Skip to content
Login Register

#resistor calculation

8 public questions tagged with this topic.

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

View Details
Potentiometer, Conductivity and Special Cases

A Wheatstone bridge has \( R_1 = 26 \, \Omega \), \( R_2 = 52 \, \Omega \), \( R_3 = 20 \, \Omega \). What is \( R_4 \)

**Potentiometer** measures potential difference without drawing current, using null deflection, principle V ∝ l, l balance length, accurate because no I r drop. Potential drop across resistor V = I R arises because electric field does work on charges, energy converted to heat, maintaining E = -dV/dx along wire. Balance condition: (R₁/R₂) = (R₃/R₄) . Substitute: (26/52) = (20/R₄) . Solve: 0.5 = (20/R₄) ⇒ R₄ = (20/0.5) = 40 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 40 Ω,

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

View Details
Temperature Dependence of Resistance and Resistivity

A \( 10 \, \text{V} \) battery with \( 2 \, \Omega \) internal resistance delivers a current of \( 2 \, \text{A} \) to a

**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. Terminal voltage: V = ε - I r = 10 - 2 × 2 = 6 V . Resistance: R = (V/I) = (6/2) = 3 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 3.0 Ω,

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

View Details
Electrical Power, Energy and Heating Effect

A \( 24 \, \text{V} \) battery with negligible internal resistance is connected to a \( 4 \, \Omega \) and \( 8 \, \Omeg

**Heating effect** depends on I² R t, explaining why high currents cause significant heating, need for thick wires, fuses. Energy supplied by battery ε I t = I²(R+r)t, split between external and internal as per resistances. Total resistance: R = 4 + 8 = 12 Ω . Current: I = (V/R) = (24/12) = 2 A . Power: P = I² R = 2² × 4 = 16 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 = I²R, evaluation yields 16 W,

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

View Details
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

View Details
Wheatstone Bridge and Meter Bridge

A Wheatstone bridge has \( R_1 = 16 \, \Omega \), \( R_2 = 32 \, \Omega \), \( R_3 = 20 \, \Omega \). 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: (16/32) = (20/R₄) . Solve: 0.5 = (20/R₄) ⇒ R₄ = (20/0.5) = 40 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 40 Ω,

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

View Details
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

View Details
EMF, Internal Resistance and Cells Combination

A \( 16 \, \text{V} \) 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 = 16 - 2 × 2 = 12 V . Resistance: R = (V/I) = (12/2) = 6 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

View Details