Electricity-QnA

Q1: What is electric current?

The rate of flow of electric charge through a conductor.

Q2: Define potential difference.

Work done to move a unit charge between two points in a circuit.

Q3: SI unit of current?

Ampere (A).

Q4: What is the unit of resistance?

Ohm (O).

Q5: State Ohm’s law.

Current is directly proportional to voltage at constant temperature.

Q6: What is a conductor?

A material that allows electric current to pass through easily.

Q7: What is an insulator?

A material that does not allow current to flow.

Q8: Define resistivity.

A property of material showing how strongly it resists current.

Q9: What is electric power?

The rate at which electrical energy is consumed.

Q10: Formula of power?

P = VI.

Q11: What is a fuse?

A safety device that melts when current exceeds limit.

Q12: What is a series circuit?

A circuit where components are connected end to end.

Q13: What is a parallel circuit?

A circuit where components are connected across common points.

Q14: State the SI unit of power.

Watt (W).

Q15: Define electric energy.

Energy consumed in doing electrical work.

Q1: Why does the filament of a bulb glow?

Because electric current heats the filament to high temperature causing it to emit light.

Q2: Why is tungsten used as bulb filament?

Tungsten has high melting point and can withstand high temperature.

Q3: Why are copper wires used in household wiring?

Copper has low resistance and high conductivity, making it efficient for current flow.

Q4: State two differences between series and parallel circuits.

Series: one path, same current. Parallel: multiple paths, same voltage.

Q5: What happens to resistance if length of wire is doubled?

Resistance becomes twice because R ? L.

Q6: What happens to resistance if area of cross-section is doubled?

Resistance becomes half because R ? 1/A.

Q7: Explain heating effect of current.

When current flows through a resistor, electrical energy converts into heat.

Q8: What is Joule’s law of heating?

Heat produced is proportional to I², R, and time t.

Q9: Why do electric wires heat up?

Due to resistance, part of electrical energy converts into heat.

Q10: Why are fuses important?

They protect circuits by breaking the flow during excessive current.

Q1: Explain Ohm’s law with a graph.

Ohm’s law states that current through a conductor is directly proportional to the voltage across it if temperature remains constant. The I–V graph is a straight line passing through the origin, showing linear relation.

Q2: Describe factors affecting resistance of a conductor.

Resistance depends on length (directly proportional), area (inversely proportional), nature of material, and temperature (resistance increases with rise in temperature for metals).

Q3: Explain series combination of resistors.

In series, resistors are connected end to end. Current remains the same, total resistance is sum of individual resistances, and voltage divides across resistors.

Q4: Explain parallel combination of resistors.

In parallel, resistors are connected across common points. Voltage remains same, current divides, equivalent resistance is less than smallest resistance.

Q5: Explain electric power and its formulas.

Power is the rate of electrical energy consumption. Formulas: P = VI, P = I²R, P = V²/R. Used to calculate power rating of appliances.

Q1: Describe in detail the heating effect of electric current with daily life examples.

When electric current flows through a resistor, heat is generated due to conversion of electrical energy. Examples: electric iron, toaster, heater, bulb filament. The heat produced depends on current, resistance, and time, following Joule’s law.

Q2: Discuss the practical applications of series and parallel circuits.

Series circuits are used in devices where same current is needed, like decorative lights. Parallel circuits are used in homes so each appliance gets full voltage and one failure doesn’t stop others.

Q3: Explain resistivity and its significance for different materials.

Resistivity is a material property indicating how strongly it opposes current. Metals have low resistivity, making them good conductors, while rubber and plastic have high resistivity, making them insulators. This helps in selecting materials for wiring and insulation.

Q4: Describe the working of an electric fuse and its importance in safety.

A fuse contains a thin metal wire that melts when excessive current flows. This breaks the circuit preventing damage to appliances and minimizing fire risk. It is essential for electrical safety.

Q5: Explain electric energy consumption in households with examples.

Electric energy consumed is calculated using E = P × t. Appliances like fans, ACs, and heaters use different amounts of power. Bills are based on total energy used in kilowatt-hour units.

Q1: What does an electric circuit mean?

Answer:

An electric circuit is a complete and continuous conducting path through which electric current can flow. Typical elements of a circuit include a power source (for example, a cell or battery), connecting wires, and electrical components such as bulbs, switches, or resistors.

