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The questions your board exam loves to ask

800 most-asked Class 11 & 12 (+1 / +2) questions across Physics, Chemistry, Maths and Biology — each with a model answer and the exact marking-scheme points examiners reward. Revise smart, walk in calm.

800 board questions•Model answers•Marking-scheme points•Every chapter•CBSE · ISC · State boards

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PhysicsClass 122 marksmedium

Current Electricity

State Kirchhoff's two laws for electrical circuits.

Reveal model answer + marking points▾

Kirchhoff's junction (current) law states that the algebraic sum of currents meeting at a junction is zero, that is, the total current entering a junction equals the total current leaving it; it is based on conservation of charge. Kirchhoff's loop (voltage) law states that the algebraic sum of the changes in potential around any closed loop of a circuit is zero; it is based on conservation of energy.

sum(I) at junction = 0; sum(V) around loop = 0

Marking-scheme points

  • ✓Junction law: sum of currents at a junction = 0 (charge conservation)
  • ✓Loop law: sum of potential changes around a loop = 0 (energy conservation)
  • ✓Used to analyse complex circuits
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PhysicsClass 123 marksmedium

Current Electricity

Draw (in words) and derive the balance condition of a Wheatstone bridge.

Reveal model answer + marking points▾

A Wheatstone bridge has four resistances P, Q, R and S arranged in a quadrilateral, with a galvanometer between the junctions of P-Q and R-S, and a cell across the other diagonal. When the bridge is balanced, no current flows through the galvanometer, so the junction connected by the galvanometer are at the same potential. Applying Kirchhoff's laws, the potential drops give P/Q = R/S. This is the balance condition; if three resistances are known, the fourth can be found.

P/Q = R/S

Marking-scheme points

  • ✓Four resistors P, Q, R, S with galvanometer across one diagonal
  • ✓At balance, no current through galvanometer
  • ✓Balance condition: P/Q = R/S
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PhysicsClass 123 marksmedium

Current Electricity

A cell of emf 2 V and internal resistance 0.5 ohm is connected to an external resistance of 4.5 ohm. Find the current in the circuit and the terminal potential difference.

Reveal model answer + marking points▾

Given emf E = 2 V, internal resistance r = 0.5 ohm, external resistance R = 4.5 ohm. Current I = E/(R + r) = 2/(4.5 + 0.5) = 2/5 = 0.4 A. Terminal potential difference V = I R = 0.4 x 4.5 = 1.8 V (equivalently V = E - I r = 2 - 0.4 x 0.5 = 1.8 V).

I = E/(R + r); V = E - I r

Marking-scheme points

  • ✓I = E/(R + r) = 2/5 = 0.4 A
  • ✓Terminal V = I R = 0.4 x 4.5 = 1.8 V
  • ✓Check: V = E - I r = 1.8 V
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