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.