When the circuit is closed, current flows from the positive terminal of the source, passes through the components, and returns to the negative terminal. If there is any break or gap, the circuit becomes open and the current stops flowing.

In simple terms, an electric circuit provides a pathway that allows electricity to travel and do useful work, such as lighting a bulb or running a fan.

Q2: Define the unit of current.

Answer:

The SI unit of electric current is the ampere, abbreviated as A. One ampere is defined as the rate of flow of electric charge in which one coulomb of charge passes through a given point in a conductor in one second.

This relation is expressed mathematically as:
\( I = \dfrac{Q}{t} \)
where \(I\) is the current in amperes, \(Q\) is the electric charge in coulombs, and \(t\) is the time in seconds.

Using the definition above, the unit can also be written as:
\(\;1\ \text{A} = 1\ \dfrac{\text{C}}{\text{s}}\).

Practical note: Current is measured using an ammeter connected in series with the circuit element through which the current is to be measured.

Q3: Calculate the number of electrons constituting one coulomb of charge.

Answer:

The quantity of charge carried by a single electron (the elementary charge) is \(e = 1.602176634 \times 10^{-19}\ \text{C}\). If n electrons together carry a total charge \(Q\), then \[ Q = n\,e. \] To find the number of electrons in one coulomb, rearrange: \[ n = \frac{Q}{e}. \]

Substitute \(Q = 1\ \text{C}\) and \(e = 1.602176634\times 10^{-19}\ \text{C}\):

\[\begin{aligned} n &= \frac{1\ \text{C}}{1.602176634\times 10^{-19}\ \text{C}}\\\\ &= \left(\frac{1}{1.602176634}\right)\times 10^{19}. \end{aligned}\]

Evaluating the numeric factor,
\[\begin{aligned}&\dfrac{1}{1.602176634}\\\\&\approx 0.6241509074460763\end{aligned}\] Therefore
\[\begin{aligned} n &\approx 0.6241509074460763\times 10^{19}\\ &= 6.241509074460763\times 10^{18}. \end{aligned}\]

Q4: Name a device that helps to maintain a potential difference across a conductor.

Answer:

A cell (or a battery, which is a combination of cells) is a device that maintains a potential difference across a conductor. Such a source supplies the necessary electromotive force (EMF) so that charges can be driven through a circuit.

In simple terms, the cell converts internal energy (for example chemical energy) into electrical energy, keeping one terminal at a higher potential and the other at a lower potential. This maintained difference in potential causes charge to flow when a closed conducting path is provided.

Notation (exam-style): The EMF of a source is usually denoted by \( \mathcal{E} \) and the potential difference across a conductor by \(V\). A cell provides \( \mathcal{E} \) so that \(V\) can be established across circuit elements when current flows.

Q5: What is meant by saying that the potential difference between two points is 1 V?

Answer:

Saying that the potential difference (p.d.) between two points is 1 volt (1 V) means that one joule of work must be done to move a unit positive charge of one coulomb from one point to the other against the electric field.

In formula form:
\[ V = \frac{W}{Q} \] where \(V\) is the potential difference in volts, \(W\) is the work done in joules, and \(Q\) is the charge in coulombs. Thus, \[ 1\ \text{V} = 1\ \frac{\text{J}}{\text{C}}. \]

Interpretation (exam-style): If the p.d. between points A and B is 1 V, moving a charge of \(1\ \text{C}\) from A to B requires \(1\ \text{J}\) of energy. Equivalently, moving \(2\ \text{C}\) would require \(2\ \text{J}\), and so on, since work scales with charge for a fixed p.d.

Q6: How much energy is given to each coulomb of charge passing through a 6 V battery?

Answer:

The energy given to each coulomb of charge by a source is equal to the potential difference (voltage) across the source expressed in joules per coulomb. Mathematically, \[ V = \frac{W}{Q}, \] where \(V\) is the potential difference in volts, \(W\) is the work (energy) in joules, and \(Q\) is the charge in coulombs.

For a 6 V battery, substituting \(V=6\ \text{V}\) and \(Q=1\ \text{C}\), \[ W = VQ = 6\times 1 = 6\ \text{J}. \]

Therefore: Each coulomb of charge is given 6 joules of energy by the 6 V battery.

Q7: On what factors does the resistance of a conductor depend?

