Question

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 2τ? 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

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Concept: use the discharging of capacitor through a resistance and the time constant of the discharging circuit as shown below

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Answer 2

Solution:

1. Time Constant (τ):

   $$\tau =R\times C=5\mathrm{\Omega }\times 120\times 10^{-6}\text{F}=0.0006\text{s}=0.6\text{ms}$$

2. Energy Discharged in 2τ:

• Initial energy stored in the capacitor:

  $$E_0=\frac{1}{2}C{V}^2=\frac{1}{2}\times 120\times 10^{-6}\times (220{)}^2=2.904\text{J}$$

• Energy remaining after 2τ (since $V=V_0{e}^{-t\mathrm{/}\tau }$, at $t=2\tau$, $V=V_0{e}^{-2}$):

  $$E=\frac{1}{2}C(V_0{e}^{-2}{)}^2=E_0{e}^{-4}\approx 2.904\times e^{-4}\approx 0.053\text{J}$$

• $$\approx 0.053\text{J}$$

• Energy discharged:

  $$E_{\text{discharged}}=E_0-E\approx 2.904-0.053=2.851\text{J}(\approx 98.2\mathrm{\%}\text{ of }{E}_0)$$

3. Voltage Drop Equation:

   $$V(t)=V_0{e}^{-t\mathrm{/}\tau }=220e^{-t\mathrm{/}0.0006}\text{V}$$

4. Graph of Voltage vs. Time:

• At $t=\tau$: $V=220e^{-1}\approx 80.9\text{V}$

• At $t=2\tau$: $V=220e^{-2}\approx 29.8\text{V}$.

• The voltage decays exponentially from 220V to near 0V over time.

• Key points: