Question

10) A camera flash consists of a 120 μF capacitor charged to 220V which discharges through a flashbulb with 5 Ω resistance. What is the time-constant τ of this RC circuit when the flash goes off? How much of the initial stored energy is discharged within 2T? What is the equation for the voltage drop across the capacitor vs. time during the discharge? Make a graph showing this voltage vs. time.

Answer 1

In the solution below, at first i found out time constant then i calculated the discharged energy and in the end i derived a formula to find the amount of discharged energy for any time t .

As there is only resistor apart from capacitor whatever energy discharged must be dissipated across the resistor according to law of conservation of energy.

Power dissipated across resistor = voltage across resistor * resistance.

Solution

In the solution below, at first i found out time constant then i calculated the discharged energy and in the end i derived a formula to find the amount of discharged energy for any time t . As there is only resistor apart from capacitor whatever energy discharged must be dissipated across the resistor according to law of conservation of energy. Power dissipated across resistor = voltage across resistor times resistance. Given R = 5 Ohm, C = 120 mu F, v_0 = 220v v_0 = initial voltage across the capacitor Time constant (tau) = R times C tau = 5 Ohm times 120 mu F tau = 600 mu s the equation for voltage drop across capacitor vs time is given by v_c = v_0 e^-t/RC -(1) t rightarrow time elapsed after switch is ON. Energy discharged within 2 tau is dissipated across the resistor E_2 tau = integral^2I_0 v^2_R/R dt V_R = voltage across resistor according to Kirchhoff�s voltage law V_R = V_c

The below graph represents the relation between voltage vs time.Where V_s = 220V and T = 600 microseconds.After 5T we may consider capacitor is fully discharged as almost all the charge is discharged(but remember that it doesnot gets completly dicharged).