Reason (R): Capacitive reactance decreases with increase in capacitance of capacitor.
Solution:
Capacitive reactance is given by \( X_C = frac{1}{omega C} \). As capacitance \( C \) increases, \( X_C \) decreases (R is true). A decrease in \( X_C \) leads to a decrease in the total circuit impedance \( Z \). With constant voltage \( V \), a lower \( Z \) results in higher current \( I = V/Z \), thus increasing the lamp's brightness (A is true). (R) correctly explains (A).
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