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 can use the formula:

C = τ / R

where C is the capacitance, τ (tau) is the time constant, and R is the total resistance.

Given that the time constant (τ) is 34.0 µs, and the total resistance (R) is 13.0 kΩ (or 13000 Ω), we can substitute these values into the formula:

C = (34.0 × 10^(-6)) / 13000

Calculating this expression, we find:

C ≈ 2.615 × 10^(-9) F (farads)

Therefore, the total capacitance of the circuit is approximately 2.615 nano-farads (nF).

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 charging or discharging of a capacitor:

V(t) = V0 × (1 - e^(-t / τ))

where V(t) is the voltage at time t, V0 is the initial voltage, τ (tau) is the time constant, and e is the base of the natural logarithm.

In this case, the initial voltage (V0) is 0 V (since the switch is closed, and the capacitor begins to charge from 0 V).

We need to find the time (t) when the voltage across the resistor (V(t)) is 11.0 V.

Rearranging the formula:

11 = 0 × (1 - e^(-t / τ))

11 = 1 - e^(-t / 34.0 × 10^(-6))

e^(-t / 34.0 × 10^(-6)) = 1 - 11

e^(-t / 34.0 × 10^(-6)) = -10

To solve for t, we need to take the natural logarithm (ln) of both sides:

-ln(e^(-t / 34.0 × 10^(-6))) = ln(-10)

-t / 34.0 × 10^(-6) = ln(-10)

t = -34.0 × 10^(-6) × ln(-10)

Please note that the result of ln(-10) is a complex number, so we cannot find a real solution for t.