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#circuit calculation

6 public questions tagged with this topic.

Sharing of Charges, Common Potential and Applications

Three capacitors \( 2 \, \text{pF} \), \( 4 \, \text{pF} \), and \( 8 \, \text{pF} \) are in parallel. What is the total

**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. C = 2 + 4 + 8 = 14 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 14 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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Wheatstone Bridge and Meter Bridge

A Wheatstone bridge with \( R_1 = 7 \, \Omega \), \( R_2 = 14 \, \Omega \), \( R_3 = 21 \, \Omega \), \( R_4 = 42 \, \Om

**Unbalanced bridge** has potential difference between galvanometer nodes, current direction determined by which node higher potential, i.e., if R₁/R₂ > R₃/R₄, left node higher, current flows one way, else opposite. Galvanometer deflection indicates imbalance magnitude. Check balance: (R₁/R₂) = (7/14) = 0.5 , (R₃/R₄) = (21/42) = 0.5 . Bridge is balanced. Since balanced, I_g = 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, V = ε - I r and P = I²R, evaluation yields 0 A,

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

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Wheatstone Bridge and Meter Bridge

A Wheatstone bridge with \( R_1 = 8 \, \Omega \), \( R_2 = 16 \, \Omega \), \( R_3 = 12 \, \Omega \), \( R_4 = 24 \, \Om

**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. Check balance: (R₁/R₂) = (8/16) = 0.5 , (R₃/R₄) = (12/24) = 0.5 . Bridge is balanced. Since balanced, I_g = 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, V = ε - I r and P = I²R, evaluation yields 0 A,

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

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Electrical Power, Energy and Heating Effect

A \( 7 \, \text{V} \) battery with \( 1 \, \Omega \) internal resistance is connected to a \( 6 \, \Omega \) 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 = 6 + 1 = 7 Ω . Current: I = (ε/Rtₒtₐl) = (7/7) = 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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EMF, Internal Resistance and Cells Combination

A cell of emf \( 10 \, \text{V} \) and internal resistance \( 1 \, \Omega \) is connected to a \( 9 \, \Omega \) resisto

**Cells combination** series ε_eq = Σ ε_i, r_eq = Σ r_i, parallel for identical cells ε_eq = ε, r_eq = r/n, n number of cells. Maximum current when external R = r_eq, power transfer theorem, explaining why matching resistances maximizes power. Total resistance: Rtₒtₐl = 9 + 1 = 10 Ω . Current: I = (ε/Rtₒtₐl) = (10/10) = 1 A . Terminal voltage: V = ε - I r = 10 - 1 × 1 = 9 V . 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 > EMF, Internal Resistance and Cells Combination

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Wheatstone Bridge and Meter Bridge

A Wheatstone bridge with \( R_1 = 10 \, \Omega \), \( R_2 = 20 \, \Omega \), \( R_3 = 15 \, \Omega \), \( R_4 = 30 \, \O

**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. Check balance: (R₁/R₂) = (10/20) = 0.5 , (R₃/R₄) = (15/30) = 0.5 . Bridge is balanced. Since balanced, I_g = 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, V = ε - I r and P = I²R, evaluation yields 0 A,

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

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