Board Previous Year Questions (PYQs)

Arranged by marks. Show the formula, substitute, and state units. Diagrams (circuits, V-I graph) earn marks.

1-Mark Questions

Q1. State the SI unit of electric charge. (Ans: coulomb, C.)

Q2. Name the physical quantity whose unit is volt. (Ans: potential difference.)

Q3. How is an ammeter connected in a circuit? (Ans: in series.)

Q4. State the SI unit of resistivity. (Ans: ohm metre, Ω m.)

Q5. Name the commercial unit of electrical energy. (Ans: kilowatt-hour, kW h.)

2-Mark Questions

Q6. State Ohm's law and write its mathematical form. Ans: At constant temperature, the potential difference across a conductor is directly proportional to the current through it: V∝IV \propto I, i.e. V=IRV = IR, where RR is the resistance.

Q7. On what factors does the resistance of a conductor depend? Ans: (i) length ll (R ∝ l), (ii) area of cross-section AA (R ∝ 1/A), and (iii) nature of the material (resistivity ρ). Combined: R=ρl/AR = \rho l/A.

Q8. Why are coils of electric toasters and irons made of an alloy rather than a pure metal? Ans: Alloys have higher resistivity (more heat) and do not oxidise (burn) readily at high temperature, so they last longer in heating devices.

3-Mark Questions

Q9. A bulb of resistance 1200 Ω and a heater of 100 Ω are connected in turn to a 220 V supply. Find the current in each. (NCERT) Ans: Bulb: I=220/1200=0.18 AI = 220/1200 = 0.18\ \text{A}. Heater: I=220/100=2.2 AI = 220/100 = 2.2\ \text{A}.

Q10. Three resistors 5 Ω, 10 Ω and 30 Ω are connected in parallel to a 12 V battery. Find the total current and equivalent resistance. (NCERT) Ans: I1=2.4I_1 = 2.4, I2=1.2I_2 = 1.2, I3=0.4I_3 = 0.4 A; total I=4 AI = 4\ \text{A}; 1/Rp=1/5+1/10+1/30=1/31/R_p = 1/5+1/10+1/30 = 1/3, so Rp=3 ΩR_p = 3\ \Omega.

Q11. Why is the series arrangement not used for domestic circuits? Give two reasons. Ans: (i) In series the same current flows, but appliances need different currents to work properly. (ii) If one component fails, the whole circuit breaks and every appliance stops. Parallel wiring avoids both problems and gives each appliance the full voltage.

5-Mark Questions

Q12. (a) Define electric power and write three expressions for it. (b) An electric iron of resistance 20 Ω takes 5 A. Find the heat developed in 30 s. (c) State Joule's law of heating. Ans: (a) Power is the rate of consumption of electrical energy: P=VI=I2R=V2/RP = VI = I^2R = V^2/R. (b) H=I2Rt=25×20×30=15000 JH = I^2Rt = 25 \times 20 \times 30 = 15000\ \text{J}. (c) Joule's law: H=I2RtH = I^2Rt - heat is proportional to I2I^2, to RR, and to tt.

Q13. (a) Draw a labelled circuit to study Ohm's law. (b) A 6 V battery is connected to a 20 Ω lamp in series with a 4 Ω conductor. Find the total resistance, the current, and the voltage across each. (NCERT) Ans: (a) Circuit: cell(s), rheostat, ammeter (series), voltmeter (parallel across the wire), plug key. (b) Rs=24 ΩR_s = 24\ \Omega; I=6/24=0.25 AI = 6/24 = 0.25\ \text{A}; V1=5 VV_1 = 5\ \text{V} (lamp), V2=1 VV_2 = 1\ \text{V} (conductor).

Q14. Explain: (a) why tungsten is used for bulb filaments, (b) why copper and aluminium are used for transmission lines, (c) how the resistance of a wire varies with its area of cross-section. Ans: (a) Tungsten has a very high melting point (3380°C), so the filament glows white-hot without melting. (b) Copper and aluminium have very low resistivity, so they conduct with little energy loss. (c) Resistance is inversely proportional to the area of cross-section (R∝1/AR \propto 1/A) - a thicker wire has less resistance.