The total resistance is 13.0 kΩ, and the battery's emf is 28.0 V. If the time constant is measured to be 34.0 µs . Calculate the total capacitance of the circuit. Calculate the time it takes for the voltage across the resistor to reach 11.0 V after the switch is closed. Set up is as follows:

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To calculate the total capacitance of the circuit, we need to use the formula for the time constant (τ) of an RC circuit. The formula is given by:

τ = RC

Where:
τ = time constant
R = total resistance
C = total capacitance

In this case, the given time constant is 34.0 µs and the total resistance is 13.0 kΩ. We can rearrange the formula to solve for the total capacitance:

C = τ / R

Substituting the given values:

C = (34.0 µs) / (13.0 kΩ)

To make the units consistent, we need to convert 34.0 µs to seconds and 13.0 kΩ to ohms:

C = (34.0 x 10^-6 s) / (13.0 x 10^3 Ω)

C = 2.615 x 10^-9 F

So the total capacitance of the circuit is approximately 2.615 nanofarads.

To calculate the time it takes for the voltage across the resistor to reach 11.0 V after the switch is closed, we can use the formula for the voltage in an RC circuit with a charging capacitor:

V(t) = V(0) * (1 - e^(-t / τ))

Where:
V(t) = voltage across the resistor at time t
V(0) = initial voltage across the resistor (0 V in this case)
t = time
τ = time constant

We need to solve for t when V(t) = 11.0 V. Rearranging the formula:

t = -τ * ln(1 - V(t) / V(0))

Substituting the given values:

t = - (34.0 x 10^-6 s) * ln(1 - 11.0 V / 0 V)

Since the natural logarithm of a negative number is undefined, we can ignore the negative sign. Also, ln(1) is equal to 0. Using these simplifications:

t = (34.0 x 10^-6 s) * ln(1)

t = 0 s

Therefore, it takes 0 seconds for the voltage across the resistor to reach 11.0 V after the switch is closed.