Maharashtra Board Class 11 Physics Chapter 11 Electric Current Through Conductors Solutions

1. Choose The Correct Alternative.

Question 1. You are given four bulbs of 25 W, 40 W, 60 W and 100 W of power, all operating a 230 V. Which of them has the lowest resistance?
(A) 25 W
(B) 40 W
(C) 60 W
(D) 100 W
Answer:
(D) 100 W
In simple words: Power \(P = V^2/R\). For a constant voltage V, resistance R is inversely proportional to power P. Therefore, the bulb with the highest power (100 W) will have the lowest resistance.

🎯 Exam Tip: Remember the relationship \(P = V^2/R\) for constant voltage applications to quickly determine resistance based on power ratings.

Question 2. Which of the following is an ohmic conductor?
(A) transistor
(B) vacuum tube
(C) electrolyte
(D) nichrome wire
Answer:
(D) nichrome wire
In simple words: An ohmic conductor is one that obeys Ohm's Law, meaning its resistance remains constant regardless of the applied voltage or current. Nichrome wire is a common example.

🎯 Exam Tip: Be familiar with examples of both ohmic and non-ohmic conductors to differentiate them based on their I-V characteristics.

Question 3. A rheostat is used
(A) to bring on a known change of resistance in the circuit to alter the current.
(B) to continuously change the resistance in any arbitrary manner and there by alter the current.
(C) to make and break the circuit at any instant.
(D) neither to alter the resistance nor the current.
Answer:
(B) to continuously change the resistance in any arbitrary manner and there by alter the current.
In simple words: A rheostat is a variable resistor designed to continuously change the resistance in a circuit, thereby altering the current flowing through it.

🎯 Exam Tip: Understand the function of common circuit components like rheostats, resistors, and switches, and their specific roles in controlling current and resistance.

Question 4. The wire of length L and resistance R is stretched so that its radius of cross-section is halved. What is its new resistance?
(A) 5R
(B) 8R
(C) 4R
(D) 16R
Answer:
(D) 16R
In simple words: When a wire is stretched, its volume remains constant. Halving the radius means the length increases significantly. Resistance depends on length and cross-sectional area (which depends on radius), leading to a much larger new resistance.

🎯 Exam Tip: For stretching/compressing wire problems, remember that volume remains constant (\(V = AL\)), and resistance is \(R = \rho L/A\). Express A in terms of r (\(A = \pi r^2\)) and find the new length/area relationship.

Question 5. Masses of three pieces of wires made of the same metal are in the ratio 1:3:5 and their lengths are in the ratio 5 : 3 : 1. The ratios of their resistances are
(A) 1:3:5
(B) 5: 3:1
(C) 1:15:125
(D) 125: 15:1
Answer:
(D) 125: 15:1
In simple words: Resistance depends on resistivity, length, and cross-sectional area. Since mass and length are given, you can find the cross-sectional area for each wire (mass = density * volume = density * area * length) and then calculate the resistance ratio.

🎯 Exam Tip: For problems involving ratios of mass, length, and resistance, use the formula \(R = \rho L/A\) and relate area (A) to mass (m), density (\(\rho_d\)), and length (L) using \(m = \rho_d A L\), so \(A = m/(\rho_d L)\). Substitute A back into the resistance formula.

Question 6. The internal resistance of a cell of emf 2 V is 0.1 Ω, it is connected to a resistance of 0.9 Ω. The voltage across the cell will be
(A) 0.5 V
(B) 1.8 V
(C) 1.95 V
(D) 3V
Answer:
(B) 1.8 V
In simple words: The voltage across the cell (terminal voltage) is the EMF minus the voltage drop across its internal resistance. First, calculate the total current, then the voltage drop, and finally the terminal voltage.

🎯 Exam Tip: The terminal voltage of a cell is given by \(V = E - Ir\), where E is EMF, I is current, and r is internal resistance. The total current in the circuit is \(I = E/(R+r)\).

Question 7. 100 cells each of emf 5 V and internal resistance 1 Ω are to be arranged so as to produce maximum current in a 25 Ω resistance. Each row contains equal number of cells. The number of rows should be
(A) 2
(B) 4
(C) 5
(D) 100
Answer:
(A) 2
In simple words: For maximum current from a combination of cells, the external resistance should be equal to the equivalent internal resistance of the combination. You'll need to set up the cells in m rows and n cells per row, then find m and n that satisfy the condition.

