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A single-phase, 230 V, 50 Hz ac mains fed step down transformer (4 : 1) is supplying power to a half-wave uncontrolled ac-dc converter used for charging a battery (12 V dc) with the series current limiting resistor being 19.04 Ω. The charging current is
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- 2.43 A
- 1.65 A
- 1.22 A
- 1.0 A
Correct Option: A
| = | ⇒ V2 = | = | = 57.5 Volts | |||||
| V2 | 1 | 4 | 4 |
Diode will conduct when V2 > E
i.e., Vm sin θ1 = E
⇒ (57.5) √2 sin θ = 12
⇒ θ1 = 8.48° or, 0.148 rad
| = | [2Vm cosθ1 - E(π - 2θ1)] | |
| 2πR |
| = | [2 × 57.5 √2 × cos 8.48° - 12(π - 2 × 0.148)] ≈ 1 A | |
| 2 π × 19.04 |
