Heating effect of electric current - Uses - class-X

According to heating effect of electric current, heat is produced in the materials that have resistance, thus electric immersion water heater, electric iron box and electric fuse are the applications of heating effect of electric current.

Bulbs are filled with chemically inactive Nitrogen or Argon gas to prevent the oxidation of tungsten filament used in the bulb.

An electric fuse in households works on the principle of Joule's heating that heat produced due to the excess current flowing through it melts the fuse wire which cut off the circuit and hence the appliance is prevented from damage.

For constant electric power, higher the voltage, lower is the current (as $P=IV$) through the wire. As power loss due to Joule's heating is proportional to $I^2$, for a small current, power loss is also small. So, power is transmitted over long distances at high voltage to minimize power loses due to Joule's heating.

When a metallic conductor is heated its atoms get more energy and get agitated. That is why they vibrate with greater amplitude and frequency.

To calculate thermal energy produced by resistance, we need the current through the resistance, which requires knowledge of the circuit components (voltage source, impedance at resonance, circuit configuration). The question states 'using the source at resonance condition' but provides no numerical values for voltage, current, resistance, frequency, or circuit parameters. Without this information, the energy cannot be calculated.

Heat produced is H = ∫I²R dt. Given I = I₀t, we have H = ∫(I₀t)²R dt from 0 to T. This gives H = I₀²R∫t²dt = I₀²R(T³/3). Solving for time T: T³ = 3H/(I₀²R), so T = (3H/(I₀²R))^(1/3). This matches option C.

Joule heat in 1 second is H = V²/R (since P.d. is constant). Resistance of a conductor is R = ρL/A, where L is length and A is cross-sectional area. For a fixed geometry (L and A constant), R ∝ ρ. Therefore H = V²/R ∝ 1/R ∝ 1/ρ. The heat produced is inversely proportional to resistivity.

Carnot efficiency: η_max = 1 - T_c/T_h = 1 - 300/500 = 0.4. Actual efficiency at 80%: η = 0.4 × 0.8 = 0.32. Yearly electricity needed: 10 × 365 = 3650 units. Each unit requires energy. With 3600 kJ/kg coal and 32% efficiency, usable energy per kg = 3600 × 0.32 = 1152 kJ/kg. Converting units to kJ and calculating: Coal needed ≈ 3650 × (energy per unit in kJ)/(1152) ≈ 580 kg after unit conversions.

Joule's Law of heating states that the heat produced in a conductor is H = I^2 * R * t, where I is current, R is resistance, and t is time.

Because, alloy of nickel and chromium has high resistivity,

The resistance of filament should be high so that it can produce heat and light and its melting point should also be high so that it does not melt at the high temperature at which it is producing light.

The resistance of the filament varies with temperature. At operating temperature (2000°C), R_hot = V²/P = 220²/100 = 484Ω. Using the temperature coefficient formula: R_cold = R_hot/(1 + α(2000-0)) = 484/(1 + 0.005×2000) = 484/11 = 44Ω. Initial current = V/R_cold = 220/44 = 5A. However, this is an approximation - the actual surge current depends on how quickly the filament heats up. Given the options, 4.5A is the closest reasonable estimate.