🎯 Exam Tip: For maximum current in a series-parallel combination of cells, the total internal resistance of the battery (\(R_{int} = (n/m)r\)) should equal the external resistance (\(R_{ext}\)), where n is cells per row and m is number of rows. Also, the total number of cells is \(N = mn\).

Question 8. Five dry cells each of voltage 1.5 V are connected as shown in diagram

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र पाँच सूखे सेल को एक विद्युत परिपथ में जोड़ना दर्शाता है। तीन सेल श्रृंखला में जुड़े हुए हैं, जहाँ प्रत्येक सेल का धनात्मक टर्मिनल अगले सेल के ऋणात्मक टर्मिनल से जुड़ा है। हालाँकि, बीच वाले दो सेल (दूसरे और तीसरे) एक दूसरे के विपरीत दिशा में जुड़े हुए हैं, जिससे उनकी EMF एक दूसरे को रद्द कर देती है, और केवल शेष श्रृंखला में जुड़े सेलों का योग कुल वोल्टेज निर्धारित करता है।
What is the overall voltage with this arrangement?
(A) 0 V
(B) 4.5 V
(C) 6.0 V
(D) 7.5 V
Answer:
(B) 4.5 V
In simple words: In the given arrangement, some cells are in series and some are connected in opposing polarity. The overall voltage is the sum of voltages of cells in series minus any opposing voltages.

🎯 Exam Tip: When cells are connected in series, their EMFs add up. If they are connected in opposition (positive to positive or negative to negative), their EMFs subtract. Carefully analyze the polarity of each cell.

2. Give Reasons / Short Answers

Question 1. In given circuit diagram two resistors are connected to a 5V supply.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक साधारण श्रृंखला परिपथ दिखाता है जिसमें दो प्रतिरोधक (8 Ω और 6 Ω) एक 5V बिजली आपूर्ति से जुड़े हुए हैं। दोनों प्रतिरोधक एक के बाद एक जुड़े हुए हैं, जिसका अर्थ है कि एक ही धारा दोनों से होकर प्रवाहित होती है, और कुल प्रतिरोध उनके व्यक्तिगत प्रतिरोधों का योग है।
i. Calculate potential difference across the 8Ω resistor.
ii. A third resistor is now connected in parallel with 6 Ω resistor. Will the potential difference across the 8 Ω resistor be larger, smaller or same as before? Explain the reason for your answer.
Answer:
Total current flowing through the circuit,
\(I = \frac{V}{R_S}\)
\( = \frac{5}{8+6}\)
\( = \frac{5}{14} = 0.36 \text{ A}\)
\( \therefore \text{ Potential difference across } 8 \Omega \text{ (Vi)} = 0.36 \times 8 \)
\( = 2.88 \text{ V}\)
ii. Potential difference across 8 Ω resistor will be larger.
Reason: As per question, the new circuit diagram will be

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक संशोधित परिपथ दिखाता है जहाँ एक तीसरा प्रतिरोधक (R Ω) को 6 Ω प्रतिरोधक के समानांतर जोड़ा गया है, जबकि 8 Ω प्रतिरोधक अभी भी श्रृंखला में है और पूरा संयोजन 5V आपूर्ति से जुड़ा है। यह व्यवस्था समानांतर संयोजन के कारण कुल प्रतिरोध में कमी और 8 Ω प्रतिरोधक पर संभावित अंतर में वृद्धि को दर्शाती है।
When any resistor is connected parallel to 6 Ω resistance. Then the resistance across that branch (6 Ω and R Ω) will become less than 6 Ω. i.e., equivalent resistance of the entire circuit will decrease and hence current will increase. Since, V = IR, the potential difference across 8 Ω resistor will be larger.
In simple words: For resistors in series, the total current is calculated using the total resistance and the supply voltage. The potential difference across a specific resistor is then found by multiplying this current by that resistor's value. When a resistor is added in parallel to an existing resistor in a series circuit, the equivalent resistance of that parallel branch decreases, reducing the overall circuit resistance, increasing the total current, which in turn increases the potential difference across the fixed series resistor.

🎯 Exam Tip: In a series circuit, the current is the same through all resistors. Use Ohm's law (\(V = IR\)) and the formula for total series resistance (\(R_{total} = R_1 + R_2 + ...\)). Adding a resistor in parallel reduces the equivalent resistance of that combination. If this combination is part of a larger series circuit, the total circuit resistance will decrease, leading to an increase in total current (assuming constant voltage) and a redistribution of potential differences.

