Chapter 3: Current Electricity Application Questions | physics Class 12 CBSE

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Chapter 3: Current Electricity Application Questions | physics Class 12 CBSE | ByteNotes.in

Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 1

Applied Concept

A copper wire of cross-sectional area 1.0 × 10⁻⁷ m² carries a current of 1.5 A. Given that the electron number density in copper is 8.5 × 10²⁸ m⁻³, calculate the drift velocity of electrons and compare it with the speed of propagation of the electric field (3 × 10⁸ m/s).

Using vd = I/(neA): vd = 1.5 / (8.5 × 10²⁸ × 1.6 × 10⁻¹⁹ × 1.0 × 10⁻⁷) = 1.1 × 10⁻³ m/s ≈ 1.1 mm/s.
The ratio of drift speed to electric field propagation speed: 1.1 × 10⁻³ / 3 × 10⁸ ≈ 3.7 × 10⁻¹², showing the drift speed is about 10¹¹ times smaller than the speed of field propagation.
This comparison explains why current is established almost instantly when a circuit is closed — it is the electric field that travels at near light speed, not the electrons; each electron merely drifts a tiny bit locally.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 2

Applied Concept

An electric toaster uses nichrome as its heating element. At room temperature (27°C) its resistance is 75.3 Ω. When connected to a 230 V supply, current settles to 2.68 A. The temperature coefficient of nichrome is 1.70 × 10⁻⁴ °C⁻¹. Find the steady operating temperature of the element.

Steady resistance R₂ = V/I = 230/2.68 = 85.8 Ω. Using R₂ = R₁[1 + α(T₂ – T₁)] with R₁ = 75.3 Ω, α = 1.70 × 10⁻⁴ °C⁻¹, T₁ = 27°C.
Solving: T₂ – T₁ = (R₂ – R₁)/(R₁α) = (85.8 – 75.3)/(75.3 × 1.70 × 10⁻⁴) = 10.5/0.01281 ≈ 820°C.
Therefore T₂ = 27 + 820 = 847°C. At this steady temperature, heat generated by the current equals heat lost to the surroundings, so both temperature and current stabilise.

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Application Question

Easy difficulty • Concept in a practical situation

Easy

Question 3

Applied Concept

A battery of EMF 10 V and internal resistance 3 Ω is connected to a resistor. The current in the circuit is 0.5 A. Find (i) the external resistance and (ii) the terminal voltage of the battery.

Using I = ε/(R + r): 0.5 = 10/(R + 3), so R + 3 = 20, giving R = 17 Ω.
Terminal voltage V = ε – Ir = 10 – (0.5 × 3) = 10 – 1.5 = 8.5 V. Alternatively, V = IR = 0.5 × 17 = 8.5 V, confirming the result.
The difference ε – V = 1.5 V is the voltage drop across the internal resistance, demonstrating that terminal voltage is always less than EMF when current flows.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 4

Applied Concept

A platinum resistance thermometer has resistance 5 Ω at 0°C and 5.23 Ω at 100°C. When inserted in a hot liquid, its resistance is 5.795 Ω. Determine the temperature of the liquid.

Using the interpolation formula: t = [(Rt – R₀)/(R₁₀₀ – R₀)] × 100°C, with R₀ = 5 Ω, R₁₀₀ = 5.23 Ω, Rt = 5.795 Ω.
t = [(5.795 – 5)/(5.23 – 5)] × 100 = (0.795/0.23) × 100 = 3.456 × 100 = 345.65°C.
This application illustrates how the nearly linear relationship between resistance and temperature in metals (platinum in this case) is used to accurately measure temperature in industrial and scientific settings.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 5

Applied Concept

A 10 V battery with negligible internal resistance is connected across opposite corners of a cube made of 12 equal resistors of 1 Ω each. Using Kirchhoff's rules and symmetry, find the equivalent resistance and the total current.

By symmetry, the current entering the cube corner splits equally into three paths of resistance 1 Ω each. From the symmetry argument and Kirchhoff's rules applied to loop ABCC'EA: –IR – (I/2)R – IR + ε = 0, giving ε = (5/2)IR.
Equivalent resistance Req = ε/(3I) = (5/2)IR / (3I) = (5/6)R = 5/6 Ω for R = 1 Ω.
Total current = V/Req = 10/(5/6) = 12 A. This problem illustrates that complex symmetric networks can be solved elegantly by exploiting symmetry along with Kirchhoff's rules, avoiding lengthy calculations.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 6

Applied Concept

With reference to power transmission in a national grid, explain why power is transmitted at 220 kV rather than at 220 V, using the formula for power loss in transmission cables.

Power wasted in cables: Pc = P²Rc/V², where P is the power transmitted, Rc is the cable resistance, and V is the transmission voltage. At V = 220 V vs V = 220 kV, the ratio of losses is (220/220,000)² = 10⁻⁶, i.e., losses at 220 V would be a million times greater.
Transmission cables connecting power stations are hundreds of kilometres long with significant resistance Rc. At low voltage, the enormous current (I = P/V) would dissipate most of the power as heat in the cables before reaching consumers.
By using transformers to step up to 220 kV at the sending end and step down at the receiving end, power utilities transmit electricity efficiently with minimal waste — a direct application of Pc ∝ 1/V².

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 7

Applied Concept

A storage battery of EMF 8.0 V and internal resistance 0.5 Ω is being charged by a 120 V DC supply using a series resistor of 15.5 Ω. Find the terminal voltage of the battery during charging and explain why the series resistor is necessary.

