A 2.0 micro Farad capacitor in a circuit in series with a resistance of 1.0 mega ohms is charged with a 6.0v battery. How long would it take to chage the capacitor to three fourths of its maximum voltage.
U =U₀exp(-t/RC)
U/U₀ =exp(-t/RC)
ln(U/U₀) = -t/RC
t= - RC ln(U/U₀) =
= - 1•10⁶•2•10⁻⁶•ln(0.75•6/6)=0.58 s
To find the time it takes to charge the capacitor to three fourths of its maximum voltage, we can use the formula for the charging of a capacitor in an RC circuit:
V = V0 * (1 - e^(-t/RC))
Where:
V is the voltage across the capacitor at time t
V0 is the maximum voltage (6.0V in this case)
C is the capacitance (2.0 microfarads = 2.0 * 10^(-6) F)
R is the resistance (1.0 megohms = 1.0 * 10^(6) Ω)
e is the base of the natural logarithm (approximately 2.71828)
We are interested in finding the time it takes for the voltage V to reach three fourths (3/4) of the maximum voltage (V0). So we can rewrite the formula as:
3/4 * V0 = V0 * (1 - e^(-t/RC))
Now, let's solve for t (time):
3/4 = 1 - e^(-t/RC)
e^(-t/RC) = 1 - 3/4
e^(-t/RC) = 1/4
Taking the natural logarithm (ln) of both sides:
ln(e^(-t/RC)) = ln(1/4)
-t/RC = ln(1/4)
Now solving for t:
t = -RC * ln(1/4)
Substituting the values:
t = - (2.0 * 10^(-6) F) * (1.0 * 10^6 Ω) * ln(1/4)
Calculating this expression mathematically:
t ≈ 0.1386 seconds
So, it would take approximately 0.1386 seconds to charge the capacitor to three fourths (3/4) of its maximum voltage.