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22 public questions tagged with this topic.

Combination of Capacitors - Series and Parallel

Why does the total energy stored in a system of capacitors connected in series decrease if one of the capacitors is remo

**Charge conservation in series** explains same Q: when battery charges first plate, it induces -Q on second plate of same capacitor, which comes from next capacitor's plate, etc., so all have same magnitude Q, potential divides as per 1/C. In series, the equivalent capacitance Cₑq decreases when a capacitor is removed ( (1/Cₑq) = sum (1/C_i) ). With a battery maintaining constant voltage V , the energy stored is U = (1/2) Cₑq V² . A smaller Cₑq reduces U , as energy is directly proportional to capacitance under constant V . Additionally, removing a capacitor redistributes charges, potentially

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Combination of Capacitors - Series and Parallel

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Potentiometer, Conductivity and Special Cases

A \( 15 \, \text{V} \) battery with \( 1 \, \Omega \) internal resistance delivers a current of \( 2.5 \, \text{A} \) to

**Potential difference drop** across resistor when current flows because charges lose potential energy qV = I² R t as heat, field E = V/l drives drift, maintaining current. At very high E, velocity saturation or heating changes τ, causing non-ohmic behaviour. Terminal voltage: V = ε - I r = 15 - 2.5 × 1 = 12.5 V . Resistance: R = (V/I) = (12.5/2.5) = 5 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 5.0 Ω,

Ref: NCERT > Physics Book > Current Electricity > Potentiometer, Conductivity and Special Cases

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Potentiometer, Conductivity and Special Cases

A circuit has a \( 10 \, \text{V} \) battery with \( 1 \, \Omega \) internal resistance and two resistors \( 5 \, \Omega

**Conductivity** σ = 1/ρ = n e² τ/m (S/m), τ relaxation time, m electron mass. Resistivity deviation at high fields occurs when τ depends on E or n changes due to impact ionization, breaking Ohm's law, seen in varistors, gas discharge. Parallel resistance: (1/R_p) = (1/5) + (1/10) = (2 + 1/10) = (3/10) ⇒ R_p = (10/3) ≈ 3.33 Ω . Total resistance: Rtₒtₐl = 1 + 3.33 = 4.33 Ω . Current: I = (ε/Rtₒtₐl) = (10/4.33) ≈ 2.31 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

Ref: NCERT > Physics Book > Current Electricity > Potentiometer, Conductivity and Special Cases

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Potentiometer, Conductivity and Special Cases

In a circuit with a battery, why does the internal resistance of the battery affect the maximum power delivered to an ex

**Potential difference drop** across resistor when current flows because charges lose potential energy qV = I² R t as heat, field E = V/l drives drift, maintaining current. At very high E, velocity saturation or heating changes τ, causing non-ohmic behaviour. Maximum power transfer occurs when the external resistance equals the internal resistance ( R = r ), as P = I² R = (ε / (R + r))² R . Internal resistance limits current, influencing the power distribution. 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 > Potentiometer, Conductivity and Special Cases

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Wheatstone Bridge and Meter Bridge

A \( 12 \, \text{V} \) battery with \( 2 \, \Omega \) internal resistance delivers a current of \( 3 \, \text{A} \) to a

**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. Terminal voltage: V = ε - I r = 12 - 3 × 2 = 6 V . Resistance: R = (V/I) = (6/3) = 2 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 2.0 Ω,

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

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Temperature Dependence of Resistance and Resistivity

Why does the terminal voltage of a battery become equal to its emf when no current flows?

**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. Terminal voltage V = ε - I r . When I = 0 (open circuit), the internal voltage drop I r = 0 , so V = ε , matching the emf. 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 No internal voltage drop,

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

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Electrical Power, Energy and Heating Effect

A \( 9 \, \text{V} \) battery with \( 1 \, \Omega \) internal resistance is connected to a \( 8 \, \Omega \) resistor. W

**Power dissipation** in resistor converts electrical energy to heat, P = V²/R inversely proportional to R for fixed V, directly proportional for fixed I. For battery with internal r, power wasted internally = I² r, useful power = I² R, efficiency η = R/(R+r). Total resistance: Rtₒtₐl = 8 + 1 = 9 Ω . Current: I = (ε/Rtₒtₐl) = (9/9) = 1 A . Power: P = I² r = 1² × 1 = 1 W . 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

Ref: NCERT > Physics Book > Current Electricity > Electrical Power, Energy and Heating Effect

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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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Electrical Power, Energy and Heating Effect

In a battery-powered circuit, if the external load resistance becomes very large, what happens to the current drawn from

**Electrical power** P = V I = I² R = V²/R (W), energy E = P t = I² R t (J), heating effect Joule's law H = I² R t. When internal r equals external R, total resistance 2R, I = ε/2R, power in external = I²R = ε²/4R, total = ε²/2R, fraction external = 1/2, illustrating maximum power transfer when R = r. Current I = ε / (R + r) . As external resistance R becomes very large, R + r ≈ R , so I ≈ ε / R , approaching zero as R to ∞ . Applying I = n

Ref: NCERT > Physics Book > Current Electricity > Electrical Power, Energy and Heating Effect

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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 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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Electric Current, Drift Velocity and Mobility

A \( 12 \, \text{V} \) battery with \( 1 \, \Omega \) internal resistance delivers a current of \( 2 \, \text{A} \) to a

**Mobility** μ = v_d/E = e τ/m, τ relaxation time (s), measures ease of electron drift under field E (V/m). Conductivity σ = n e μ = 1/ρ, linking microscopic τ to macroscopic resistivity, explaining why metals conduct well due to large n and τ. Terminal voltage: V = ε - I r = 12 - 2 × 1 = 10 V . Resistance: R = (V/I) = (10/2) = 5 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 5.0

Ref: NCERT > Physics Book > Current Electricity > Electric Current, Drift Velocity and Mobility

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