Answer :
The voltage on the capacitor will continuously increase, but it will never reach the desired voltage of 6V. The circuit consists of a resistor, a capacitor, a switch, and a battery. The resistor has a resistance of 18,895Ω, and the capacitor has a capacitance of 8,632μF. The circuit is connected in series to a 14V battery.
To find out how long it will take for the voltage on the capacitor to reach 6V after the switch is closed, we need to use the concept of the time constant (τ) for an RC circuit. The time constant is given by the product of the resistance and capacitance in the circuit, τ = R * C.
In this case, the resistance (R) is 18,895Ω and the capacitance (C) is 8,632μF. However, we need to convert the capacitance from microfarads (μF) to farads (F) for the calculation. 1μF is equal to 1 x 10^-6 F, so the capacitance becomes 8,632 x 10^-6 F.
Now we can calculate the time constant:
τ = R * C
= 18,895Ω * 8,632 x 10^-6 F
= 162.8 seconds
The time constant represents the time it takes for the voltage on the capacitor to reach approximately 63.2% of the final voltage. To find the time it takes for the voltage on the capacitor to reach 6V, we can use the equation:
V = V0 * (1 - e^(-t/τ))
Where:
V is the final voltage (6V in this case)
V0 is the initial voltage (0V at the beginning)
t is the time in seconds
τ is the time constant (162.8 seconds)
Substituting the values into the equation:
6V = 0V * (1 - e^(-t/162.8))
Simplifying the equation:
1 - e^(-t/162.8) = 1
Taking the natural logarithm of both sides to isolate the exponent:
e^(-t/162.8) = 0
Solving for t:
-t/162.8 = ln(0)
Since the natural logarithm of 0 is undefined, it means that the voltage on the capacitor will never reach 6V in this circuit. The voltage on the capacitor will continuously increase, but it will never reach the desired voltage of 6V.
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