Answer:

The resistance of a conductor is the opposition offered to the flow of electric current. Its value depends on several physical factors that determine how easily electrons can move through the material. The key factors are:

• Length of the conductor (L): Resistance increases with length. A longer wire provides more obstruction to the movement of electrons.
  Mathematically, \( R \propto L \).
• Area of cross-section (A): Resistance decreases as the cross-sectional area increases. A thicker wire allows more electrons to pass simultaneously.
  Thus, \( R \propto \frac{1}{A} \).
• Nature (material) of the conductor: Different materials offer different degrees of opposition to current. This property is expressed by their resistivity (\( \rho \)). Metals generally have low resistivity, whereas insulators have very high resistivity.
• Temperature: For most metallic conductors, resistance increases with rise in temperature due to increased vibration of atoms. In some materials like semiconductors, resistance decreases with temperature.

Combining these factors, the resistance of a uniform conductor is given by:
\[ R = \rho \, \frac{L}{A}, \] where \( \rho \) is the resistivity of the material, \(L\) is the length, and \(A\) is the area of cross-section.

Q8: 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:

Current will flow more easily through a thick wire. Both wires being of the same material and length means their resistivity \( \rho \) and length \( L \) are identical, but the thick wire has a larger cross-sectional area \( A \). Since resistance of a uniform conductor is

\[ R = \rho\,\dfrac{L}{A}, \]

a larger area \(A\) gives a smaller resistance \(R\). When both wires are connected to the same source (same potential difference \(V\)), the current through each wire is given by Ohm’s law:

\[ I = \dfrac{V}{R}. \]

Because the thick wire has lower resistance, its current \(I\) will be larger for the same \(V\). In simple words: a thicker conductor offers less obstruction to the flow of electrons, so more charge passes per second.

Q9: Let the resistance of an electrical component remains 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:

Since the resistance of the component does not change, the current in the circuit will respond only to the change in potential difference. According to Ohm’s law:

\[ I = \dfrac{V}{R} \]

If the potential difference is reduced to half, say from \(V\) to \(\dfrac{V}{2}\), and the resistance remains the same, the new current becomes:

\[\begin{aligned} I' &= \dfrac{\frac{V}{2}}{R}\\\\ I' &= \dfrac{1}{2}\left( \dfrac{V}{R} \right) \\\\&= \dfrac{I}{2} \end{aligned}\]

Thus, the current becomes half of its original value. A smaller potential difference provides less “push” to the charges, causing fewer charges to pass per second through the conductor.

Q10: Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?

Answer: Heating elements in electric appliances like toasters and irons are commonly made using alloys (e.g., nichrome) instead of pure metals. This is because:

• Higher Resistivity: Alloys have much greater electrical resistivity (\(\rho\)) than pure metals. When current (\(I\)) passes through a coil of resistance (\(R\)), the heat generated (\(H\)) given by \[H = I^2Rt\] is more significant in an alloy due to its larger \(R\) value.
• High Melting Point & Stability: Alloys do not melt, deform, or oxidize easily at elevated temperatures, making them ideal for consistent and safe heating.
• Durability: The mechanical strength and resistance to chemical change of alloys ensure the heating coil lasts longer and requires less maintenance.

Hence, coils made of alloys deliver efficient, reliable heating and withstand repeated use without quick deterioration.

Q11: Judge the equivalent resistance when the following are connected in parallel – (a) 1 ? and 106 ?, (b) 1 ? and 103 ?, and 106 ?

Answer:

1. (a) 1 Ω and 106 Ω in parallel:
   When resistances are connected in parallel, equivalent resistance (\(R_{\text{eq}}\)) is given by:
   Formula: \[\frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2}\]
   Substituting values: \[\frac{1}{R_{\text{eq}}} = \frac{1}{1} + \frac{1}{106}\] \[\frac{1}{R_{\text{eq}}} \\\\\approx 1 + 0.0094 = 1.0094\] \[R_{\text{eq}}\\\\ \approx \frac{1}{1.0094} \approx 0.99\;\Omega\] Since the smallest resistance dominates, \(R_{\text{eq}}\) is just less than 1 Ω.
2. (b) 1 Ω, 10³ Ω, and 10⁶ Ω in parallel:
   Formula: \[\frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}\]
   Substituting values: \[\frac{1}{R_{\text{eq}}} = \frac{1}{1} + \frac{1}{10^3} + \frac{1}{10^6}\] \[\begin{aligned}\frac{1}{R_{\text{eq}}} &\approx 1 + 0.001 + 0.000001 \\\\&= 1.001001\\\\\implies R_{\text{eq}} &\approx \frac{1}{1.001001} \\\\&\approx 0.999\;\Omega\end{aligned}\] Again, the smallest resistance controls the equivalent resistance, which is very close to 1 Ω.

