A 12 V battery is connected to 5 mega ohm resistor and a 6 micro farad capacitor and an open switch. After how many seconds will the voltage across the resistor reach 9 V?

Suppose the battery is replaced by 12 V ac with a frequency of 60 Hz.Calculate the Irms in the circuit

If the resistor is 9v, the capacitor is 3 volts.

Vc=q/C=12/(5E6*6E-6)*e^-t/rc

3*30/12=e^-t/30
take the ln of each side, solve for t.
check my typing, my eyes are blurry this am.

Now for the ac part, Z=5E6+1/(2PI60)*6E-6 figure that, it is nearly all resistive.
Then I=12/Z
I can check your work on this.

a. RC = 5*10^6 * 6*10^-6 = 30 s.

Vr = 12/e^x = 9
e^x = 12/9 = 1.333
X = 0.288 = t/RC.
t/30 = 0.288
t = 8.63 s.

b. Xc = 1/(2pi*F*C) = 1/(6.28*60*6*10^-6) = 442 ohms.
Z = 5*10^6 - j442 = 5*10^6 0hms.
I = E/Z = 12/(5*10^6) = 2.4*10^-6 A = 2.4 uA.

To find out how long it takes for the voltage across the resistor to reach 9 V, we need to analyze the behavior of the RC circuit.

In an RC circuit, the voltage across the capacitor (Vc) changes over time according to the equation:

Vc(t) = V0 * (1 - e^(-t/RC))

Where:
- Vc(t) is the voltage across the capacitor at time t
- V0 is the initial voltage across the capacitor (in this case, 12 V)
- e is Euler's number, approximately 2.71828
- t is the time in seconds
- RC is the time constant, equal to the resistance (R) multiplied by the capacitance (C)

In this case, the voltage across the resistor (Vr) is equal to the difference between the battery voltage (12 V) and the voltage across the capacitor (Vc). So,

Vr(t) = V0 - Vc(t)

We want to find the value of t when Vr(t) is equal to 9 V. So, we can set up the following equation:

9 = 12 - 12 * (1 - e^(-t/RC))

Now we can solve for t:

3 = 12 * e^(-t/RC)

Divide both sides by 12:

1/4 = e^(-t/RC)

Since e is a constant, we can take the natural logarithm (ln) of both sides:

ln(1/4) = -t/RC

Rearrange the equation to solve for t:

t = -ln(1/4) * RC

Now, let's substitute the given values: R = 5 mega ohms (5 * 10^6 ohms) and C = 6 microfarads (6 * 10^-6 farads):

t = -ln(1/4) * (5 * 10^6 ohms) * (6 * 10^-6 farads)

Calculate the value:

t ≈ 612.1 seconds

So, it will take approximately 612.1 seconds for the voltage across the resistor to reach 9 V.

Moving on to the second part of your question, to calculate the Irms in the circuit when a 12 V AC source with a frequency of 60 Hz is used, we need to consider the impedance (Z) of the circuit.

The impedance of an RC circuit is given by:

Z = √(R^2 + (1/(ωC))^2)

Where:
- Z is the impedance
- R is the resistance
- ω is the angular frequency, given by 2πf (where f is the frequency in Hz)
- C is the capacitance

In this case, the resistance is still 5 mega ohms and the capacitance is 6 microfarads. Let's calculate the angular frequency:

ω = 2πf = 2π * 60 Hz ≈ 377 rad/s

Now, we can calculate the impedance:

Z = √((5 * 10^6 ohms)^2 + (1/(377 rad/s * 6 * 10^-6 farads))^2)

Calculating the value:

Z ≈ √(2.5 * 10^13 ohms^2 + 4.2 * 10^7 ohms^2)

Z ≈ √(2.5 * 10^13 + 4.2 * 10^7) ohms

Z ≈ √(2.500042 * 10^13) ohms

Z ≈ 5 * 10^6 ohms

Since the Irms (root mean square current) in the circuit is given by the ratio of the voltage (12 V) to the impedance (Z), we can calculate:

Irms = Vrms / Z

Vrms is the root mean square voltage, which is the peak voltage divided by √2. In this case, the peak voltage is 12 V, so:

Vrms = 12 V / √2 ≈ 8.49 V

Now we can calculate the Irms:

Irms ≈ 8.49 V / (5 * 10^6 ohms)

Irms ≈ 1.697 * 10^-6 A ≈ 1.7 μA

Therefore, the Irms in the circuit is approximately 1.7 microamperes.