Question

A capacitor with C = 0.2 µF is initially charged to a potential of 6.1 V. The capacitor is then connected across a resistor and allowed to discharge. After a time of 5.00 10-3 s, the potential across the capacitor has dropped to 1.16 V. What value of the resistance can you deduce from this?

Answer 1

To determine the resistance R where a capacitor with capacitance C = 0.2 µF discharges from 6.1 V to 1.16 V in 5.00 × 10^-3 s, use the formula for voltage across a discharging capacitor and rearrange it to solve for RC. Then divide by C to find R.

Step-by-step explanation:

To find the resistance through which a 0.2 µF capacitor discharges, we can use the formula for the voltage across a discharging capacitor, which is V(t) = V_0 e^{-t/(RC)}, where V(t) is the potential at time t, V_0 is the initial potential, and RC is the time constant for the circuit.

Given that V_0 = 6.1 V, V(t) is 1.16 V after 5.00 × 10^-3 s, and C = 0.2 µF, we can solve for R.

First, rearrange the formula to solve for RC: RC = -t / ln(V(t)/V_0).

Now substitute the known values: RC = -(5.00 × 10^-3 s) / ln(1.16 V / 6.1 V).

After calculating the natural logarithm and multiplying by -5.00 × 10^-3 s, we will find the value of RC. Finally, divide this result by the capacitance C to get the resistance R.

The time constant τ (tau) is equivalent to the product RC, and it represents the time it takes for the potential to drop to approximately 36.8% of its initial value. This concept is heavily utilized in the field of electronics, particularly when designing circuits that require a certain timing behavior, such as in defibrillators or flash circuits for cameras.