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#electric circuits

18 public questions tagged with this topic.

Combination of Capacitors - Series and Parallel

In a series combination of capacitors, why does each capacitor have the same charge but different potential differences?

**Series combination** of capacitors has same charge Q on each because connected end-to-end, single path for charge flow, induced charges equal, total voltage V = Σ V_i = Q Σ 1/C_i, so 1/C_eq = Σ 1/C_i. Different potential differences V_i = Q/C_i inversely proportional to C_i, smaller C gets larger V. In a series combination, capacitors are connected end-to-end, forming a single path for charge flow. When a voltage is applied, the same charge Q accumulates on each capacitor (as charge conservation ensures the same Q passes through each during charging). However, the potential difference across

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

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Potentiometer, Conductivity and Special Cases

What happens to the current in a conductor if its cross-sectional area is doubled while keeping the potential difference

**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. Resistance R = rho l / A . If A doubles, R becomes R/2 . Current I = V / R , so if R halves and V is constant, I doubles. 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 It

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

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Temperature Dependence of Resistance and Resistivity

In a circuit, a \( 20 \, \text{V} \) battery with negligible internal resistance is connected across a cubical network o

**Conductivity** σ=1/ρ decreases with temperature for metals, σ = n e² τ/m, τ ∝1/T due to lattice vibrations. For semiconductors, n increases exponentially with T, so σ increases, opposite to metals, explaining why metallic resistance rises with temperature. Equivalent resistance of cube network: Rₑq = (5/6) R = (5/6) × 2 = (10/6) = (5/3) Ω . Total current: I = (V/Rₑq) = (20/(5/3)) = 20 × (3/5) = 12 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 12 A,

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 \( 0.5 \, \Omega \) internal resistance is connected to a \( 8.5 \, \Omega \) resisto

**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 = 8.5 + 0.5 = 9 Ω . Current: I = (ε/Rtₒtₐl) = (9/9) = 1 A . Power: P = I² R = 1² × 8.5 = 8.5 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 \( 12 \, \text{V} \) battery with \( 3 \, \Omega \) internal resistance and two resistors \( 6 \, \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) = (1/6) + (1/3) = (1 + 2/6) = (3/6) = 0.5 ⇒ R_p = 2 Ω . Total resistance: Rtₒtₐl = 3 + 2 = 5 Ω . Current: I = (ε/Rtₒtₐl) = (12/5) = 2.4 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,

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 \( 9 \, \text{V} \) battery with \( 1 \, \Omega \) internal resistance and two resistors \( 5 \, \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) = (1/5) + (1/3) = (3 + 5/15) = (8/15) ⇒ R_p = (15/8) = 1.875 Ω . Total resistance: Rtₒtₐl = 1 + 1.875 = 2.875 Ω . Current: I = (ε/Rtₒtₐl) = (9/2.875) ≈ 3.13 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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Kirchhoff's Laws and Combination of Resistors

A circuit has a \( 14 \, \text{V} \) battery with \( 2 \, \Omega \) internal resistance and two resistors \( 8 \, \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) = (1/8) + (1/4) = (1 + 2/8) = (3/8) ⇒ R_p = (8/3) ≈ 2.67 Ω . Total resistance: Rtₒtₐl = 2 + 2.67 = 4.67 Ω . Current: I = (ε/Rtₒtₐl) = (14/4.67) ≈ 3.0 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,

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

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Resistance, Resistivity and Ohm's Law

A \( 27 \, \text{V} \) battery with negligible internal resistance is connected to a cubical network of 12 resistors, ea

**Resistance** R = ρ l/A, ρ resistivity (Ω·m), l length (m), A area (m²), ρ = m/(n e² τ) from Drude model, τ average collision time. Ohm's law V = I R holds when ρ constant, independent of V. Volume constant stretching l→2l implies A→A/2, so R' = ρ·2l/(A/2)=4R, resistance quadruples when length doubles at constant volume. Equivalent resistance: Rₑq = (5/6) R = (5/6) × 3 = (15/6) = 2.5 Ω . Total current: I = (V/Rₑq) = (27/2.5) = 10.8 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 =

Ref: NCERT > Physics Book > Current Electricity > Resistance, Resistivity and Ohm's Law

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EMF, Internal Resistance and Cells Combination

A \( 14 \, \text{V} \) battery with \( 2 \, \Omega \) internal resistance delivers a current of \( 2 \, \text{A} \) to a

**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. Terminal voltage: V = ε - I r = 14 - 2 × 2 = 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 =

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

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Electrical Power, Energy and Heating Effect

A \( 4 \, \Omega \) resistor dissipates \( 16 \, \text{W} \) of power. What is the voltage across it?

**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. Power: P = (V²/R) . Rearrange: V = √(P R) . Substitute: V = √(16 × 4) = √(64) = 8 V . Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT], Kirchhoff's ΣI=0, ΣV=0,

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

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PHYSICS

A 10 Ω resistor dissipates 40 W of power. What is the voltage across it?

Given: A 10 Ω resistor dissipates 40 W of power. What is the voltage across it? These values define the system as per NCERT data. Formula: Power: P = V²/R. This is standard NCERT relation. Substitution & Calculation: Rearrange: V = sqrtP R . Substitute: V = sqrt40 × 10 = sqrt400 = 20 V . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Relevant Physics topic covering fundamental principles,

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PHYSICS

Two capacitors of 15 pF each are connected in series. What is the equivalent capacitance?

Given: Two capacitors of 15 pF each are connected in series. What is the equivalent capacitance? These values define the system as per NCERT data. Formula: 1/C = 1/15 + 1/15 = 2/15. This is standard NCERT relation. Substitution & Calculation: C = 15/2 = 7.5 pF . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Electrostatic Potential and Capacitance, Topic: Capacitors in series, equivalent capacitance 1/C_eq = 1/C₁ + 1/C₂. The section explains definitions, governing laws, formulas like μ₀ = 4π × 10⁻⁷ T·m/A, units and illustrative examples.

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