By Fatskills Exam Guides Team — the exam nerds behind 28,500+ quizzes and 2.1M practice questions across 500+ global exams.
Kirchhoff’s first law (junction rule) states that the algebraic sum of currents at a junction is zero; based on conservation of charge. Example: If 3 A enters a junction and splits into two branches, the sum of outgoing currents must be 3 A.
Kirchhoff’s second law (loop rule) states that the algebraic sum of potential differences in any closed loop is zero; based on conservation of energy. Example: In a loop with a 6 V battery and two resistors, the total voltage drop across resistors equals 6 V.
In applying Kirchhoff’s laws, current entering a junction is taken as positive and leaving as negative (or vice versa), but consistency is mandatory.
The Wheatstone bridge is balanced when no current flows through the galvanometer, i.e., ( \frac{P}{Q} = \frac{R}{S} ), where P, Q, R, S are resistances in the four arms.
A meter bridge is a practical form of Wheatstone bridge; it uses a 1 m long uniform wire, so resistance is proportional to length: ( \frac{R}{S} = \frac{l}{100 - l} ), where ( l ) is the balancing length in cm.
In a meter bridge experiment, if R = 10 Ω and balancing length l = 40 cm, then unknown resistance S = ( \frac{(100 - 40)}{40} \times 10 = 15 \, \Omega ).
The material of the meter bridge wire is usually constantan or manganin due to low temperature coefficient of resistance.
The null point in a meter bridge is found by sliding the jockey on the wire until the galvanometer shows zero deflection.
End resistances (due to copper strips and connections) introduce errors in meter bridge; they are minimized by interchanging resistances and taking mean.
To reduce end error, the experiment is repeated with R and S swapped, and the corrected value is ( R = \sqrt{R_1 R_2} ), where ( R_1 ) and ( R_2 ) are two calculated values.
The sensitivity of a Wheatstone bridge increases when all four resistances are of comparable magnitude.
If the galvanometer and battery are interchanged in a balanced Wheatstone bridge, the balance condition remains unchanged.
Internal resistance of a cell can be determined using a meter bridge with the formula: ( r = R \left( \frac{l_1}{100 - l_1} - 1 \right) ), where ( l_1 ) is the balancing length in open circuit.
The potential gradient along the wire of a potentiometer is ( k = \frac{V}{L} ), where V is the voltage across the wire and L is its length in meters.
Kirchhoff’s laws are applicable to both AC and DC circuits, provided instantaneous values are used for AC.
The Wheatstone bridge cannot measure very low resistances accurately due to lead and contact resistances.
For high resistance measurement, Wheatstone bridge becomes insensitive; hence, other methods like leakage method are used.
In a balanced bridge, the power consumed in the galvanometer branch is zero because current through it is zero.
The meter bridge wire must be uniform in cross-section; otherwise, resistance will not be proportional to length.
Verify from NCERT: The standard resistance in meter bridge experiments is usually placed in the right gap.
Intermediate — Requires conceptual clarity in circuit analysis and application of proportionality in wire resistance, but numericals are formula-based and repetitive.
Trap: Assuming that the balancing length in meter bridge is always less than 50 cm. Avoid: Balancing length depends on the ratio of resistances; it can be less than, equal to, or greater than 50 cm.
Trap: Using total wire length as 100 m instead of 100 cm in meter bridge formulas. Avoid: The wire is 1 meter = 100 cm long; lengths in formulas are in cm, so ( l ) and ( 100 - l ) are in cm.
Trap: Thinking that Kirchhoff’s laws apply only to linear circuits. Avoid: Kirchhoff’s laws apply to any circuit (linear or nonlinear) as they are based on conservation laws, but solutions may be complex in nonlinear cases.
Q1. In a meter bridge, the balancing length for a resistance of 4 Ω in the left gap is 40 cm. What is the value of the resistance in the right gap? A) 6 Ω B) 4 Ω C) 2.67 Ω D) 10 Ω
Answer: A) 6 Ω Explanation: Using ( \frac{R}{S} = \frac{l}{100 - l} ), ( \frac{4}{S} = \frac{40}{60} \Rightarrow S = 6 \, \Omega ). Why others fail: Option C comes from incorrectly using ( \frac{60}{40} \times 4 ) instead of ( \frac{60}{40} \times 4 = 6 ).
Q2. Which law is the basis of the junction rule in Kirchhoff’s laws? A) Conservation of energy B) Conservation of charge C) Ohm’s law D) Faraday’s law
Answer: B) Conservation of charge Explanation: The junction rule (sum of currents = 0) is based on conservation of charge. Why others fail: Option A is tempting because loop rule is based on energy, but junction rule is about charge.
Q3. In a Wheatstone bridge, P = 2 Ω, Q = 3 Ω, R = 4 Ω. What should be the value of S for balance? A) 5 Ω B) 6 Ω C) 8 Ω D) 10 Ω
Answer: B) 6 Ω Explanation: For balance, ( \frac{P}{Q} = \frac{R}{S} \Rightarrow \frac{2}{3} = \frac{4}{S} \Rightarrow S = 6 \, \Omega ). Why others fail: Option C comes from cross-multiplying incorrectly as ( 2S = 4 \times 3 \Rightarrow S = 6 ), not 8.
Q4. In a meter bridge experiment, the null point is obtained at 60 cm. If the resistances in the gaps are interchanged, the new balancing length will be: A) 40 cm B) 50 cm C) 60 cm D) 100 cm
Answer: A) 40 cm Explanation: On interchanging, new length ( l' = 100 - l = 100 - 60 = 40 \, \text{cm} ). Why others fail: Students often assume the length remains same, but it shifts symmetrically.
Q5. A Wheatstone bridge is balanced with P = Q = R = S = 10 Ω. If the battery and galvanometer are swapped, the bridge will: A) Become unbalanced B) Remain balanced C) Show maximum current in galvanometer D) Depend on internal resistance
Answer: B) Remain balanced Explanation: Balance condition depends only on resistance ratios, not on position of battery or galvanometer. Why others fail: Option A is tempting due to confusion with circuit symmetry, but the condition is unchanged.
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