Question

In one camera, the resistance of the bulb is 0.1 Ω, and the capacitor has a value of 4700 μF. How long will it take to discharge the capacitor to just 5% of its original voltage?

Answer 1

Let's first find the time constant:

$$\tau=RC=0.1\ \Omega\cdot4700\cdot10^{-6}\ F=4.7\cdot10^{-4}\ s.$$

Let's write the equation for voltage across the capacitor:

$$V=V_0e^{-\dfrac{t}{\tau}},$$ $$0.05V_0=V_0e^{-\dfrac{t}{\tau}},$$ $$ln(0.05)=ln(e^{-\dfrac{t}{\tau}}),$$ $$-\dfrac{t}{\tau}=ln(0.05),$$ $$t=-\tau ln(0.05)=-4.7\cdot10^{-4}\cdot ln(0.05)=1.4\cdot10^{-3}\ s.$$