Conclusion:
Whenever one resistor in a parallel group has a much smaller value than the others, the equivalent resistance is just less than the smallest resistance. This property helps ensure that current prefers the path of least resistance in parallel circuits.

Q12: An electric lamp of 100 ?, a toaster of resistance 50 ?, and a water filter of resistance 500 ? 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?

Answer:

1. Find current through each appliance:
   \[I = \frac{V}{R}\]
   Lamp: \(I_1 = \frac{220}{100} = 2.2\,A\)
   Toaster: \(I_2 = \frac{220}{50} = 4.4\,A\)
   Water filter: \(I_3 = \frac{220}{500} = 0.44\,A\)
2. Total current drawn by all:
   \[\begin{aligned}I_{total} &= I_1 + I_2 + I_3 \\&= 2.2 + 4.4 + 0.44 \\&= 7.04\,A\end{aligned}\]
3. Resistance of iron for same current:
   Iron is connected to 220 V and must draw \(7.04\,A\).
   \[R_{iron} = \frac{V}{I_{total}} = \frac{220}{7.04} \approx 31.25\,\Omega\]
4. Current through the iron:
   \(I_{iron} = I_{total} = 7.04\,A\)

Summary:
The electric iron must have resistance \(31.25\,\Omega\) and will draw current \(7.04\,A\) when connected to a 220 V source, matching the total current taken by the three parallel appliances.

Q13: What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?

Answer:

• Same Voltage Across Every Device: In parallel, every device receives full battery voltage, allowing each appliance to operate at its rated power regardless of the total number of devices.
• Independent Operation: Devices can be switched on or off individually without affecting the others, making practical usage flexible and convenient.
• Uninterrupted Function: If one device gets disconnected or fails, the rest continue working since each has a direct path to the battery.
• Optimal Power Distribution: Parallel connections allow devices of different ratings to draw required currents (\(I\)) independently through their respective resistances (\(R\)), following the relation \[I = \frac{V}{R}\].

In summary: Parallel connections support household convenience, reliable appliance performance, and safety — key reasons why practical wiring is never done in series for homes.

Q14: How can three resistors of resistances 2 ?, 3 ?, and 6 ? be connected to give a total resistance of (a) 4 ?, (b) 1 ??

Answer:

1. (a) Total resistance = 4 Ω:
   Connect the 6 Ω resistor in parallel with the 3 Ω resistor, and this combination in series with the 2 Ω resistor.
   Stepwise Solution:
   Parallel part:
   \[\begin{aligned}R_{p} &= \frac{R_2 \times R_3}{R_2 + R_3} \\\\&= \frac{3 \times 6}{3 + 6} \\\\&= \frac{18}{9} \\&= 2\,\Omega\end{aligned}\] Series with 2 Ω:
   \[\begin{aligned}R_{total} &= R_1 + R_p \\&= 2 + 2 \\&= 4\,\Omega\end{aligned}\]
2. (b) Total resistance = 1 Ω:
   Connect all three resistors in parallel.
   Stepwise Solution:
   \[\frac{1}{R_{total}} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6}\] Calculate: \[\begin{aligned}\frac{1}{R_{total}} &= \frac{3 + 2 + 1}{6} \\\\&= 1\end{aligned}\] \[R_{total} = 1\,\Omega\]

Conclusion:
To achieve 4 Ω, use (6 Ω || 3 Ω) + 2 Ω. To get 1 Ω, join all three in parallel.

Q15: What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistance 4 ?, 8 ?, 12 ?, 24 ??

Answer:

1. (a) Highest total resistance:
   To get the highest value, connect all coils in series. The resistances simply add:
   \[\begin{aligned}R_{max} &= 4 + 8 + 12 + 24 \\&= 48\,\Omega\end{aligned}\] All resistances combine to give the largest possible value.
2. (b) Lowest total resistance:
   To get the lowest total, connect all coils in parallel:
   \[\frac{1}{R_{min}} = \frac{1}{4} + \frac{1}{8} + \frac{1}{12} + \frac{1}{24}\] Calculate numerator:
   The common denominator is 24. \[\begin{aligned} \frac{1}{R_{min}} &= \frac{6}{24} + \frac{3}{24} + \frac{2}{24} + \frac{1}{24} \\\\&= \frac{12}{24} \\\\&= \frac{1}{2}\end{aligned}\] Therefore, \[R_{min} = 2\,\Omega\] Parallel connection yields the least resistance.

