Concept:Capacitors block steady DC current (\( X_C = \frac{1}{2\pi f C} \to \infty \text{ as } f \to 0 \)) while providing a low-impedance path to ground for AC ripple components.
Formula:$$X_C(f = 0\text{ Hz}) = \infty \quad (\text{Open circuit to DC})$$
Solution:- Connecting a capacitor in series with the load creates an open circuit for DC, preventing direct current from reaching the load.
- Connecting the capacitor in parallel (shunt) allows DC to pass freely through the load while routing AC ripple through the capacitor to ground.
Why other options are incorrect:- Option A: A series capacitor blocks DC rather than increasing it.
- Option C: Shunt capacitors attenuate (smooth) ripple rather than amplifying it.
- Option D: A series capacitor acts as an open circuit to DC and does not short-circuit the transformer.
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