#electric circuit

Applications of EM Waves in Communication and Medicine

A capacitor in a circuit has a conduction current of \( 1.8 \, \text{A} \) in the wires. What is the displacement curren

**Microwaves in ovens** cause water molecules to rotate at 2.45 GHz, friction heats food, penetration depth few cm, efficient heating, also radar uses reflection of microwaves from objects, Doppler shift measures speed, medical diathermy uses microwaves for tissue heating. The document states that in a charging capacitor, the displacement current between the plates equals the conduction current in the wires, so i_d = 1.8 A . Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields 1.8 A, illustrating EM wave transverse nature and Maxwell's displacement current concept.

Ref: NCERT > Physics Book > Electromagnetic Waves > Applications of EM Waves in Communication and Medicine

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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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Wheatstone Bridge and Meter Bridge

A \( 21 \, \text{V} \) battery with negligible internal resistance is connected to a cubical network of 12 resistors, ea

**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. Equivalent resistance: Rₑq = (5/6) R = (5/6) × 2.5 = (12.5/6) ≈ 2.08 Ω . Total current: I = (V/Rₑq) = (21/(12.5/6)) = 21 × (6/12.5) = 10.08 A ≈ 10.1 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 10.1 A,

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

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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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EMF, Internal Resistance and Cells Combination

A \( 10 \, \text{V} \) battery with negligible internal resistance is connected to a cubical network of 12 resistors, ea

**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. Equivalent resistance: Rₑq = (5/6) R = (5/6) × 1.5 = 1.25 Ω . Total current: Itₒtₐl = (V/Rₑq) = (10/1.25) = 8 A . Corner current: I = (Itₒtₐl/3) = (8/3) ≈ 2.67 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 =

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

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