During charging, current flows into the battery's positive terminal. Current I = (120 – 8)/(15.5 + 0.5) = 112/16 = 7 A. During charging, terminal voltage = ε + Ir = 8 + (7 × 0.5) = 8 + 3.5 = 11.5 V.
The series resistor limits the charging current to a safe value. Without it, current = (120 – 8)/0.5 = 224 A — far exceeding the battery's safe charging rate and potentially causing damage or hazard.
The series resistor also provides a controlled voltage drop (7 × 15.5 = 108.5 V), ensuring the remaining voltage (11.5 V) is applied across the battery terminals for proper charging.

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Application Question

Easy difficulty • Concept in a practical situation

Easy

Question 8

Applied Concept

A heating element of resistance 100 Ω at 27°C has a temperature coefficient α = 1.70 × 10⁻⁴ °C⁻¹. What is the resistance at 1027°C, and by what factor has the resistance increased?

Using RT = R₀[1 + α(T – T₀)]: R = 100 × [1 + 1.70 × 10⁻⁴ × (1027 – 27)] = 100 × [1 + 1.70 × 10⁻⁴ × 1000].
R = 100 × [1 + 0.17] = 100 × 1.17 = 117 Ω.
The resistance has increased by a factor of 1.17 (a 17% increase). This modest increase illustrates why alloys like nichrome are used for heating elements — their resistance doesn't change drastically with temperature, maintaining a relatively stable current and power output.

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Application Question

Easy difficulty • Concept in a practical situation

Easy

Question 9

Applied Concept

A wire of length 15 m and cross-sectional area 6.0 × 10⁻⁷ m² has a measured resistance of 5.0 Ω at a certain temperature. Calculate the resistivity of its material at that temperature.

Using R = ρl/A: ρ = RA/l = 5.0 × (6.0 × 10⁻⁷)/15 = 5.0 × 4.0 × 10⁻⁸ = 2.0 × 10⁻⁷ Ωm.
This value (2.0 × 10⁻⁷ Ωm) lies in the metallic range (10⁻⁸ to 10⁻⁶ Ωm), confirming the material is a conductor — consistent with the relatively low resistance of 5 Ω over 15 m.
Resistivity is a material property independent of shape; knowing ρ, the resistance of any length and cross-section of this material at the same temperature can be calculated using R = ρl/A.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 10

Applied Concept

In a Wheatstone bridge, AB = 100 Ω, BC = 10 Ω, CD = 5 Ω, DA = 60 Ω, and a galvanometer of 15 Ω resistance is connected across BD with 10 V across AC. Using Kirchhoff's rules, find the galvanometer current.

Applying Kirchhoff's loop rules to the three meshes with currents I₁ (through 60 Ω and 100 Ω arms), I₂, and Ig: 100I₁ + 15Ig – 60I₂ = 0; 10(I₁ – Ig) – 15Ig – 5(I₂ + Ig) = 0; 60I₂ + 5(I₂ + Ig) = 10.
Simplifying and solving: 63Ig – 2I₂ = 0 and 13I₂ + Ig = 2. From first: I₂ = 31.5Ig. Substituting: 13(31.5Ig) + Ig = 2, giving 410.5Ig = 2.
Ig = 4.87 mA. This non-zero galvanometer current confirms the bridge is unbalanced (check: 100/60 ≠ 10/5), demonstrating Kirchhoff's method for analysing unbalanced bridge circuits.

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Chapter 3: Current Electricity | Application Questions

A copper wire of cross-sectional area 1.0 × 10⁻⁷ m² carries a current of 1.5 A. Given that the electron number density in copper is 8.5 × 10²⁸ m⁻³, calculate the drift velocity of electrons and compare it with the speed of propagation of the electric field (3 × 10⁸ m/s).

An electric toaster uses nichrome as its heating element. At room temperature (27°C) its resistance is 75.3 Ω. When connected to a 230 V supply, current settles to 2.68 A. The temperature coefficient of nichrome is 1.70 × 10⁻⁴ °C⁻¹. Find the steady operating temperature of the element.

A battery of EMF 10 V and internal resistance 3 Ω is connected to a resistor. The current in the circuit is 0.5 A. Find (i) the external resistance and (ii) the terminal voltage of the battery.

A platinum resistance thermometer has resistance 5 Ω at 0°C and 5.23 Ω at 100°C. When inserted in a hot liquid, its resistance is 5.795 Ω. Determine the temperature of the liquid.

A 10 V battery with negligible internal resistance is connected across opposite corners of a cube made of 12 equal resistors of 1 Ω each. Using Kirchhoff's rules and symmetry, find the equivalent resistance and the total current.

With reference to power transmission in a national grid, explain why power is transmitted at 220 kV rather than at 220 V, using the formula for power loss in transmission cables.

A storage battery of EMF 8.0 V and internal resistance 0.5 Ω is being charged by a 120 V DC supply using a series resistor of 15.5 Ω. Find the terminal voltage of the battery during charging and explain why the series resistor is necessary.

A heating element of resistance 100 Ω at 27°C has a temperature coefficient α = 1.70 × 10⁻⁴ °C⁻¹. What is the resistance at 1027°C, and by what factor has the resistance increased?

A wire of length 15 m and cross-sectional area 6.0 × 10⁻⁷ m² has a measured resistance of 5.0 Ω at a certain temperature. Calculate the resistivity of its material at that temperature.

In a Wheatstone bridge, AB = 100 Ω, BC = 10 Ω, CD = 5 Ω, DA = 60 Ω, and a galvanometer of 15 Ω resistance is connected across BD with 10 V across AC. Using Kirchhoff's rules, find the galvanometer current.

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