MP Board · Class 12 · Physics · Semiconductor ElectronicsIn a full-wave rectifier circuit operating from 50 Hz mains, the transformer used has a turns ratio of 2:1. The resistance of the load is 1000 $\Omega$. The peak value of the input AC voltage across the secondary is 50 V. Calculate: (i) The peak value of the output current. (ii) The DC value of the output current. (iii) The RMS value of the output current. (iv) The ripple frequency of the output.

Given Data:

• Frequency of input AC mains ($f$) = 50 Hz
• Turns ratio = 2:1
• Load resistance ($R_L$) = 1000 $\Omega$
• Peak value of secondary AC voltage ($V_m$) = 50 V

(i) Peak value of the output current ($I_m$):\nThe peak current depends on the peak secondary voltage and the total resistance in the circuit. Assuming ideal diodes with zero forward resistance:

$$I_m = \frac{V_m}{R_L}$$ $$I_m = \frac{50 \text{ V}}{1000 , \Omega} = 0.05 \text{ A} = 50 \text{ mA}$$

(ii) DC value of the output current ($I_{dc}$):\nFor a full-wave rectifier, the DC output current is given by the formula:

$$I_{dc} = \frac{2 I_m}{\pi}$$\nSubstituting the value of $I_m$: $$I_{dc} = \frac{2 \times 0.05}{\pi} = \frac{0.1}{3.1416} \approx 0.0318 \text{ A} = 31.8 \text{ mA}$$

(iii) RMS value of the output current ($I_{rms}$):\nThe root-mean-square value of the output current for a full-wave rectifier is:

$$I_{rms} = \frac{I_m}{\sqrt{2}}$$\nSubstituting the value of $I_m$: $$I_{rms} = \frac{0.05}{1.414} \approx 0.0354 \text{ A} = 35.4 \text{ mA}$$

(iv) Ripple frequency of the output ($f_{ripple}$):\nIn a full-wave rectifier, both the positive and the negative half-cycles of the input AC are rectified into output pulses. Therefore, the frequency of the output ripple is twice the frequency of the input AC mains.

$$f_{ripple} = 2 \times f$$ $$f_{ripple} = 2 \times 50 \text{ Hz} = 100 \text{ Hz}$$