Medical College Admission Test (MCAT) Practice Exam

In a parallel circuit, the total capacitance is calculated by simply adding the capacitances of each individual capacitor present in the circuit. This is expressed by the formula \( C_{total} = C_1 + C_2 + C_3 + ... + C_n \), where each \( C \) represents the capacitance of an individual capacitor.

When a capacitor is removed from a parallel circuit, the total capacitance decreases. This is because the overall capacitance depends on the sum of the capacitances of all the capacitors in the circuit. By removing a capacitor, you are effectively subtracting its capacitance from the total, resulting in a lower overall capacitance.

To further consider the implications, removing a capacitor not only decreases total capacitance but also affects how the circuit stores charge. The total charge stored in a parallel circuit is related to the total capacitance and the voltage across the capacitors, following the formula \( Q = C \times V \). When total capacitance decreases, the ability of the circuit to store charge also decreases.

In summary, the correct answer clearly reflects the foundational principle that the total capacitance in a parallel circuit is cumulative; thus, the removal of a capacitor leads to a decrease in total capacitance