Question 2. Prove that the current density of a metallic conductor is directly proportional to the drift speed of electrons.
Answer:
i. Consider a part of conducting wire with its free electrons having the drift speed \(v_d\) in the direction opposite to the electric field \(E\).
ii. All the electrons move with the same drift speed \(v_d\) and the current I is the same throughout the cross section (A) of the wire.
iii. Let L be the length of the wire and n be the number of free electrons per unit volume of the wire. Then the total number of free electrons in the length L of the conducting wire is nAL.
iv. The total charge in the length L is,
q = nAle .......... (1)
where, e is the charge of electron.
v. Equation (1) is total charge that moves through any cross section of the wire in a certain time interval t.
\( \therefore t = \frac{L}{v_d} \) .......... (2)
vi. Current is given by,
\( I = \frac{q}{t} = \frac{nALe}{L/v_d} \) [From Equations (1) and (2)]
\( = n A v_d e \)
Hence
\( v_d = \frac{I}{nAe} \)
\( \implies \frac{J}{ne} \) \( (\because J = \frac{I}{A}) \)
Hence for constant 'ne', current density of a metallic conductor is directly proportional to the drift speed of electrons, \(J \propto V_d\).
In simple words: Current density (J) is the current per unit cross-sectional area. It is derived from the total charge flowing per unit time through an area, which in turn depends on the number of charge carriers, their charge, and their average drift speed (\(v_d\)).

🎯 Exam Tip: Remember the fundamental relation \(I = nAv_d e\). Current density \(J = I/A\), so dividing the fundamental relation by A directly gives \(J = n v_d e\). Since n and e are constants for a given conductor, J is proportional to \(v_d\).

3. Answer The Following Questions.

Question 1. Distinguish between ohmic and non ohmic substances; explain with the help of example.
Answer:

In simple words: Ohmic substances follow Ohm's Law, meaning their resistance is constant and their V-I graph is a straight line. Non-ohmic substances do not follow Ohm's Law; their resistance changes with voltage or current, and their V-I graph is not linear.

🎯 Exam Tip: Clearly state the definition, I-V characteristics (linear vs. non-linear), and provide specific examples for both ohmic and non-ohmic conductors. Understanding the underlying physical reasons for non-ohmic behavior (e.g., temperature changes, semiconductor properties) is also helpful.

Question 2. DC current flows in a metal piece of non uniform cross-section. Which of these quantities remains constant along the conductor: current, current density or drift speed?
Answer:
Drift velocity and current density will change as it depends upon area of cross-section whereas current will remain constant.
In simple words: When current flows through a conductor with non-uniform cross-section, the total current remains constant everywhere due to charge conservation. However, current density and drift speed will change inversely with the cross-sectional area.

🎯 Exam Tip: Apply the principle of conservation of charge. The total current entering a section must equal the total current leaving it. Current density \(J = I/A\) and drift velocity \(v_d = J/(ne)\) are both inversely proportional to the area, A.

4. Solve The Following Problems.

Question 1. What is the resistance of one of the rails of a railway track 20 km long at 20°C? The cross-section area of rail is 25 cm² and the rail is made of steel having resistivity at 20°C as \(6 \times 10^{-8} \Omega \text{ m}\).
Answer:
Given: \(I = 20 \text{ km} = 20 \times 10^3 \text{ m}\),
\(A = 25 \text{ cm}^2 = 25 \times 10^{-4} \text{ m}^2\),
\(\rho = 6 \times 10^{-8} \Omega \text{ m}\)
To find: Resistance of rail (R)
Formula: \( \rho = \frac{RA}{l} \)
Calculation: From formula.
\( R = \rho \frac{l}{A} \)
\( \therefore R = \frac{6 \times 10^{-8} \times 20 \times 10^3}{25 \times 10^{-4}} \)
\( = \frac{6 \times 4}{5} \times 10^{-1} \)
\( = 0.48 \Omega \)
In simple words: To find the resistance, use the formula \(R = \rho L/A\). Remember to convert all units to a consistent system (like meters for length, square meters for area).

🎯 Exam Tip: Always ensure consistent units (SI units are preferred) before plugging values into formulas. Pay attention to conversions like km to m and cm² to m².

