--- title: "Full wave bridge rectifier load power" subject: "Electronics Devices and Circuit" chapter: 6 key: "semester_3/electronics_devices_and_circuit:bridge-rectifier-power-5ohm" source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/bridge-rectifier-power-5ohm site: BCT Engineering --- # Full wave bridge rectifier load power Electronics Devices and Circuit · Chapter 6: Power Supplies, Breakdown Diodes, and Voltage Regulators - Asked in: 2069 Chaitra, 4 marks - Video: [Full Wave Rectifiers](https://www.youtube.com/watch?v=pb5nzUBehyY) — The Organic Chemistry Tutor - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/bridge-rectifier-power-5ohm - Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-6 ## Question Draw full wave bridge rectifier circuit with 5 ohm load resistor connected at its output. If input ac voltage is 10V, calculate the power dissipation in the load resistor (Assume diodes operate at forward voltage of 0.7V). ## Short answer For 10 V rms input, peak is 14.14 V. Subtract two diode drops (1.4 V) giving peak load 12.74 V. RMS load voltage is 12.74/√2 ≈ 9.01 V. Power dissipation P = (9.01)²/5 Ω ≈ 16.2 W. ## Answer #### Circuit diagram ```text D1 D2 A o---+---|>|---+---|<|---+---o B | | | | (+) | | | | | [RL = 5 Ω] | | | | | (−) | | | | +---|<|---+---|>|---+ D3 D4 10 V ac input applied between A and B ``` Bridge: four diodes D1 to D4, with the ac input at A and B and RL between the (+) and (−) corners. During the half-cycle when A is positive, D1 and D4 conduct (A → D1 → RL → D4 → B). During the other half-cycle, D2 and D3 conduct (B → D2 → RL → D3 → A). In both half-cycles, the current flows through RL in the same direction, from (+) to (−). #### Calculation **Assumption:** the 10 V is the rms value of the ac input to the bridge. (If 10 V is the peak value, see the note below.) 1. Peak input voltage: Vm = 10 √2 = 14.14 V 2. Two diodes conduct in series in each half-cycle, so the drop is 2 × 0.7 = 1.4 V. Peak load voltage: VL(p) = 14.14 − 1.4 = 12.74 V 3. For a full-wave rectified sine wave, the rms load voltage is VL(rms) = VL(p)/√2 = 12.74/1.414 = 9.01 V (Average voltage Vdc = 2VL(p)/π = 8.11 V) 4. Power dissipated in the load: P = VL(rms)²/RL = (9.01)²/5 = 81.2/5 = **16.2 W** (Average power uses the rms value, not the dc value: using Vdc would give only 13.2 W, which is not the power dissipated.) Peak load current = 12.74/5 = 2.55 A; each diode must be rated for at least 2.55 A peak and a PIV above 14.14 V. **Note:** If 10 V is the *peak* ac voltage, VL(p) = 8.6 V, VL(rms) = 6.08 V, and P = 6.08²/5 = 7.4 W. **Answer: P ≈ 16.2 W (for 10 V rms input).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How many diodes conduct in series in each half-cycle of a bridge rectifier, and what drop results with 0.7 V diodes? A. One diode, so the drop is 0.7 V B. Four diodes, so the drop is 2.8 V C. Two diodes, so the drop is 1.4 V D. Two diodes, but they are in parallel so the drop is 0.7 V Answer: C. Current passes through one diode on the way to the load and one on the way back. (Key point: Diode drop in bridge) 2. For 10 V rms input, what is the peak load voltage? A. 10 x sqrt2 - 1.4 = 12.74 V B. 10 - 1.4 = 8.6 V C. 14.14 - 0.7 = 13.44 V D. 10 x sqrt2 = 14.14 V Answer: A. Peak input is 14.14 V and 1.4 V is lost in two diodes. (Key point: Peak load voltage) 3. Which value is used to calculate the average power in the load? A. The dc (average) voltage 8.11 V: P = 13.2 W B. The peak voltage 12.74 V: P = 32.5 W C. The input rms voltage 10 V: P = 20 W D. The rms load voltage: P = (9.01)^2/5 = 16.2 W Answer: D. Average power equals Vrms^2/RL. Using the dc value would underestimate the power. (Key point: Load power) 4. How does the bridge give load current in the same direction in both half-cycles? A. All four diodes conduct together in each half-cycle B. D1 and D2 conduct in one half-cycle and D3 and D4 in the other, each pair sending current through RL the same way C. Only D1 conducts in the positive half-cycle and only D4 in the negative D. The load current reverses direction in each half-cycle Answer: B. Two opposite diodes steer each half-cycle through RL in the same direction, giving full-wave rectification. 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