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Current Electricity - Wheatstone Bridge

Grade 12CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A Wheatstone Bridge is an arrangement of four resistors P,Q,R,P, Q, R, and SS connected to form a quadrilateral, with a galvanometer GG connected across one diagonal and a battery across the other.

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The bridge is said to be 'balanced' when the potential difference across the galvanometer is zero, meaning no current flows through it (Ig=0I_g = 0).

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In the balanced condition, the ratio of the resistances in the adjacent arms is equal: \frac{P}{Q} = rac{R}{S}.

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The sensitivity of the Wheatstone bridge is maximum when all four resistances P,Q,R,P, Q, R, and SS are of the same order of magnitude.

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The Meter Bridge is a practical application of the Wheatstone Bridge used to find an unknown resistance. It consists of a wire of length 100 cm100\text{ cm} with uniform cross-sectional area.

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In a Meter Bridge, if the null point is obtained at a distance ll from the zero end, the unknown resistance SS is calculated using the known resistance RR in the other gap.

📐Formulae

PQ=RS\frac{P}{Q} = \frac{R}{S}

Ig=0 (Condition for Balance)I_g = 0 \text{ (Condition for Balance)}

S=(100−ll)RS = \left( \frac{100 - l}{l} \right) R

ρ=S⋅πr2L (Resistivity calculation using Meter Bridge where S is resistance, r is radius, and L is length of wire)\rho = \frac{S \cdot \pi r^2}{L} \text{ (Resistivity calculation using Meter Bridge where } S \text{ is resistance, } r \text{ is radius, and } L \text{ is length of wire)}

💡Examples

Problem 1:

In a Wheatstone bridge, the four arms have resistances P=100 ΩP = 100\, \Omega, Q=10 ΩQ = 10\, \Omega, R=50 ΩR = 50\, \Omega, and SS is unknown. If the bridge is balanced, calculate the value of SS.

Solution:

Given: P=100 ΩP = 100\, \Omega, Q=10 ΩQ = 10\, \Omega, R=50 ΩR = 50\, \Omega. Using the balance condition PQ=RS\frac{P}{Q} = \frac{R}{S}, we have: 10010=50S  ⟹  10=50S  ⟹  S=5010=5 Ω\frac{100}{10} = \frac{50}{S} \implies 10 = \frac{50}{S} \implies S = \frac{50}{10} = 5\, \Omega.

Explanation:

The balanced Wheatstone bridge condition states that the product of opposite arm resistances is equal, or the ratios of adjacent arms are equal. Here, we solve for the unknown arm SS by substituting the known values into the ratio formula.

Problem 2:

In a Meter Bridge experiment, the null point is found at a distance of 40 cm40\text{ cm} from end AA when a resistance R=12 ΩR = 12\, \Omega is connected in the left gap. Find the value of the unknown resistance SS in the right gap.

Solution:

Given: l=40 cml = 40\text{ cm}, R=12 ΩR = 12\, \Omega. The length of the remaining wire is (100−l)=100−40=60 cm(100 - l) = 100 - 40 = 60\text{ cm}. Using the formula S=100−ll×RS = \frac{100 - l}{l} \times R: S=6040×12=32×12=18 ΩS = \frac{60}{40} \times 12 = \frac{3}{2} \times 12 = 18\, \Omega.

Explanation:

The Meter Bridge works on the principle of the Wheatstone bridge. The ratio of the resistances in the two gaps equals the ratio of the lengths of the two segments of the wire at the balance point.

Wheatstone Bridge - Revision Notes & Key Formulas | CBSE Class 12 Physics