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#resistor network

7 public questions tagged with this topic.

Kirchhoff's Laws and Combination of Resistors

A \( 30 \, \text{V} \) battery with negligible internal resistance is connected to a cubical network of 12 resistors, ea

**Kirchhoff's loop rule** (energy conservation) states algebraic sum of potential differences around closed loop zero, Σ ε - Σ I R =0, ensures total voltage drop across series resistors equals source emf. Junction rule (charge conservation) Σ I_in = Σ I_out ensures current continuity. Equivalent resistance: Rₑq = (5/6) R = (5/6) × 5 = (25/6) ≈ 4.17 Ω . Total current: I = (V/Rₑq) = (30/(25/6)) = 30 × (6/25) = 7.2 A . Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT], Kirchhoff's ΣI=0, ΣV=0, R_eq series/parallel, V = ε - I r and P = I²R, evaluation

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Kirchhoff's Laws and Combination of Resistors

A \( 36 \, \text{V} \) battery with negligible internal resistance is connected to a cubical network of 12 resistors, ea

**Series combination** R_eq = R₁+R₂+..., same current I through each, voltage divides proportionally V_i = I R_i. Parallel combination 1/R_p = 1/R₁+1/R₂+..., same voltage V across each, current divides inversely, equivalent R_p = (R₁ R₂)/(R₁+R₂) for two resistors. Equivalent resistance: Rₑq = (5/6) R = (5/6) × 6 = 5 Ω . Total current: I = (V/Rₑq) = (36/5) = 7.2 A . Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT], Kirchhoff's ΣI=0, ΣV=0, R_eq series/parallel, V = ε - I r and P = I²R, evaluation yields 7.2 A,

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Wheatstone Bridge and Meter Bridge

A Wheatstone bridge has \( R_1 = 30 \, \Omega \), \( R_2 = 60 \, \Omega \), \( R_3 = 25 \, \Omega \). What is \( R_4 \)

**Wheatstone bridge balance** condition R₁/R₂ = R₃/R₄, R₄ = R₂ R₃/R₁, when galvanometer current zero, potentials at midpoints equal. At balance, no current through galvanometer, enabling precise resistance measurement independent of source voltage. Balance condition: (R₁/R₂) = (R₃/R₄) . Substitute: (30/60) = (25/R₄) . Solve: 0.5 = (25/R₄) ⇒ R₄ = (25/0.5) = 50 Ω . Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT], Kirchhoff's ΣI=0, ΣV=0, R_eq series/parallel, V = ε - I r and P = I²R, evaluation yields 50 Ω,

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EMF, Internal Resistance and Cells Combination

A \( 10 \, \text{V} \) battery with negligible internal resistance is connected to a cubical network of 12 resistors, ea

**Cells combination** series ε_eq = Σ ε_i, r_eq = Σ r_i, parallel for identical cells ε_eq = ε, r_eq = r/n, n number of cells. Maximum current when external R = r_eq, power transfer theorem, explaining why matching resistances maximizes power. Equivalent resistance: Rₑq = (5/6) R = (5/6) × 1.5 = 1.25 Ω . Total current: Itₒtₐl = (V/Rₑq) = (10/1.25) = 8 A . Corner current: I = (Itₒtₐl/3) = (8/3) ≈ 2.67 A . Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT], Kirchhoff's ΣI=0, ΣV=0, R_eq series/parallel, V = ε - I r and P =

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EMF, Internal Resistance and Cells Combination

A \( 24 \, \text{V} \) battery with negligible internal resistance is connected to a cubical network of 12 resistors, ea

**Internal resistance** causes voltage drop I r inside battery, so V = ε - I r decreases with I. For 16 V battery, r=2 Ω, I=2 A, V=16-4=12 V, external R = V/I =6 Ω. Measurement of V and I yields r = (ε - V)/I. Equivalent resistance: Rₑq = (5/6) R = (5/6) × 4 = (20/6) = (10/3) ≈ 3.33 Ω . Total current: I = (V/Rₑq) = (24/(10/3)) = 24 × (3/10) = 7.2 A . Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT], Kirchhoff's ΣI=0, ΣV=0, R_eq series/parallel, V = ε - I r and

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EMF, Internal Resistance and Cells Combination

A circuit has a \( 16 \, \text{V} \) battery with \( 2 \, \Omega \) internal resistance and three resistors \( 2 \, \Ome

**EMF ε** is work done by non-electrostatic forces per unit charge, terminal voltage V = ε - I r, r internal resistance (Ω), I current (A). When external R = r, total resistance 2r, current I = ε/2r, power in external R is I²R = ε²/4r, total power ε²/2r, so half power dissipated externally, half internally. Parallel resistance: (1/R_p) = (1/2) + (1/4) + (1/8) = (4 + 2 + 1/8) = (7/8) ⇒ R_p = (8/7) ≈ 1.14 Ω . Total resistance: Rtₒtₐl = 2 + 1.14 = 3.14 Ω . Total current: I = (ε/Rtₒtₐl) = (16/3.14) ≈ 5.10 A . Voltage across

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EMF, Internal Resistance and Cells Combination

A circuit has a \( 18 \, \text{V} \) battery with \( 3 \, \Omega \) internal resistance and three resistors \( 4 \, \Ome

**Cells combination** series ε_eq = Σ ε_i, r_eq = Σ r_i, parallel for identical cells ε_eq = ε, r_eq = r/n, n number of cells. Maximum current when external R = r_eq, power transfer theorem, explaining why matching resistances maximizes power. Parallel resistance: (1/R_p) = (1/4) + (1/8) + (1/16) = (4 + 2 + 1/16) = (7/16) ⇒ R_p = (16/7) ≈ 2.29 Ω . Total resistance: Rtₒtₐl = 3 + 2.29 = 5.29 Ω . Total current: I = (ε/Rtₒtₐl) = (18/5.29) ≈ 3.4 A . Voltage across parallel: V = I R_p = 3.4 × 2.29 ≈ 7.79 V . Current through

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