Question 2. A battery after a long use has an emf 24 V and an internal resistance 380 Ω. Calculate the maximum current drawn from the battery. Can this battery drive starting motor of car?
Answer:
\(E = 24 \text{ V}, r = 380 \Omega\)
i. Maximum current (\(I_{max}\))
ii. Can battery start the motor?
Formula: \(I_{max} = \frac{E}{r}\)
Calculation:
From formula,
\(I_{max} = \frac{24}{380}\)
\( = 0.063 \text{ A}\)
As, the value of current is very small compared to required current to run a starting motor of a car, this battery cannot be used to drive the motor.
In simple words: Maximum current from a battery occurs when the external resistance is zero, in which case it is limited only by its internal resistance (\(I_{max} = E/r\)). Compare this value with the typical current needed for a car starter.

🎯 Exam Tip: Maximum current from a source is obtained when the external resistance is negligible (short circuit), given by \(I_{max} = E/r\). Car starter motors require very high currents (hundreds of amperes), so a battery with high internal resistance cannot supply this.

Question 3. A battery of emf 12 V and internal resistance 3 Ω is connected to a resistor. If the current in the circuit is 0.5 A,
i. Calculate resistance of resistor.
ii. Calculate terminal voltage of the battery when the circuit is closed.
Answer:
Given: \(E = 12 \text{ V}, r = 3 \Omega, I = 0.5 \text{ A}\)
To find:
i. Resistance (R)
ii. Terminal voltage (V)
Formulae:
i. \(E = I (r + R)\)
ii. \(V = IR\)
Calculation: From formula (i),
\(E = Ir + IR\)
\( \therefore R = \frac{E-Ir}{I} \)
\( = \frac{12-0.5 \times 3}{0.5} \)
\( = 21 \Omega \)
From formula (ii),
\(V = 0.5 \times 21\)
\( = 10.5 \text{ V}\)
In simple words: The total resistance in the circuit is the sum of external and internal resistance. Use Ohm's law to find the total resistance, then subtract internal resistance to get external. Terminal voltage is the voltage across the external resistor, or \(E - Ir\).

🎯 Exam Tip: Remember \(E = I(R+r)\) and \(V = IR\). These two formulas are crucial for solving problems involving EMF, internal resistance, external resistance, current, and terminal voltage.

Question 4. The magnitude of current density in a copper wire is 500 A/cm². If the number of free electrons per cm³ of copper is \(8.47 \times 10^{22}\), calculate the drift velocity of the electrons through the copper wire (charge on an \(e = 1.6 \times 10^{-19} \text{ C}\))
Answer:
Given: \(J = 500 \text{ A/cm}^2 = 500 \times 10^4 \text{ A/m}^2\),
\(n = 8.47 \times 10^{22} \text{ electrons/cm}^3\)
\( = 8.47 \times 10^{28} \text{ electrons/m}^3\)
\(e = 1.6 \times 10^{-19} \text{ C}\)
To Find: Drift velocity (\(v_d\))
Formula: \(v_d = \frac{J}{ne}\)
Calculation:
From formula,
\(v_d = \frac{500 \times 10^4}{8.47 \times 10^{28} \times 1.6 \times 10^{-19}}\)
\( = \frac{500}{8.47 \times 1.6} \times 10^{-5}\)
\( = \{\text{antilog } [\log 500 - \log 8.47 - \log 1.6]\} \times 10^{-5}\)
\( = \{\text{antilog } [2.6990 - 0.9279 - 0.2041]\} \times 10^{-5}\)
\( = \{\text{antilog } [1.5670]\} \times 10^{-5}\)
\( = 3.690 \times 10^1 \times 10^{-5}\)
\( = 3.69 \times 10^{-4} \text{ m/s}\)
In simple words: Drift velocity is directly related to current density, number of charge carriers, and elementary charge. Convert all given units to SI before applying the formula \(J = n e v_d\).

🎯 Exam Tip: The formula for current density \(J = nev_d\) is fundamental. Ensure all units are consistent (e.g., A/m² for J, electrons/m³ for n) before calculation. Pay attention to unit conversions from cm² to m² and cm³ to m³.

