Problem 92 - Physics

Question

A \(3.00-\mu F\) and a \(5.00-\mu\) F capacitor are connected in series across a \(30.0-V\) battery. A \(7.00-\mu \mathrm{F}\) capacitor is then connected in parallel across the \(3.00-\mu \mathrm{F}\) capacitor. Determine the voltage across the \(7.00-\mu \mathrm{F}\) capacitor.

Step-by-Step Solution

Answer

Voltage across the 7.00-μF capacitor is 18.75 V.

Step 1: Find the equivalent capacitance for capacitors in series

The capacitance for capacitors in series is given by the formula: \[ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} \] where \( C_1 = 3.00\, \mu F \) and \( C_2 = 5.00\, \mu F \). So, \[ \frac{1}{C_{eq}} = \frac{1}{3.00} + \frac{1}{5.00} \].Calculating further, \[ \frac{1}{C_{eq}} = \frac{5 + 3}{15} = \frac{8}{15} \] hence, \( C_{eq} = \frac{15}{8} \mu F \approx 1.875 \mu F \).

Step 2: Calculate total voltage across the equivalent system

Since capacitors in series share the same charge, the total voltage across them is the sum of individual voltages. The voltage across this equivalent capacitance from the battery is given as 30.0 V.

Step 3: Determine voltage across each capacitor in series

The voltage \( V_1 \) across the \(3.00\, \mu F\) capacitor and \( V_2 \) across the \(5.00\, \mu F\) capacitor in a series can be calculated using the formula: \[ V = Q/C \].The total charge on each capacitor is the same: \( Q = C_{eq} \times V_{\text{total}} = 1.875 \times 30.0 \). This determines the voltages: \[ V_1 = \frac{Q}{C_1} \text{ and } V_2 = \frac{Q}{C_2} \].

Step 4: Solve for the voltage across the 3.00-μF capacitor

We already have that \( Q = C_{eq} \times V_{\text{total}} = 1.875 \times 30.0 = 56.25 \, \mu C \).Now substitute \( C_1 = 3.00 \, \mu F \) into \[ V_1 = \frac{Q}{C_1} = \frac{56.25}{3.00} \approx 18.75 \text{ V} \].

Step 5: Determine the voltage across the 7.00-μF capacitor

The 7.00-μF capacitor is connected in parallel with the 3.00-μF capacitor, so it shares the same voltage. Thus, the voltage across the 7.00-μF capacitor is also 18.75 V.

Key Concepts

Series Capacitors

When capacitors are connected in series, they share the same charge but divide the voltage supply among them. This is because the charge has only one path to take, moving from one capacitor to the next. The equivalent capacitance (C_{eq}) of capacitors in series is lower than any individual capacitor in the system. This is calculated using the formula: \[ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots + \frac{1}{C_n} \].
The reciprocal nature of the formula emphasizes that as more capacitors are added in series, the equivalent capacitance decreases. Understanding the combination of capacitors in series is crucial as it allows us to predict how the voltage will be distributed across each component.

Parallel Capacitors

In parallel capacitor arrangements, each capacitor connects directly to the voltage source. Therefore, they all experience the same voltage across their terminals. The equation for the total capacitance of parallel capacitors is quite different from series capacitors: \[ C_{eq} = C_1 + C_2 + \cdots + C_n \].
Notice that the capacitance simply adds up to give a larger total capacitance. This results in a system that can store more charge. It's important to remember that, in this configuration, the system maintains the same voltage across all capacitors.
Because they share the voltage but are able to individually handle part of the stored energy, parallel capacitors can be useful in circuits requiring larger energy storage.

Voltage Calculation

Calculating voltage in capacitor circuits can be simplified by understanding their arrangements. For series circuits, although the charge on each capacitor is the same, the voltage across each is different, depending on its capacitance. The voltage is calculated using \[ V = \frac{Q}{C} \].
In such circuits, the total voltage across the system is the sum of individual voltages: \[ V_{total} = V_1 + V_2 + \cdots + V_n \].
This approach becomes straightforward when dealing with parallel configurations because all capacitors share the same voltage: \[ V_1 = V_2 = \cdots = V_n \].
Knowing these rules helps in predicting how voltage distributes in a network, which is essential when designing or analyzing capacitive circuits.