Question

Please give answer with steps, thanks

Consider a resistor (R=1000 kΩ) and a capacitor (C=1μF) connected in series. This configuration is connected in series to a battery with an emf of 10 V.

After the capacitor is charged to its maximum value, the capacitor and resistor are disconnected from the battery and from each other. Then they are connected in parallel, so that the capacitor discharges through the resistor. How long will it take for the voltage across the capacitor to drop to 2.50 V?

Answer 1

Answer: Iit will take approximately 0.916 seconds for the voltage across the capacitor to drop to 2.50 V when the capacitor and resistor are connected in parallel.

Explanation: We can use the equation for the voltage across a capacitor in a discharging RC circuit, which is:

V = V₀ * e^(-t / RC)

where V₀ is the initial voltage across the capacitor, V is the voltage across the capacitor at any time t, R is the resistance of the resistor, C is the capacitance of the capacitor, and e is the mathematical constant e (~2.718).

First, we need to find the initial voltage across the capacitor when it is fully charged in series with the battery:

V₀ = 10 V

Next, we can use the values given for R and C to calculate the time constant of the circuit:

τ = RC = 1000 kΩ * 1 μF = 1 second

Finally, we can use the equation for the voltage across the capacitor to find the time it takes for the voltage to drop to 2.50 V:

2.50 V = 10 V * e^(-t / 1 s)

Taking the natural logarithm of both sides and solving for t:

ln(2.50 V / 10 V) = -t / 1 s

t = -1 s * ln(2.50 V / 10 V) = 0.916 s (rounded to three significant figures)

Therefore, it will take approximately 0.916 seconds for the voltage across the capacitor to drop to 2.50 V when the capacitor and resistor are connected in parallel.