ou Answered orrect Answer An R=88MN resistor and a C=43μF capacitor are connected in series with a open switch and an V = 45.4V potential source. The capacitor is initially uncharged. The switch is closed at t = 0. Determine the current in the circuit when the charge on the capacitor is 1/9 of its maximum value (in μA). 47.6 0.4586 margin of error +/- 1%
ou Answered orrect Answer An R=88MN resistor and a C=43μF capacitor are connected in series with a open switch and an V = 45.4V potential source. The capacitor is initially uncharged. The switch is closed at t = 0. Determine the current in the circuit when the charge on the capacitor is 1/9 of its maximum value (in μA). 47.6 0.4586 margin of error +/- 1%
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Question
An R=88MS resistor and a C =43. capacitor are connected in
series with a open switch and an V = 45.4V potential source. The
capacitor is initially uncharged. The switch is closed at t = 0. Determine
the current in the circuit
when the charge on the capacitor is 1/9 of its maximum value (in p.A).
I need to look. At correct answer and show all work please
![### Problem Description
An \( R = 88 \, M\Omega \) resistor and a \( C = 43 \, \mu F \) capacitor are connected in series with an open switch and a \( V = 45.4 \, V \) potential source. The capacitor is initially uncharged. The switch is closed at \( t = 0 \). Determine the current in the circuit when the charge on the capacitor is \( \frac{1}{9} \) of its maximum value (in \( \mu A \)).
### Your Answer
47.6
### Correct Answer
0.4586 with a margin of error of +/- 1%
### Explanation
This problem deals with a basic RC (resistor-capacitor) charging circuit. When the switch is closed, the capacitor begins to charge through the resistor until it reaches its maximum charge, \( Q_{\text{max}} = C \cdot V \). The charge at any time \( t \) is given by \( Q(t) = Q_{\text{max}} \cdot (1 - e^{-t/RC}) \).
To find the current when the charge is \( \frac{1}{9} \) of \( Q_{\text{max}} \), you can use the relationship:
\[ Q(t) = \frac{1}{9} Q_{\text{max}} \]
The current \( I(t) \) in the circuit at time \( t \) is:
\[ I(t) = \frac{V}{R} \cdot e^{-t/RC} \]
Through solving, you find:
1. \( t = RC \cdot \ln(9) \)
2. \( I(t) = \frac{V}{R} \cdot \left(\frac{1}{9}\right) = \frac{45.4}{88 \times 10^6} \cdot \frac{1}{9} \)
This results in the correct answer approximately 0.4586 \( \mu A \).
The discrepancy between the given answer of 47.6 and the correct answer 0.4586 likely stems from a miscalculation involving either the natural logarithm, unit conversion, or arithmetic.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fee22cf2f-b974-4b00-a3cf-09b388e7d65d%2Fcbc98d4d-e1a7-4c0b-8bdf-19ec2ffc2ef0%2Fbxaipce_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem Description
An \( R = 88 \, M\Omega \) resistor and a \( C = 43 \, \mu F \) capacitor are connected in series with an open switch and a \( V = 45.4 \, V \) potential source. The capacitor is initially uncharged. The switch is closed at \( t = 0 \). Determine the current in the circuit when the charge on the capacitor is \( \frac{1}{9} \) of its maximum value (in \( \mu A \)).
### Your Answer
47.6
### Correct Answer
0.4586 with a margin of error of +/- 1%
### Explanation
This problem deals with a basic RC (resistor-capacitor) charging circuit. When the switch is closed, the capacitor begins to charge through the resistor until it reaches its maximum charge, \( Q_{\text{max}} = C \cdot V \). The charge at any time \( t \) is given by \( Q(t) = Q_{\text{max}} \cdot (1 - e^{-t/RC}) \).
To find the current when the charge is \( \frac{1}{9} \) of \( Q_{\text{max}} \), you can use the relationship:
\[ Q(t) = \frac{1}{9} Q_{\text{max}} \]
The current \( I(t) \) in the circuit at time \( t \) is:
\[ I(t) = \frac{V}{R} \cdot e^{-t/RC} \]
Through solving, you find:
1. \( t = RC \cdot \ln(9) \)
2. \( I(t) = \frac{V}{R} \cdot \left(\frac{1}{9}\right) = \frac{45.4}{88 \times 10^6} \cdot \frac{1}{9} \)
This results in the correct answer approximately 0.4586 \( \mu A \).
The discrepancy between the given answer of 47.6 and the correct answer 0.4586 likely stems from a miscalculation involving either the natural logarithm, unit conversion, or arithmetic.
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