistance (R) 1/R = 1/176 + 1/176 …… times = n/176 or R = 176/n Ω
It is given that V= 220 V and I = 5 A
R = V/I or 176/n = 220/5 = 44 Ω
n = 176/44 = 4, 4 resistors should be joined in parallel.
Q24. Show how you would connect three resistors, each of resistance 6 Ω so that the combination has resistance of (i) 9 Ω (ii) 4 Ω.
Ans. It is given here that R1 = R2 = 6 Ω.
- To get net resistance of 9 Ω we should join three resistors as below:
- To get 4 Ω net resistance we should join three resistors as below:
Q25. Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A?
Ans. Each bulb is rated as 10 W, 220 V, It draws a current (I) = P/V = 10 W/220 V
= 1/22 A.
As the maximum allowable current is 5 A and all lamps are connected in parallel, hence maximum number of bulbs joined in parallel with each other = 5 x 22 = 110.
Q26. A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B. Each of 24 Ω resistances, which may be used separately, in series or in parallel. What are the currents in the three cases?
Ans. It is given that potential difference (V) = 220 V.
Resistance of coil (A) = Resistance of coil (B) = 24 Ω
- When either coil is used separately, the circuit (I) = V/R = 220 V/ 24 Ω
= 9.2 A.
- When two coils are used in series total resistance (R)
= R1 + R2 = 24 +24 = 48 Ω
Current flowing (I) = V/ R = 220 V/ 48 Ω = 4.6 A.
- When two coils are joined in parallel. Total resistance (R) = 1/24 + 1/24
= 2/24, R = 12 Ω.
Current (I) = V/R = 220 V / 12 Ω = 18.3 A.
Q27. Compare the power used in the 2 Ω resistor in each of the following circuits:
- a 6-volt battery in series with 1 Ω and 2 Ω resistors and,
- a 4 V battery in parallel with 12 Ω and Ω resistors.
Ans. (i) When a 2 Ω resistor is joined t a 6 V battery in series with 1 Ω and 2 Ω resistors. Total resistance (R) = 2 + 1 + 2 = 5 Ω.
Current (I) = 6 V/5 Ω = 1.2 A
Power used in 2 A resistor = I2R = 2.88 W
When 2 Ω resistor is joined to a 4 V battery in parallel with 12 Ω resistor and 2 Ω resistors, the current flowing in 2 Ω = 4 V/ 2 Ω = 2 A/.
Power used in 2 Ω resistor = I2R = 8 W Ratio = 2.88/8 = 0.36: 1.
Q28. Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is 220 V?
Ans. Current drawn by 1st lamp rated 100 W at 220 V = P/V = 100/ 220 = 5/11 A. Current drawn by 2nd lamp rated 60 W at 220 V = 60/220 = 3/11 A.
In parallel arrangement the total current = I1 +I2 = 3/11+ 5/11 = 8/11 = 0.73 A.
Q29. Which uses more energy, a 250 W TV set in 1 hour, or a 1200 W toaster in 10 minutes? Ans. Energy used by a TV set of power 250 W in 1 hour = P x t = 250 Wh.
Energy used by toaster of power 1200 W in 10 minute (10/60 h)
= P x t = 1200 W x 10/60 h = 200 Wh.
Q30. An electric heater of resistance 8 draws 15 A from the service mains for 2 hours. Calculate the rate at which heat is developed in the heater.
Ans. Resistance of electric heater (R) = 8 Ω, current (I) = 15 A.
Rate at which heat developed in the heater = I2Rt/t = 15 x 15 x 8 = 1800 W.
Q31. Explain the following.
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- Why is the tungsten used almost exclusively for filament of electric lamps?
- Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal?
- Why is the series arrangement not used for domestic circuits?
- How does the resistance of a wire vary with its area of cross-section?
- Why copper and aluminium wires are usually employed for electricity transmission?
Ans. a. The resistivity and melting point of tungsten is very high. Due to this property, it doesn’t burn readily when heated. Electric lamps operate at high temperature. Hence, tungsten is a choice of metal for the filament of electric lamps.
- The conductors of electric heating devices are alloys because of their high resistivity. Due to its high resistivity it produces large amount of heat.
- The voltage is divided in series circuit as result each component in the circuit receives a small voltage because of which the amount of current decreases and the device gets hot and does not work properly. This is the reason why series circuits are not used in domestic circuits.
- Resistance is inversely proportional to the area of cross section. When the area of cross section increases the resistance decreases and vice versa.
- Copper and aluminium are good conductors of electricity and have low resistivity because of which they are usually employed for electricity transmission.