Question

The flash on a compact camera stores energy in a $120\mu F$capacitor that is charged to $220V$. When the flash is fired, the capacitor is quickly discharged through a lightbulb with$5.0\Omega$of resistance.

a. Light from the flash is essentially finished after two time constants have elapsed. For how long does this flash illuminate the scene?

b. At what rate is the lightbulb dissipating energy $250\mu s$after the flash is fired?

c. What total energy is dissipated by the lightbulb?

Short answer

$$\begin{gathered} \left(a\right)t=1.2ms. \\ \left(b\right)P_R=4205W. \\ \left(c\right)E=2.9J \end{gathered}$$.

Step-by-step solution

Part (a) Step 1 :Given information. 

We have to find for how long does this flash illuminate the scene.

Part (a) Step 2: Simplify

When the switch is off, the capacitors discharge and the current flows through the two resistors. The tine taken to discharge the capacitor is called the time constant $\tau$and it is given by equation (28.30) in the form

$\tau =RC$ (1)

Where R is the resistance, and C is the capacitance in the circuit. For time equals two of the time constant, we get6 this time by

$t=2_T=2RC$ (2)

Now, we plug the values for R and C into equation $\left(2\right)$to get t

$t=2RC=2\left(5\Omega \right)\left(120\mu F\right)=1.2\times {10}^{-3}s=1.2ms$

Part (b) Step 1 : Given information 

We should find what total energy is dissipated by the lightbulb.

Part (b) Step 2 : Simply 

When the switch is off, the capacitor discharge and the current flows through the resistor. To discharge the capacitor, the charge flow through the circuit in time r.The intial voltage $∆V_o$is realated to the final voltage $∆V$by

$∆V=∆\hspace{0.17em}{V}_o{e}^{-t/RC}$ (3)

plug the values for $∆V_o$,t and $RC==1.2ms/2=0.6ms$in to eauation $\left(3\right)$to get $∆V$

$∆V=∆V_o{e}^{t/RC}$

$=145V$

As the current flows through resistors, it dissipates energy, the rate of the dissipated energy is the power. Thuis rate where the energy is transfered from the current to the resistor is

$$\begin{gathered} P_R=\frac{{(∆V)}^2}{R} \\ =\frac{{\left(145V\right)}^2}{5\Omega } \\ =4205W \end{gathered}$$.

Part (c) Step 1 : Given information

We should find What total energy is dissipated by the lightbulb.

Part (c) Step 2: Solution

When the capacitor is fully charged,it stores energy due to its capacitance and the stored energy in the capacitor is related to the voltage across the capacitor in the form

$E=\frac{1}{2}C{V}^2$ (4)

Now we plug the values for C and V into equation $\left(1\right)$to get E

localid="1649099920140" $$\begin{gathered} E=\frac{1}{2}C{V}^2 \\ =\frac{1}{2}(120\times {10}^{-5}F){(220V)}^2 \\ =2.9J \end{gathered}$$.