Question 5. Three resistors 10 Ω, 20 Ω and 30 Ω are connected in series combination.
i. Find equivalent resistance of series combination.
ii. When this series combination is connected to 12 V supply, by neglecting the value of internal resistance, obtain potential difference across each resistor.
Answer:
Given: \(R_1 = 10 \Omega, R_2 = 20 \Omega\),
\(R_3 = 30 \Omega, V = 12 \text{ V}\)
To Find: i. Series equivalent resistance(\(R_S\))
ii. Potential difference across each resistor (\(V_1, V_2, V_3\))
Formula: i. \(R_S = R_1 + R_2 + R_3\)
ii. \(V = IR\)
Calculation:
From formula (i),
\(R_S = 10 + 20 + 30 = 60 \Omega\)
From formula (ii),
\(I = \frac{V}{R_S} = \frac{12}{60} = 0.2 \text{ A}\)
\( \therefore \text{ Potential difference across } R_1, \)
\(V_1 = I \times R_1 = 0.2 \times 10 = 2 \text{ V}\)
\( \therefore \text{ Potential difference across } R_2, \)
\(V_2 = 0.2 \times 20 = 4 \text{ V}\)
\( \therefore \text{ Potential difference across } R_3, \)
\(V_3 = 0.2 \times 30 = 6 \text{ V}\)
In simple words: For resistors in series, the equivalent resistance is simply their sum. The current flowing through each resistor is the same, calculated by dividing the total voltage by the equivalent resistance. The potential difference across each resistor is then found using Ohm's law.

🎯 Exam Tip: In a series circuit, \(R_{eq} = R_1 + R_2 + R_3\), and current \(I\) is constant throughout. The voltage drop across each resistor is \(V_x = I \times R_x\). Always verify that the sum of individual voltage drops equals the total supply voltage.

Question 6. Two resistors 1 kΩ and 2 kΩ are connected in parallel combination.
i. Find equivalent resistance of parallel combination.
ii. When this parallel combination is connected to 9 V supply, by neglecting internal resistance, calculate current through each resistor.
Answer:
Given: \(R_1 = 1 \text{ k}\Omega = 10^3 \Omega\),
\(R_2 = 2 \text{ k}\Omega = 2 \times 10^3 \Omega, V = 9 \text{ V}\)
To find:
i. Parallel equivalent resistance (\(R_p\))
ii. Current through 1 kΩ and 2 kΩ (\(I_1\) and \(I_2\))
Formula:
i. \( \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} \)
ii. \(V = IR\)
Calculation: From formula (i),
\( \frac{1}{R_p} = \frac{1}{10^3} + \frac{1}{2 \times 10^3} \)
\( = \frac{3}{2 \times 10^3} \)
\( \therefore R_P = \frac{2 \times 10^3}{3} = 0.66 \text{ k}\Omega \)
From formula (ii),
\( I_1 = \frac{V}{R_1} = \frac{9}{10^3} \)
\( = 9 \times 10^{-3} \text{ A}\)
\( = 9 \text{ mA}\)
\( I_2 = \frac{V}{R_2} = \frac{9}{2 \times 10^3} \)
\( = 4.5 \times 10^{-3} \text{ A}\)
\( = 4.5 \text{ mA}\)
In simple words: For resistors in parallel, the reciprocal of the equivalent resistance is the sum of the reciprocals of individual resistances. When connected to a voltage supply, the voltage across each parallel branch is the same, allowing individual currents to be calculated using Ohm's law.

🎯 Exam Tip: For parallel resistors, \(1/R_{eq} = 1/R_1 + 1/R_2 + ...\). The voltage across each parallel resistor is the same as the supply voltage. Current through each branch is \(I_x = V/R_x\). Remember to convert kΩ to Ω for calculations.

Question 7. A silver wire has a resistance of 4.2 Ω at 27°C and resistance 5.4 Ω at 100°C. Determine the temperature coefficient of resistance.
Answer:
Given: \(R_1 = 4.2 \Omega, R_2 = 5.4 \Omega\),
\(T_1 = 27^\circ \text{ C}, T_2 = 100 ^\circ \text{C}\)
To find: Temperature coefficient of resistance (\(\alpha\))
Formula: \( \alpha = \frac{R_2-R_1}{R_1 (T_2-T_1)} \)
Calculation:
From Formula
\( \alpha = \frac{5.4-4.2}{4.2(100-27)} = 3.91 \times 10^{-3}/^\circ \text{C} \)
In simple words: The temperature coefficient of resistance describes how much a material's resistance changes per degree Celsius. It can be calculated using the resistances at two different temperatures and the temperature difference.

🎯 Exam Tip: The formula for temperature coefficient of resistance is \(\alpha = (R_2 - R_1) / (R_1 (T_2 - T_1))\). Ensure consistent units for temperature (Celsius or Kelvin difference is the same) and resistance.

