PhysicsCBSE Class 12 · Sample paper, session 2024-25 · 5 marks · Long answer

CBSE Class 12 Physics sample paper 2024-25: question 31(II)

From the official CBSE sample question paper for 2024-25, section E, question 31(II), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q31(II) · 5 marks · Long answerCurrent Electricity

a) Using Kirchhoff's laws obtain the equation of the balanced state in Wheatstone bridge.
b) A wire of uniform cross-section and resistance of 12 ohm is bent in the shape of circle as shown in the figure. A resistance of 10 ohms is connected to diametrically opposite ends C and D. A battery of emf 8V is connected between A and B. Determine the current flowing through arm AD.

Figure in the paper: Circle of wire with centre O; C at top and D at bottom, joined by a vertical diameter containing a 10 Ω resistor. A and B are points on the circle on left and right, each making 60° with the vertical diameter at O (angle AOD = 60°, BOD = 60°). An 8 V battery is connected between A and B. See it in the official paper.

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2024-25.

Hints

Open one hint at a time and try again before the next.

Hint 1

Write loop equations around the two meshes that contain the galvanometer, then set the galvanometer current to zero.

Hint 2

For the circle, resistance of each arc is proportional to the angle it subtends at O.

Hint 3

Check whether the four arcs form a balanced bridge; if so, the 10 Ω arm carries no current.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

(a) Circuit: four resistances P (AB), Q (BC), R (AD) and S (DC) form a quadrilateral. A cell is connected across A and C, and a galvanometer of resistance G across B and D. The current $I_{1}$ flows through P, $I_{2}$ through R, and $I_{g}$ through G from B to D.
1 mark
Kirchhoff's loop law for mesh ABDA: $I_{1} P + I_{g} G - I_{2} R = 0$ …(1)
½ mark
For mesh BCDB: $(I_{1} - I_{g}) Q - (I_{2} + I_{g}) S - I_{g} G = 0$ …(2)
½ mark
At balance $I_{g} = 0$ . Then (1) gives $I_{1} P = I_{2} R$ and (2) gives $I_{1} Q = I_{2} S$ . Dividing: $\frac{P}{Q} = \frac{R}{S}$ , the balance condition.
1 mark
(b) The wire is uniform, so resistance ∝ arc length; 12 Ω for 360°. Arc AD = arc DB = 60°, so each is $12 \times \frac{60}{360} = 2 Ω$ . Arc AC = arc CB = 120°, so each is $4 Ω$ . Since $\frac{R _{A C}}{R _{C B}} = \frac{4}{4} = \frac{R _{A D}}{R _{D B}} = \frac{2}{2}$ , the bridge is balanced and no current flows through the 10 Ω resistor in CD.
1 mark
The circuit reduces to two parallel branches between A and B: ACB = 8 Ω and ADB = 4 Ω, each with 8 V across it. Current through arm AD: $I = \frac{8}{4} = 2$ A.
1 mark

Answer: (a) $P / Q = R / S$ ; (b) current through AD = 2 A (no current in the 10 Ω arm)

Where the marks are

Labelled Wheatstone bridge circuit (1 mark)
Loop equation for ABDA (½ mark)
Loop equation for BCDB (½ mark)
Puts $I_{g} = 0$ and derives $P / Q = R / S$ (1 mark)
Arc resistances 2, 2, 4, 4 Ω and recognises a balanced bridge (1 mark)
Current in AD = 8/4 = 2 A (1 mark)

Revise the chapter

Chapter: Current Electricity (notes, formulas, where marks are lost and every past question)
Topic: Kirchhoff's Rules and Wheatstone Bridge

Practise this chapterBoard-style questions from Current Electricity, one at a time, marked against the answer key and marking points.Back to the full paperSample paper 2024-25: every question, by section.

The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.

CBSE Class 12 Physics sample paper 2024-25: question 31(II)

Hints

Step-by-step solution

Revise the chapter

More questions from Current Electricity