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#voltage source

3 public questions tagged with this topic.

Kirchhoff's Laws and Combination of Resistors

A circuit has a \( 12 \, \text{V} \) battery with \( 2 \, \Omega \) internal resistance and three resistors \( 3 \, \Ome

**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. Parallel resistance: (1/R_p) = (1/3) + (1/6) + (1/9) = (6 + 3 + 2/18) = (11/18) ⇒ R_p = (18/11) ≈ 1.64 Ω . Total resistance: Rtₒtₐl = 2 + 1.64 = 3.64 Ω . Total current: I = (ε/Rtₒtₐl) = (12/3.64) ≈ 3.3 A . Voltage across parallel: V = I R_p = 3.3 × 1.64 ≈ 5.41 V . Current through 9 Ω :

Ref: NCERT > Physics Book > Current Electricity > Kirchhoff's Laws and Combination of Resistors

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Kirchhoff's Laws and Combination of Resistors

A circuit has a \( 6 \, \text{V} \) battery with \( 1 \, \Omega \) internal resistance and two resistors \( 2 \, \Omega

**Total voltage drop** across series equals source voltage because loop rule Σ V = ε, with internal resistance r included V = ε - I r. For 4 Ω,8 Ω,16 Ω parallel, 1/R_p =1/4+1/8+1/16=7/16, R_p=16/7≈2.29 Ω, then total resistance with internal 3 Ω is 5.29 Ω, current I=18/5.29≈3.4 A. Parallel resistance: (1/R_p) = (1/2) + (1/3) = (3 + 2/6) = (5/6) ⇒ R_p = (6/5) = 1.2 Ω . Total resistance: Rtₒtₐl = 1 + 1.2 = 2.2 Ω . Current: I = (ε/Rtₒtₐl) = (6/2.2) ≈ 2.73 A . Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT],

Ref: NCERT > Physics Book > Current Electricity > Kirchhoff's Laws and Combination of Resistors

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EMF, Internal Resistance and Cells Combination

A \( 15 \, \text{V} \) battery with negligible internal resistance is connected to a \( 3 \, \Omega \) and \( 9 \, \Omeg

**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. Total resistance: R = 3 + 9 = 12 Ω . Current: I = (V/R) = (15/12) = 1.25 A . Power: P = I² R = (1.25)² × 9 = 1.5625 × 9 = 14.06 W ≈ 14 W . Applying I = n e A v_d, R

Ref: NCERT > Physics Book > Current Electricity > EMF, Internal Resistance and Cells Combination

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