#RC circuit

Rectifiers, Filters and Applications

The time constant of a capacitor filter in a rectifier depends on:

**Rectifier** converts AC to DC using diode unilateral conduction, half-wave conducts only during positive half-cycle when forward biased, output pulsating DC with frequency equal to input 50 Hz, full-wave with centre-tap uses two diodes conducting alternate half-cycles, output frequency 100 Hz for 50 Hz input, 120 Hz for 60 Hz input, output voltage closer to peak with capacitor filter. The time constant ( tau = R_L C ) in a capacitor filter determines the discharge rate, depending inversely on the product of capacitance ( C ) and load resistance ( R_L ). Substituting values gives Capacitance

Ref: NCERT > Physics Book > Electronic Devices > Rectifiers, Filters and Applications

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AC Through Resistor - Phasor and Power

A \( 50 \, \Omega \) resistor and \( 20 \, \mu\text{F} \) capacitor are in series with a \( 110 \, \text{V} \), \( 50 \,

**AC through resistor** voltage and current in phase, φ=0°, I = V/R instantaneously, I(t)=I_peak sin ωt, V(t)=V_peak sin ωt, phasor diagram V and I same direction, power instantaneous P = V I = V_peak I_peak sin² ωt, average P_avg = V_rms I_rms = V_rms²/R, always positive, energy dissipated as heat. X_C = (1/ω C) = (1/314 × 20 × 10⁻⁶) ≈ 159.2 Ω . Z = √(R² + X_C²) = √(50² + 159.2²) ≈ 166.6 Ω . RMS current: I = (V/Z) = (110/166.6) ≈ 0.66 A . Voltage across resistor: V_R = I R = 0.66 × 50 ≈ 33 V . Applying X_L

Ref: NCERT > Physics Book > Alternating Currents > AC Through Resistor - Phasor and Power

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AC Through Capacitor - Capacitive Reactance

In an AC circuit with a series combination of resistor and capacitor, what determines whether the circuit is predominant

**AC through capacitor** voltage lags current by 90°, Q= C V, I= dQ/dt = C dV/dt, V(t)=V_peak sin ωt, I(t)=I_peak sin(ωt+90°), average power zero because energy stored in electric field ½ C V² returned each cycle, capacitor blocks DC but passes AC, X_C decreases with f. In an RC series circuit, the circuit is predominantly capacitive if the capacitive reactance ( X_C = (1/ω C) ) is greater than the resistance ( R ). This makes the impedance Z = √(R² + X_C²) dominated by X_C , and the current leads the voltage significantly. Applying X_L = ωL, X_C = 1/ωC, Z = √(R² +

Ref: NCERT > Physics Book > Alternating Currents > AC Through Capacitor - Capacitive Reactance

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AC Through Resistor - Phasor and Power

In an AC circuit with a capacitor and resistor in series, how does the phase difference between voltage and current chan

**Resistor in AC** behaves as DC, no reactance, impedance Z=R, current follows voltage exactly, average power over cycle V_rms I_rms, for 200 V rms, 80 Ω, P=500 W? Actually 200²/80=500 W, peak current √2×2.5=3.535 A, average power ½ V_peak I_peak. In an RC series circuit, the phase angle Φ = tan⁻¹ ( (X_C/R) ) , where X_C = (1/ω C) . Increasing capacitance decreases X_C , reducing the phase angle, meaning the current leads the voltage by a smaller angle. Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²), I_rms = V_rms/Z, P = V_rms I_rms cosφ, V_s/V_p = N_s/N_p,

Ref: NCERT > Physics Book > Alternating Currents > AC Through Resistor - Phasor and Power

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AC Through Resistor - Phasor and Power

In an AC circuit with a series combination of resistor and capacitor, what happens to the phase difference if the freque

**Current relative to voltage** in resistor in phase, φ=0°, power factor cos φ=1, maximum power, unlike inductor/capacitor where average power zero due to 90° phase shift, explaining why resistor heats while pure L/C does not. In an RC series circuit, Φ = tan⁻¹ ( (X_C/R) ) , where X_C = (1/ω C) . As frequency ( ω ) approaches infinity, X_C approaches zero, making Φ approach 0°, so the circuit becomes nearly resistive. Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²), I_rms = V_rms/Z, P = V_rms I_rms cosφ, V_s/V_p = N_s/N_p, calculation gives It approaches 0°, consistent with

Ref: NCERT > Physics Book > Alternating Currents > AC Through Resistor - Phasor and Power

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AC Through Resistor - Phasor and Power

A \( 60 \, \Omega \) resistor and \( 40 \, \mu\text{F} \) capacitor are in series with a \( 230 \, \text{V} \), \( 50 \,

**Resistor in AC** behaves as DC, no reactance, impedance Z=R, current follows voltage exactly, average power over cycle V_rms I_rms, for 200 V rms, 80 Ω, P=500 W? Actually 200²/80=500 W, peak current √2×2.5=3.535 A, average power ½ V_peak I_peak. X_C = (1/ω C) = (1/314 × 40 × 10⁻⁶) ≈ 79.6 Ω . Z = √(R² + X_C²) = √(60² + 79.6²) = √(3600 + 6336.16) ≈ 99.8 Ω . Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²), I_rms = V_rms/Z, P = V_rms I_rms cosφ, V_s/V_p = N_s/N_p, calculation gives 99.8 Ω, consistent with phasor analysis and resonance condition X_L = X_C.

Ref: NCERT > Physics Book > Alternating Currents > AC Through Resistor - Phasor and Power

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Energy Stored in Capacitor and Energy Density

An isolated charged capacitor is connected across a resistor momentarily. Why does the energy stored in the capacitor de

**Work done charging capacitor** is integral ∫ V dQ = ∫ Q/C dQ = Q²/2C, stored as electrostatic energy. When capacitor discharges, energy released as heat or work, explaining spark when shorted, energy proportional to V². When a charged capacitor is connected across a resistor, it discharges through the resistor, forming an RC circuit. The charge Q on the capacitor decreases exponentially ( Q(t) = Q₀ e⁻t/RC ). After a long time ( t to ∞ ), Q to 0 , so the energy U = (Q²/2C) to 0 . The energy is dissipated as heat in the resistor during discharge, leaving no energy stored

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Energy Stored in Capacitor and Energy Density

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