Conclusion:
The greatest resistance possible is \(48\,\Omega\) by series connection. The lowest possible is \(2\,\Omega\) when in parallel.

Q16: Why does the cord of an electric heater not glow while the heating element does?

Answer:

• Low Resistance of Cord: The cord is made of thick copper or aluminum wire with very low resistance (\(R_{cord}\)). According to Joule’s law, heat produced (\(H\)) in a component is \(H = I^2 R t\). Since \(R_{cord}\) is very small, it produces negligible heat.
• High Resistance of Element: The heating element is made of an alloy (often nichrome) with much higher resistance (\(R_{element}\)). This causes substantial heat generation, enough to make the element red-hot and visible as it glows.
• Same Current, Different Effects: Both cord and element carry the same electric current (\(I\)), but the amount of heat—and thus glowing—is controlled by their resistances.

Summary: The heater’s cord does not glow because its resistance is much smaller, so it cannot heat up enough to emit light. The heating element glows due to its large resistance and concentrated heat.

Q17: Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.

Answer:

1. Given Data:
   • Charge transferred, \(Q = 96000\,C\)
   • Time, \(t = 1\,\text{hour} = 3600\,s\)
   • Potential difference, \(V = 50\,V\)
2. Formula to use:
   • Heat produced (\(H\)), using Joule’s law: \[H = V \cdot I \cdot t\]
   • Current (\(I\)), where \[\begin{aligned}I &= \frac{Q}{t} \\\\&= \frac{96000}{3600} \\\\&= 26.67\,A\end{aligned}\]
3. Calculation: \[\begin{aligned} H &= V \cdot I \cdot t \\&= 50 \times 26.67 \times 3600 \\&= 4,800,600\,\text{J} \end{aligned}\]
   Therefore, the heat generated is 4,800,600 joules.

The heat produced when 96,000 coulombs of charge pass through a 50V potential in one hour is \(4,800,600\,J\).

Q18: An electric iron of resistance 20 ? takes a current of 5 A. Calculate the heat developed in 30 s.

Answer:

1. Given Data:
   • Resistance, \(R = 20\,\Omega\)
   • Current, \(I = 5\,A\)
   • Time, \(t = 30\,\text{s}\)
2. Formula to use: \[H = I^2 R t\]
3. Calculation: \[\begin{aligned} H &= (5)^2 \times 20 \times 30 \\&= 25 \times 20 \times 30 \\&= 500 \times 30 \\&= 15,000\,J \end{aligned}\]
   Therefore, the heat developed is 15,000 joules.

The iron develops \(15,000\,J\) of heat in 30 seconds.

Q19: What determines the rate at which energy is delivered by a current?

Answer:

• Potential Difference: The rate of energy delivery (electric power) depends on the voltage (\(V\)) across the device or circuit.
• Electric Current: It also depends on the amount of current (\(I\)) flowing through the circuit.
• Power Formula: The relationship is given by: \[P = V \times I\] Power (\(P\)) is measured in watts (\(W\)), representing energy delivered per second.

Summary: The product of current and voltage sets the rate at which electrical energy is supplied or consumed in any device.

Q20: An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.

Answer:

1. Given Data:
   • Current, \(I = 5\,A\)
   • Voltage, \(V = 220\,V\)
   • Time, \[\begin{aligned}t &= 2\,\text{hours} \\&= 2 \times 3600\,s \\&= 7200\,s\end{aligned}\]
2. Power of the motor: \[\begin{aligned}P &= V \times I \\&= 220 \times 5 \\&= 1100\,W\end{aligned}\] The motor’s power is 1100 watts (or 1.1 kilowatts).
3. Energy consumed in 2 hours: \[\begin{aligned}E &= P \times t \\&= 1100 \times 7200 \\&= 7,920,000\,J\end{aligned}\] In kilowatt-hours: \[\begin{aligned}E &= \frac{1100}{1000} \times 2 \\\\&= 2.2\,kWh\end{aligned}\]

The power of the motor is \(1100\,W\) and the energy consumed in 2 hours is \(7,920,000\,J\) or \(2.2\,kWh\).