**Problem:**A 10.0 µF capacitor is charged by a 14.0 V battery through a resistance \( R \). The capacitor reaches a potential difference of 4.00 V at a time 3.00 s after charging begins. Find \( R \).**Solution:**To solve this problem, we use the formula for the voltage across a charging capacitor:\[ V(t) = V_0 (1 - e^{-t/RC}) \]Where:- \( V(t) \) is the voltage across the capacitor at time \( t \),- \( V_0 \) is the battery voltage (14.0 V),- \( R \) is the resistance,- \( C \) is the capacitance (10.0 µF),- \( t \) is the time (3.00 s).Given:- \( V(t) = 4.00 \) V- \( C = 10.0 \times 10^{-6} \) F- \( V_0 = 14.0 \) V- \( t = 3.00 \) sUsing the formula:\[ 4.00 = 14.0 \times (1 - e^{-3.00/(R \times 10.0 \times 10^{-6})}) \]Solve for \( R \).

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A 10.0 µF capacitor is charged by a 14.0 V battery through a resistance R. The capacitor reaches a potential difference of 4.00 V at a time 3.00 s after charging begins. Find R. k82 Additional Materials eBook A

A 10.0 µF capacitor is charged by a 14.0 V battery through a resistance R. The capacitor reaches a potential difference of 4.00 V at a time 3.00 s after charging begins. Find R. k82 Additional Materials eBook A

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Question

**Problem:**

A 10.0 µF capacitor is charged by a 14.0 V battery through a resistance \( R \). The capacitor reaches a potential difference of 4.00 V at a time 3.00 s after charging begins. Find \( R \).

**Solution:**

To solve this problem, we use the formula for the voltage across a charging capacitor:

\[ V(t) = V_0 (1 - e^{-t/RC}) \]

Where:
- \( V(t) \) is the voltage across the capacitor at time \( t \),
- \( V_0 \) is the battery voltage (14.0 V),
- \( R \) is the resistance,
- \( C \) is the capacitance (10.0 µF),
- \( t \) is the time (3.00 s).

Given:
- \( V(t) = 4.00 \) V
- \( C = 10.0 \times 10^{-6} \) F
- \( V_0 = 14.0 \) V
- \( t = 3.00 \) s

Using the formula:

\[ 4.00 = 14.0 \times (1 - e^{-3.00/(R \times 10.0 \times 10^{-6})}) \]

Solve for \( R \).

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