### Capacitors in Series: Problem and Solution#### Problem Statement:A 15-μF capacitor and a 30-μF capacitor are connected in series and charged to a potential difference of 50 V. What is the resulting charge on the 30-μF capacitor?#### Multiple Choice Options:- O 0.80 mC- O 0.60 mC- O 0.50 mC- O 0.70 mC #### Explanation:When capacitors are connected in series, the charge (Q) on each capacitor is the same. The total potential difference (V_total) across the series combination is distributed across the individual capacitors inversely proportional to their capacitances.For capacitors in series:\[ \frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} \]Where:- $C_{\text{eq}}$ is the equivalent capacitance.- $C_1 = 15 \, \mu \text{F}$- $C_2 = 30 \, \mu \text{F}$First, calculate the equivalent capacitance:\[ \frac{1}{C_{\text{eq}}} = \frac{1}{15 \, \mu \text{F}} + \frac{1}{30 \, \mu \text{F}} \]\[ \frac{1}{C_{\text{eq}}} = \frac{2}{30 \, \mu \text{F}} + \frac{1}{30 \, \mu \text{F}} = \frac{3}{30 \, \mu \text{F}} = \frac{1}{10 \, \mu \text{F}} \]\[ C_{\text{eq}} = 10 \, \mu \text{F} \]Next, use the total voltage across the capacitors to find the charge:\[ V_{\text{total}} = 50 \text{ V} \]Using the formula $Q = C_{\text{eq}} \cdot V_{\text{total}}$:\[ Q = 10 \, \mu \text{F} \cdot 50 \text{ V} = 500 \, \mu \text{C} = 0.50 \, \text{mC

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A 15-μF capacitor and a 30-μF capacitor are connected in series, and charged to a potential difference of 50 V. What is the resulting charge on the 30-μF capacitor? O 0.80 mC O 0.60 mC O 0.50 mc O 0.70 MC

A 15-μF capacitor and a 30-μF capacitor are connected in series, and charged to a potential difference of 50 V. What is the resulting charge on the 30-μF capacitor? O 0.80 mC O 0.60 mC O 0.50 mc O 0.70 MC

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ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

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Concept explainers

Combination of Capacitors

The connection of capacitors can be established in a circuit in two ways.

Question

### Capacitors in Series: Problem and Solution

#### Problem Statement:
A 15-μF capacitor and a 30-μF capacitor are connected in series and charged to a potential difference of 50 V. What is the resulting charge on the 30-μF capacitor?

#### Multiple Choice Options:
- O 0.80 mC
- O 0.60 mC
- O 0.50 mC
- O 0.70 mC

#### Explanation:
When capacitors are connected in series, the charge (Q) on each capacitor is the same. The total potential difference (V_total) across the series combination is distributed across the individual capacitors inversely proportional to their capacitances.

For capacitors in series:
\[ \frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} \]

Where:
- $C_{\text{eq}}$ is the equivalent capacitance.
- $C_1 = 15 \, \mu \text{F}$
- $C_2 = 30 \, \mu \text{F}$

First, calculate the equivalent capacitance:
\[ \frac{1}{C_{\text{eq}}} = \frac{1}{15 \, \mu \text{F}} + \frac{1}{30 \, \mu \text{F}} \]
\[ \frac{1}{C_{\text{eq}}} = \frac{2}{30 \, \mu \text{F}} + \frac{1}{30 \, \mu \text{F}} = \frac{3}{30 \, \mu \text{F}} = \frac{1}{10 \, \mu \text{F}} \]
\[ C_{\text{eq}} = 10 \, \mu \text{F} \]

Next, use the total voltage across the capacitors to find the charge:
\[ V_{\text{total}} = 50 \text{ V} \]

Using the formula $Q = C_{\text{eq}} \cdot V_{\text{total}}$:
\[ Q = 10 \, \mu \text{F} \cdot 50 \text{ V} = 500 \, \mu \text{C} = 0.50 \, \text{mC

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