How does the addition of a third capacitor in series affect the voltage drop across capacitor 1 and the charge stored on it?

How does the addition of a third capacitor in series affect the voltage drop across capacitor 1 and the charge stored on it?

• A. Increases both
• B. Decreases both
• C. No change
• D. Increases voltage, decreases charge

Answer: (B)

When a third capacitor is added in series to a circuit that already includes two capacitors, the total capacitance of the series combination decreases. This is because the formula for total capacitance \( C_{total} \) in series is given by \[ \frac{1}{C_{total}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \] As the total capacitance decreases, the overall ability of the circuit to store charge at a given voltage is diminished. When capacitors are connected in series, they all share the same charge. However, the voltage drop across each capacitor depends on its capacitance. The voltage drop across a capacitor in series can be determined with the formula: \[ V_i = \frac{Q}{C_i} \] where \( V_i \) is the voltage across capacitor \( i \), \( Q \) is the charge (which remains the same across all capacitors in series), and \( C_i \) is the capacitance of capacitor \( i \). Since adding the third capacitor decreases the total capacitance, the charge stored in the system is reduced. Therefore, the voltage

When a third capacitor is added in series to a circuit that already includes two capacitors, the total capacitance of the series combination decreases. This is because the formula for total capacitance ( C_{total} ) in series is given by

[

\frac{1}{C_{total}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}

]

As the total capacitance decreases, the overall ability of the circuit to store charge at a given voltage is diminished.

When capacitors are connected in series, they all share the same charge. However, the voltage drop across each capacitor depends on its capacitance. The voltage drop across a capacitor in series can be determined with the formula:

[

V_i = \frac{Q}{C_i}

]

where ( V_i ) is the voltage across capacitor ( i ), ( Q ) is the charge (which remains the same across all capacitors in series), and ( C_i ) is the capacitance of capacitor ( i ).

Since adding the third capacitor decreases the total capacitance, the charge stored in the system is reduced. Therefore, the voltage