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Current Electricity - Kirchhoff's Laws and Electrical Measurements

Grade 12ICSEPhysics

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

🔑Concepts

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Kirchhoff's First Law (Junction Rule): The algebraic sum of currents meeting at any junction in a circuit is zero. It is based on the law of conservation of charge: ∑I=0\sum I = 0.

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Kirchhoff's Second Law (Loop Rule): The algebraic sum of the e.m.f.s in any closed loop is equal to the algebraic sum of the products of current and resistance in that loop. It is based on the law of conservation of energy: ∑E=∑IR\sum E = \sum IR.

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Wheatstone Bridge: A circuit arrangement of four resistors P,Q,R,SP, Q, R, S used to measure an unknown resistance. At balance, the galvanometer current is zero (Ig=0I_g = 0).

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Meter Bridge: A practical application of the Wheatstone Bridge. It uses a uniform wire of length 100 cm100 \text{ cm}. The unknown resistance XX is calculated using the null point ll from one end.

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Potentiometer: A device used to measure EMF or potential difference without drawing current from the source. Its principle is that the potential drop across a uniform wire is directly proportional to its length (V∝lV \propto l) provided the current is constant.

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Potential Gradient (kk): The potential drop per unit length of the potentiometer wire, given by k=VLk = \frac{V}{L}.

📐Formulae

∑I=0\sum I = 0

∑E=∑IR\sum E = \sum IR

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

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

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

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

k=IρAk = \frac{I \rho}{A}

💡Examples

Problem 1:

In a Meter Bridge, the null point is found at a distance of 40 cm40 \text{ cm} from end AA when a resistor R=12 ΩR = 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=12 ΩR = 12 \, \Omega. Using the Meter Bridge formula: S=R(100−ll)S = R \left( \frac{100 - l}{l} \right). Substituting the values: S=12×(100−4040)=12×6040=12×1.5=18 ΩS = 12 \times \left( \frac{100 - 40}{40} \right) = 12 \times \frac{60}{40} = 12 \times 1.5 = 18 \, \Omega.

Explanation:

The unknown resistance SS is found by applying the balanced Wheatstone Bridge condition adapted for the lengths of the wire.

Problem 2:

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

Solution:

Total length L=4 mL = 4 \text{ m}, Resistance R=8 ΩR = 8 \, \Omega, Voltage V=2 VV = 2 \text{ V}. Potential gradient k=VLk = \frac{V}{L}. So, k=2 V4 m=0.5 V/mk = \frac{2 \text{ V}}{4 \text{ m}} = 0.5 \text{ V/m}.

Explanation:

Potential gradient is the potential drop per unit length of the potentiometer wire, which determines the sensitivity of the instrument.

Problem 3:

In a potentiometer experiment, the balancing length for a cell in open circuit is 350 cm350 \text{ cm}. When a resistor of 10 Ω10 \, \Omega is connected across the cell, the balancing length shifts to 300 cm300 \text{ cm}. Find the internal resistance rr of the cell.

Solution:

Given l1=350 cml_1 = 350 \text{ cm}, l2=300 cml_2 = 300 \text{ cm}, and R=10 ΩR = 10 \, \Omega. Using the formula r=R(l1l2−1)r = R \left( \frac{l_1}{l_2} - 1 \right): r=10(350300−1)=10(76−1)=10×16≈1.67 Ωr = 10 \left( \frac{350}{300} - 1 \right) = 10 \left( \frac{7}{6} - 1 \right) = 10 \times \frac{1}{6} \approx 1.67 \, \Omega.

Explanation:

The internal resistance is calculated by comparing the balancing lengths of the cell in an open circuit (EMF) and a closed circuit (Terminal Voltage).

Kirchhoff's Laws and Electrical Measurements Revision - Class 12 Physics ICSE