Question 8. A 6 m long wire has diameter 0.5 mm. Its resistance is 50 Ω. Find the resistivity and conductivity.
Answer:
Given: \(I = 6 \text{ m}, D = 0.5 \text{ mm}\),
\(r = 0.25 \text{ mm} = 0.25 \times 10^{-3} \text{ m}, R = 50 \Omega\)
To find:
i. Resistivity (\(\rho\))
ii. Conductivity (\(\sigma\))
Formulae:
i. \( \rho = \frac{RA}{l} = \frac{R \pi r^2}{l} \)
ii. \( \sigma = \frac{1}{\rho} \)
Calculation:
From formula (i),
\( \rho = \frac{50 \times 3.142 \times (0.25 \times 10^{-3})^2}{6} \)
\( = \{\text{antilog } [\log 50 + \log 3.142 + 2 \log 0.25 - \log 6]\} \times 10^{-6}\)
\( = \{\text{antilog } [1.6990 + 0.4972 + 2(\overline{1}.3979) - 0.7782]\} \times 10^{-6}\)
\( = \{\text{antilog } [2.1962 + \overline{2}.7958 - 0.7782]\} \times 10^{-6}\)
\( = \{\text{antilog } [0.9920 - 0.7782]\} \times 10^{-6}\)
\( = \{\text{antilog } [0.2138]\} \times 10^{-6}\)
\( = 1.636 \times 10^{-6} \Omega/\text{m}\)
From formula (ii),
\( \sigma = \frac{1}{1.636 \times 10^{-6}} \)
\( = 0.6157 \times 10^6 \)
....(Using reciprocal from log table)
\( = 6.157 \times 10^5 \text{ m/}\Omega \)
In simple words: Resistivity is an intrinsic property of a material, calculated using resistance, length, and cross-sectional area. Conductivity is simply the reciprocal of resistivity.

🎯 Exam Tip: Resistivity (\(\rho\)) and conductivity (\(\sigma\)) are inversely related (\(\sigma = 1/\rho\)). Use the formula \(R = \rho L/A\) to find \(\rho\), remembering to calculate the area (A) from the diameter, \(A = \pi (D/2)^2 = \pi r^2\), and convert all units to SI.

Question 9. Find the value of resistances for the following colour code.
i. Blue Green Red Gold
ii. Brown Black Red Silver
iii. Red Red Orange Gold
iv. Orange White Red Gold
v. Yellow Violet Brown Silver
Answer:
i. Given: Blue – Green – Red – Gold
To find: Value of resistance
Formula: Value of resistance
\( = (\text{xy} \times 10^z \pm \text{T}\%)\Omega\)
Calculation:

From formula,
Value of resistance \( = (65 \times 10^2 \pm 5\%)\Omega\)
Value of resistance \( = 6.5 \text{ k}\Omega \pm 5\%\)
ii. Given: Brown – Black – Red – Silver
To find: Value of resistance
Formula: Value of resistance
\( = (\text{xy} \times 10^z + \text{T}\%)\Omega\)
Calculation:

From formula,
Value of resistance \( = (10 \times 10^2 \pm 10\%)\Omega\)
Value of resistance \( = 1.0 \text{ k}\Omega \pm 10\%\)
iii. Given: Red – Red – Orange – Gold
To find: Value of the resistance
Formula: Value of the resistance
\( = (\text{xy} \times 10^z \pm \text{T}\%)\)
Calculation:

From formula,
Value of resistance \( = (22 \times 10^3 \pm 5\%)\Omega\)
Value of resistance \( = 22 \text{ k}\Omega \pm 5\%\)
[Note: The answer given above is presented considering correct order of magnitude.]
iv. Given: Orange – White – Red – Gold
To find: Value of the resistance
Formula: Value of the resistance
\( = (\text{xy} \times 10^z \pm \text{T}\%)\)
Calculation:

From formula,
Value of resistance \( = (39 \times 10^2 \pm 5\%)\Omega\)
Value of resistance \( = 3.9 \text{ k}\Omega \pm 5\%\)
v. Given: Yellow-Violet-Brown-Silver
To find: Value of the resistance
Formula: Value of the resistance
\( = (\text{xy} \times 10^z + \text{T}\%)\)
Calculation:

From formula,
Value of resistance \( = (47 \times 10 \pm 10\%)\Omega\)
Value of resistance \( = 470 \Omega \pm 10\%\)
[Note: The answer given above is presented considering correct order of magnitude.]
In simple words: Resistor color codes use bands to represent the first two significant digits, a multiplier, and tolerance. Each color corresponds to a specific number or percentage. To find the color code, break down the resistance value into its significant digits, multiplier, and tolerance. Match these values to their corresponding colors using the standard resistor color code chart.

