Final answer:
The time until a 7 µF capacitor discharges to 37% of its initial charge in series with a 4 MΩ resistor is found using the natural logarithm associated with the RC time constant formula and the result is approximately 12.936 seconds.
Step-by-step explanation:
To find the time until a capacitor with a 7 microfarads (µF) capacitance reaches 37% of its initial charge after disconnecting from a battery, when it's in series with a 4 Megohms (MΩ) resistor, we need to use the concept of an RC time constant. The RC time constant (τ) for an RC circuit is given by the product of the resistance (R) and the capacitance (C), τ = R × C. To find the time (t) at which the charge (q) on the capacitor is a certain percentage of the initial charge (q_0), we use the formula q = q_0 × e-(t/τ), where e is the base of the natural logarithm. For 37% of the initial charge, this equation simplifies to t = τ × ln(1/(1-0.37)), as ln(1/0.37) is the natural logarithm of the inverse of the percentage we are interested in.
Calculating the RC time constant using the given values, we have τ = 7 × 10-6 F × 4 × 106 Ω = 28 seconds. Now, calculate the time for the capacitor to discharge to 37% using the formula mentioned: t = τ × ln(1/(1-0.37)) = 28 × ln(1/0.63) ≈ 28 × 0.462 = 12.936 seconds.
Therefore, the capacitor will take approximately 12.936 seconds to reach 37% of its initial charge upon discharging through the 4 MΩ resistor.