What happens to the charge drawn from the battery and the total capacitance when capacitor 2 is removed from a parallel circuit with two capacitors?

In a parallel circuit with capacitors, the total capacitance is the sum of the individual capacitances. When two capacitors are connected in parallel, their capacitance adds together, providing a greater total capacitance than each individual capacitor. The relationship can be described by the equation ( C_{\text{total}} = C_1 + C_2 ).

When capacitor 2 is removed from the circuit, the only capacitor left is capacitor 1. This act reduces the total capacitance of the circuit since the overall capacitance is now solely dependent on capacitor 1.

With the removal of capacitor 2, the total charge stored in the circuit is given by the equation ( Q = C_{\text{total}} \cdot V ), where ( V ) is the voltage across the capacitors. If the total capacitance decreases due to the removal of one of the capacitors, the total charge drawn from the battery must also decrease, assuming the voltage remains constant.

Therefore, with the reduction in total capacitance resulting from the removal of capacitor 2, both the charge drawn from the battery and the total capacitance decrease. This leads to the conclusion that both values go down when one of the capacitors in a parallel configuration is