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#parallel circuit

21 public questions tagged with this topic.

Sharing of Charges, Common Potential and Applications

When two identical capacitors, one charged and one uncharged, are connected in parallel, why does the total energy decre

**Applications** include charge sharing for voltage division, Van de Graaff generator accumulates charge on spherical dome to high potential V = k Q/R, up to MV, using belt to transport charge, capacitance of sphere C=4π ε₀ R ≈10 pF for R=0.1 m, so Q= C V ≈10⁻⁸ C for 1000 V. When a charged capacitor ( C , charge Q , voltage V ) is connected in parallel with an uncharged capacitor ( C ), the total capacitance becomes 2C , and the charge redistributes to a final voltage V' = Q/(2C) = V/2 . Initial energy is U_i = (Q²/2C) , while final energy

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Sharing of Charges, Common Potential and Applications

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Sharing of Charges, Common Potential and Applications

Three capacitors \( 10 \, \text{pF} \), \( 20 \, \text{pF} \), and \( 40 \, \text{pF} \) are in parallel. What is the to

**Common potential** after connection is weighted average of initial potentials by capacitances. Energy loss ΔU = ½ C₁ C₂ (V₁-V₂)²/(C₁+C₂) dissipated as heat and spark, always positive unless V₁=V₂, explaining why energy reduces after sharing. C = 10 + 20 + 40 = 70 pF . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq series, C_eq = ΣC parallel, U = ½ C V² and common potential V = Q_total/C_total, result 70 pF follows, reflecting potential-capacitance relations.

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Sharing of Charges, Common Potential and Applications

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Energy Stored in Capacitor and Energy Density

In an isolated system of two capacitors initially charged and then connected in parallel with opposite polarities, what

**Energy density** in electric field u = ½ ε E², ε = K ε₀, E = V/d, total energy U = u·volume = ½ ε E²·A d =½ ε A d·(V/d)²=½ ε A V²/d=½ C V², consistent. For parallel plate, E = V/d ≈10⁶ V/m for 400 V across 0.4 mm, u≈½×8.85×10⁻¹²×10¹²≈4.4 J/m³. When two capacitors are connected in parallel with opposite polarities, charges redistribute such that the net charge on the positive plates (and negative plates) adjusts to a new equilibrium. The initial energy stored in the capacitors ( U_i = (1/2) C₁ V₁² + (1/2) C₂ V₂² ) is greater than the final

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Energy Stored in Capacitor and Energy Density

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Wheatstone Bridge and Meter Bridge

Why does the equivalent resistance of two resistors in parallel always lie between zero and the smallest individual resi

**Meter bridge** uses uniform wire of length 1 m, balance length l gives R_unknown = R_known·l/(100-l). Principle same as Wheatstone, with wire resistances proportional to lengths, allowing unknown resistance determination from length ratio. For parallel resistors, 1/Rₑq = 1/R₁ + 1/R₂ , so Rₑq = R₁ R₂ / (R₁ + R₂) . This value is less than the smaller resistance (e.g., if R₁ < R₂ , Rₑq < R₁ ) but greater than zero, as additional paths reduce resistance. 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

Ref: NCERT > Physics Book > Current Electricity > Wheatstone Bridge and Meter Bridge

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Temperature Dependence of Resistance and Resistivity

Two cells in parallel have emf \( 15 \, \text{V} \) and \( 5 \, \text{V} \) with internal resistances \( 3 \, \Omega \)

**Resistivity temperature variation** ρ_t = ρ₀[1+α(T-T₀)], α ≈4×10⁻³ /°C for copper, 1.7×10⁻⁴ /°C for nichrome. Given R=60 Ω at 30°C, α=1.7×10⁻⁴ /°C, T=330°C, ΔT=300°C, R_t=60[1+1.7×10⁻⁴×300]=60×1.051=63.06 Ω, modest increase for nichrome due to small α. For parallel: (1/rₑq) = (1/r₁) + (1/r₂) = (1/3) + (1/1) = (1 + 3/3) = (4/3) . rₑq = (3/4) = 0.75 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 0.75 Ω,

Ref: NCERT > Physics Book > Current Electricity > Temperature Dependence of Resistance and Resistivity

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Kirchhoff's Laws and Combination of Resistors

A circuit has a \( 8 \, \text{V} \) battery with \( 2 \, \Omega \) internal resistance and two resistors \( 4 \, \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) = (1/4) + (1/4) = (2/4) = 0.5 ⇒ R_p = 2 Ω . Total resistance: Rtₒtₐl = 2 + 2 = 4 Ω . Current: I = (ε/Rtₒtₐl) = (8/4) = 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

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

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Temperature Dependence of Resistance and Resistivity

In a circuit with parallel branches, why does the branch with the lowest resistance carry the highest current?

**Temperature dependence** of resistance R_t = R₀[1+α(T-T₀)], α temperature coefficient (per °C), R₀ resistance at T₀ (Ω). For metals α positive ≈10⁻³ /°C, resistance increases with temperature because τ decreases due to increased phonon scattering, n nearly constant. In parallel, voltage across each branch is the same ( V ). Current I = V / R , so the branch with the lowest R has the highest I , as current is inversely proportional to resistance. 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,

Ref: NCERT > Physics Book > Current Electricity > Temperature Dependence of Resistance and Resistivity

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Kirchhoff's Laws and Combination of Resistors

A circuit has a \( 10 \, \text{V} \) battery with \( 2 \, \Omega \) internal resistance and two resistors \( 4 \, \Omega

**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) = (1/4) + (1/6) = (3 + 2/12) = (5/12) ⇒ R_p = (12/5) = 2.4 Ω . Total resistance: Rtₒtₐl = 2 + 2.4 = 4.4 Ω . Current: I = (ε/Rtₒtₐl) = (10/4.4) ≈ 2.27 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

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 \( 10 \, \text{V} \) battery with \( 2 \, \Omega \) internal resistance is connected to two resistors \( 4 \, \Omega \

**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. Parallel resistance: (1/R_p) = (1/4) + (1/6) = (3 + 2/12) = (5/12) ⇒ R_p = (12/5) = 2.4 Ω . Total resistance: Rtₒtₐl = 2 + 2.4 = 4.4 Ω . Total current: I = (10/4.4) ≈ 2.27 A . Voltage across parallel: V = I R_p = 2.27 × 2.4 ≈ 5.45 V . Current through 6 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 \( 24 \, \text{V} \) battery with \( 3 \, \Omega \) internal resistance and three resistors \( 6 \, \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) = (1/6) + (1/12) + (1/24) = (4 + 2 + 1/24) = (7/24) ⇒ R_p = (24/7) ≈ 3.43 Ω . Total resistance: Rtₒtₐl = 3 + 3.43 = 6.43 Ω . Total current: I = (ε/Rtₒtₐl) = (24/6.43) ≈ 3.73 A . Voltage across parallel: V = I R_p = 3.73 × 3.43 ≈ 12.79 V . Current through 12 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 \( 12 \, \text{V} \) 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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