🎯 Exam Tip: Memorize the resistor color code sequence (BB ROY G B V G W for digits, Gold/Silver/None for tolerance) and their corresponding values. Practice reading various combinations to avoid errors. When reversing the process (value to code), identify the first two digits, the power of ten (multiplier), and the tolerance percentage. Remember that for ±10% tolerance, the fourth band is silver. For values like 330 Ω, the multiplier is 10¹ (brown).

Question 10. Find the colour code for the following value of resistor having tolerance ± 10%.
i. 330 Ω
ii. 100 Ω
iii. 47 KΩ
iv. 160 Ω
v. 1 KΩ
Answer:

In simple words: To find the color code, break down the resistance value into its significant digits, multiplier, and tolerance. Match these values to their corresponding colors using the standard resistor color code chart.

🎯 Exam Tip: When reversing the process (value to code), identify the first two digits, the power of ten (multiplier), and the tolerance percentage. Remember that for ±10% tolerance, the fourth band is silver. For values like 330 Ω, the multiplier is 10¹ (brown).

Question 11. A current of 4 A flows through an automobile headlight. How many electrons flow through the headlight in a time of 2 hrs?
Answer:
Given: \(I = 4 \text{ A}, t = 2 \text{ hrs} = 2 \times 60 \times 60 \text{ s}\)
To find: Number of electrons (N)
Formula: \(I = \frac{q}{t} = \frac{Ne}{t}\)
Calculation: As we know, \(e = 1.6 \times 10^{-19} \text{ C}\)
From formula,
\( N = \frac{It}{e} = \frac{4 \times 2 \times 60 \times 60}{1.6 \times 10^{-19}} = 18 \times 10^{23} \)
In simple words: The total charge flowing is current multiplied by time. The number of electrons is found by dividing this total charge by the charge of a single electron.

🎯 Exam Tip: Use the fundamental relation \(Q = It\) to find total charge. Then, \(N = Q/e\), where \(e\) is the elementary charge (\(1.6 \times 10^{-19}\) C). Ensure time is in seconds.

Question 12. The heating element connected to 230 V draws a current of 5 A. Determine the amount of heat dissipated in 1 hour (J = 4.2 J/cal).
Answer:
Given: \(V = 230 \text{ V}, I = 5 \text{ A}\),
At \(t = 1 \text{ hour} = 60 \times 60 \text{ sec}\)
To find: Heat dissipated (H)
Formula: \(H = \Delta U = I \Delta t V\)
Calculation: From formula,
\(H = 5 \times 60 \times 60 \times 230\)
\( = 4.14 \times 10^6 \text{ J}\)
Heat dissipated in calorie,
\( H = \frac{4.14 \times 10^6}{4.2} = 985.7 \times 10^3 \text{ cal}\)
\( = 985.7 \text{ kcal}\)
In simple words: Heat dissipated in a resistive element is given by the power dissipated multiplied by time. It can be calculated using voltage, current, and time, or converted from Joules to calories using the given conversion factor.

🎯 Exam Tip: Use the formula \(H = VIt\) (or \(I^2Rt\), or \(V^2t/R\)). Ensure all units are in SI (Volts, Amperes, Seconds) to get heat in Joules. For conversion to calories, use the provided conversion factor.

Can You Recall? (Textbookpage No. 207)

Question. An electric current in a metallic conductor such as a wire is due to flow of electrons, the negatively charged particles in the wire. What is the role of the valence electrons which are the outermost electrons of an atom?
Answer:
(i) The valence electrons become de-localized when large number of atoms come together in a metal.
(ii) These electrons become conduction electrons or free electrons constituting an electric current when a potential difference is applied across the conductor.
In simple words: In metals, valence electrons are not bound to individual atoms; they become delocalized and form a 'sea' of mobile electrons. These free electrons are responsible for carrying electric current when a potential difference is applied across the conductor.

🎯 Exam Tip: When discussing metallic conductors, always emphasize the 'delocalized' nature of valence electrons and their transformation into 'conduction electrons' as the key to understanding electrical conductivity.