Ans: Potential difference V = 120 - 110 = 10 V. Work done W = Q × V = 3 × 10 = 30 J. Power P = W/t = 30/4 = 7.5 W. Alternatively, Power P = QV/t = (3 × 10)/4 = 7.5 W. This problem illustrates that potential difference is the work done per unit charge: V = W/Q. The higher potential point (120V) means more work is needed to push charge there against the electric field.

Ans: Parallel: 1/R = 1/5 + 1/10 + 1/30 = 6/30 + 3/30 + 1/30 = 10/30. R = 3Ω. Current through each (V same = 12V): I₁ = 12/5 = 2.4 A; I₂ = 12/10 = 1.2 A; I₃ = 12/30 = 0.4 A. Total current I = I₁ + I₂ + I₃ = 2.4 + 1.2 + 0.4 = 4.0 A. Check: I = V/R = 12/3 = 4.0 A ✓. Note: the equivalent resistance (3Ω) is less than the smallest individual resistor (5Ω) — always the case in parallel.

Ans: Tungsten is used for electric bulb filaments because: (1) Very high melting point (~3422°C) — can withstand the extremely high temperatures (about 2500–3000°C) needed to produce visible light without melting. (2) High resistivity — offers significant resistance, causing the filament to heat up intensely when current flows. (3) Ductile — can be drawn into thin wires. (4) Low vapor pressure at high temperatures — does not evaporate quickly inside the bulb. The bulb is filled with inert gas (argon) to prevent oxidation of the filament at high temperatures, extending its life.

Ans: Ohm's Law: At constant temperature, the current (I) through a conductor is directly proportional to the potential difference (V) across it: V ∝ I, i.e., V = IR, where R is resistance. Resistance is the opposition to the flow of electric current. Unit: Ohm (Ω). Dependence of resistance: R ∝ l (length) — longer wire has more resistance. R ∝ 1/A (area of cross-section) — thicker wire has less resistance. R depends on resistivity (ρ) of material: R = ρl/A. Good conductors (copper, silver) have very low resistivity. Poor conductors (nichrome, tungsten) have high resistivity — used in heating elements. Temperature: Resistance of metals increases with temperature; of semiconductors decreases with temperature.

Ans: Parallel: 1/R = 1/R₁ + 1/R₂ + 1/R₃ = 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1. So R = 1Ω. Equivalent resistance (1Ω) is less than the smallest resistance (2Ω). Advantages of parallel connection in household wiring: (1) Each appliance gets the same supply voltage (230 V). (2) If one appliance fails, others continue to work independently. (3) Total resistance decreases as more appliances are added, allowing more current to flow. (4) Individual on/off control of each appliance is possible. (5) Overall power consumed can be calculated by adding individual powers.

Ans: Coulomb (C).

Ans: The thin wire has more resistance, as resistance is inversely proportional to the area of cross-section.

Ans: An ammeter, connected in series in the circuit.

Ans: The heat produced in a resistor is directly proportional to the square of the current, to the resistance and to the time for which the current flows: H = I²Rt.

Ans: The same current flows through both, but the heating element is made of an alloy of very high resistance, so heat produced (I²R) is large and it glows. The cord has low resistance, so very little heat is produced.

Ans: Current in each resistor = V/R = 220/176 = 1.25 A. Number required = 5 ÷ 1.25 = 4 resistors.

Ans: P = VI = 220 × 0.5 = 110 W. Energy = P × t = 0.110 kW × 2 h = 0.22 kWh (= 0.22 units).

Ans: A voltmeter is connected in parallel across the component, and an ammeter in series. An ammeter has a very low resistance so that it does not change the current in the circuit. A voltmeter has a very high resistance so that it draws negligible current from the circuit and reads the true potential difference.

Ans: Ohm's law: at constant temperature, the potential difference across a conductor is directly proportional to the current through it, V = IR. Activity: connect a nichrome wire, an ammeter in series, a voltmeter in parallel across the wire, a key and a battery of increasing cells (or a rheostat). For each setting, note the current and the potential difference, and compute V/I. V/I stays constant. The V–I graph is a straight line through the origin; its slope is the resistance R. This shows that current is proportional to potential difference, provided the temperature does not change.

Ans: Parallel combination: R = (12 × 6)/(12 + 6) = 72/18 = 4 Ω. Total resistance = 4 + 4 = 8 Ω. Main current I = V/R = 12/8 = 1.5 A. Potential difference across the parallel part = 1.5 × 4 = 6 V. Current in 12 Ω = 6/12 = 0.5 A; current in 6 Ω = 6/6 = 1.0 A (total 1.5 A).