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Current Electricity - Potentiometer and Wheatstone Bridge

Grade 12ICSEPhysics

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

🔑Concepts

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The Wheatstone Bridge is an arrangement of four resistances P,Q,R,SP, Q, R, S used to measure an unknown resistance. When the bridge is balanced, no current flows through the galvanometer (Ig=0I_g = 0), and the condition is PQ=RS\frac{P}{Q} = \frac{R}{S}.

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A Metre Bridge (or slide wire bridge) is a practical application of the Wheatstone Bridge. It uses a uniform wire of length 100 cm100\text{ cm}. The unknown resistance SS is calculated using the balance point ll from the formula S=(100−l)lRS = \frac{(100-l)}{l} R.

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The Potentiometer is an instrument used to measure the EMF or potential difference without drawing any current from the source. It works on the principle that the potential drop across any portion of a uniform wire is directly proportional to the length of that portion, provided a constant current flows through it: V∝lV \propto l.

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Potential Gradient (kk): It is defined as the potential drop per unit length of the potentiometer wire. It is given by k=VL=IρAk = \frac{V}{L} = \frac{I \rho}{A}, where ρ\rho is resistivity and AA is the area of cross-section.

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A potentiometer is more sensitive than a voltmeter because it uses the null-deflection method, meaning it does not draw current from the circuit being measured (Rin→∞R_{in} \to \infty).

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Sensitivity of a potentiometer can be increased by decreasing the potential gradient (kk). This can be done by increasing the length of the wire or decreasing the current in the primary circuit using a rheostat.

📐Formulae

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

V=klV = k l

k=VL=IRhLk = \frac{V}{L} = \frac{I R_h}{L}

E1E2=l1l2\frac{E_1}{E_2} = \frac{l_1}{l_2}

r=R(l1−l2l2)=R(l1l2−1)r = R \left( \frac{l_1 - l_2}{l_2} \right) = R \left( \frac{l_1}{l_2} - 1 \right)

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

💡Examples

Problem 1:

In a Metre Bridge, the balance point is found at a distance of 40 cm40\text{ cm} from end AA when a resistor R=12ΩR = 12 \Omega is in the left gap and an unknown resistor SS is in the right gap. Calculate the value of SS.

Solution:

Given l=40 cml = 40\text{ cm} and R=12ΩR = 12 \Omega. Using the Metre Bridge formula: S=(100−l)lRS = \frac{(100 - l)}{l} R. Substituting the values: S=(100−40)40×12=6040×12=1.5×12=18ΩS = \frac{(100 - 40)}{40} \times 12 = \frac{60}{40} \times 12 = 1.5 \times 12 = 18 \Omega.

Explanation:

The Metre Bridge works on the Wheatstone principle where the ratio of resistances equals the ratio of the lengths of the wire segments.

Problem 2:

A potentiometer wire has a length of 4 m4\text{ m} and a resistance of 8Ω8 \Omega. A cell of EMF 2 V2\text{ V} is connected across it. Calculate the potential gradient.

Solution:

Length L=4 mL = 4\text{ m}, Resistance Rw=8ΩR_w = 8 \Omega, V=2 VV = 2\text{ V}. Potential gradient k=VL=2 V4 m=0.5 V/mk = \frac{V}{L} = \frac{2\text{ V}}{4\text{ m}} = 0.5\text{ V/m}. To convert to V/cm\text{V/cm}: k=0.5100=0.005 V/cmk = \frac{0.5}{100} = 0.005\text{ V/cm}.

Explanation:

Potential gradient is the potential drop divided by the total length of the potentiometer wire.

Problem 3:

With a cell of EMF EE in the secondary circuit of a potentiometer, the balance point is 200 cm200\text{ cm}. When a resistor of 5Ω5 \Omega is connected across the cell, the balance point shifts to 150 cm150\text{ cm}. Find the internal resistance of the cell.

Solution:

Given l1=200 cml_1 = 200\text{ cm}, l2=150 cml_2 = 150\text{ cm}, and external resistance R=5ΩR = 5 \Omega. Internal resistance r=R(l1l2−1)r = R \left( \frac{l_1}{l_2} - 1 \right). r=5(200150−1)=5(43−1)=5×13≈1.67Ωr = 5 \left( \frac{200}{150} - 1 \right) = 5 \left( \frac{4}{3} - 1 \right) = 5 \times \frac{1}{3} \approx 1.67 \Omega.

Explanation:

The internal resistance is calculated by comparing the balancing length of the cell in open circuit (l1l_1) and closed circuit (l2l_2).

Potentiometer and Wheatstone Bridge Revision - Class 12 Physics ICSE