Topic: Analog Circuits
If the transformer and diodes in the following circuit are ideal, Find out the value of Capacitor (Co) provides 5% ripple voltage across RL. (Vin = 100sin(2\pi × 1000 × t), Np = 10, Ns1 = 5, Ns2 = 5, RL = 1kΩ)
- A. 4.7 μF
- B. 8.7 μF
- C. 10 μF
- D. 12 μF
Answer: C. 10 μF
Explanation: For a Full Wave Rectifier with a C-filter, the ripple factor is given by $\gamma = \frac{1}{4\sqrt{3}fCR_{L}}$. Alternatively, using peak values, $V_{r(rms)} = \frac{V_{max}}{4\sqrt{3}fCR_L}$. Given $f = 1000$ Hz, $R_{L} = 1000 \Omega$, and $\gamma = 5\%$. The capacitance is calculated as $C = \frac{50}{2 \times 100 \times 2.5 \times 1000} = 10 \mu F$ based on the voltage drop parameters.