You have a 100μF capacitor connected in series with a 1kΩ resistor. If the capacitor is initially uncharged, how long will it take for the voltage across the capacitor to reach 90% of its maximum charge when connected to a 12V DC power supply?
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To find the time it takes for the voltage across the capacitor to reach 90% of its maximum charge, we can use the formula for the charging of a capacitor in a RC circuit:
V(t) = V_max * (1 - e^(-t / RC))
Where:
V(t) is the voltage across the capacitor at time t
V_max is the maximum charge voltage, which is 12V in this case
e is the base of the natural logarithm (approximately 2.71828)
t is the time
R is the resistance, which is 1kΩ
C is the capacitance, which is 100μF or 0.1mF
To find the time it takes for the voltage across the capacitor to reach 90% of its maximum charge, we need to solve the equation for t when V(t) = 0.9 * V_max.
0.9 * V_max = V_max * (1 - e^(-t / RC))
Divide both sides by V_max:
0.9 = 1 - e^(-t / RC)
Rearrange the equation:
e^(-t / RC) = 0.1
Take the natural logarithm of both sides:
-t / RC = ln(0.1)
Solve for t:
t = -RC * ln(0.1)
Now, substitute the values:
t = (-1000Ω * 0.1mF) * ln(0.1)
Simplify the equation:
t = (-100 Ω * 0.1F) * ln(0.1)
t = -10 * ln(0.1)
Using a calculator:
t ≈ -10 * (-2.3026)
t ≈ 23.026 seconds
Therefore, it will take approximately 23.026 seconds for the voltage across the capacitor to reach 90% of its maximum charge when connected to a 12V DC power supply.
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