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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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.
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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