consider an open switch rc series circuit with r the capacitor is charged with a 24.0 v battery as in figure below, obtain the equation of the charge growth in the capacitor as a function of time. if the capacitor is at a voltage 18.0 v, when the switch is transferred from point a to point b, how, long does it take the capacitor to discharge to 90.0% of its original voltage? 4. 17.5 k? and c-231 ? f. if
Consider an open switch RC series circuit with a capacitor charged by a 24.0 V battery. Obtain the equation of the charge growth in the capacitor as a function of time, and determine the time it takes for the capacitor to discharge to 90.0% of its original voltage when the capacitor is at 18.0 V.
Equation of Charge Growth in the Capacitor:
In an RC series circuit, the voltage across a charging capacitor as a function of time is given by the equation:
V(t) = V_0(1 - e^{-\frac{t}{RC}})
where:
- ( V(t) ) is the voltage across the capacitor at time ( t ),
- ( V_0 ) is the initial voltage across the capacitor (24.0 V in this case),
- ( R ) is the resistance in ohms,
- ( C ) is the capacitance in farads, and
- ( e ) is the base of the natural logarithm (approximately 2.71828).
Time taken to discharge to 90.0% of its original voltage:
Given that the capacitor is at 18.0 V and we need to find the time taken to discharge to 90% of its original voltage, we can use the equation above to find the time (( t )) when the voltage (( V(t) )) across the capacitor reaches 90% of the initial voltage:
0.90 \times 24.0 = 24.0 \times (1 - e^{-\frac{t}{RC}})
21.6 = 24.0 \times (1 - e^{-\frac{t}{RC}})
\frac{21.6}{24.0} = 1 - e^{-\frac{t}{RC}}
0.9 = 1 - e^{-\frac{t}{RC}}
e^{-\frac{t}{RC}} = 0.1
-\frac{t}{RC} = ln(0.1)
t = -RC \times ln(0.1)
Substitute the values of the resistance (17.5 kΩ) and capacitance (231 μF) into the equation above and calculate the time it takes for the capacitor to discharge to 90.0% of its original voltage.