Current Electricity

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Kirchhoff's First Law states that the algebraic sum of currents at any junction in an electrical network is zero: ∑Ii = 0. Sign convention: Currents arriving at junction are positive, currents leaving

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Kirchhoff's Voltage Law states that the algebraic sum of potential differences and emfs in any closed loop is zero: ∑IR + ∑ε = 0. Sign conventions: (1) For resistors: Potential drop is negative when t

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Given: Currents entering = 8A, 3A; Currents leaving = 5A, X A Applying Kirchhoff's Current Law: ∑I = 0 Current entering - Current leaving = 0 (8 + 3) - (5 + X) = 0 11 - 5 - X = 0 6 - X = 0 Therefore,

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Condition: P/Q = S/R or PS = QR Derivation: For balanced bridge, galvanometer current Ig = 0 From Kirchhoff's laws: Loop ABDA: -I₁P + I₂S = 0 → I₁P = I₂S ... (1) Loop BCDB: -I₁Q + I₂R = 0 → I₁Q = I₂R

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Given: P = 100Ω, Q = 200Ω, R = 150Ω For balance condition: P/Q = S/R 100/200 = S/150 1/2 = S/150 S = 150/2 = 75Ω Verification: P×R = 100×150 = 15000, Q×S = 200×75 = 15000 ✓ Therefore, S = 75Ω for bala

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Principle: Wheatstone bridge using uniform wire of 1m length. Unknown resistance X in left gap, known resistance R in right gap. Derivation: For balance, X/R = RAD/RDB For uniform wire: RAD = ρℓx/A, R

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Given: Balance point at 40cm from left, R = 6Ω ℓx = 40cm, ℓR = 100-40 = 60cm Using metre bridge formula: X = R × (ℓx/ℓR) X = 6 × (40/60) = 6 × (2/3) = 4Ω Verification: X/R = 4/6 = 2/3, ℓx/ℓR = 40/60 =

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Principle: Potential difference between two points on a uniform wire carrying steady current is directly proportional to the length between those points. Mathematically: VAC = K×ℓ, where K is potentia