Q85P - Fundamentals Of Physics

Question

A 30 $\mathbf{\mu F}$ capacitor is connected across a programmed power supply. During the interval from t = 0 to t = 0.300s the output voltage of the supply is given by $\mathbf{V}\left(t\right)\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{00}\mathbf{+}\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{t}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{00}{\mathbf{t}}^{\mathbf{2}}\mathbf{volts}$ volts. At $\mathbf{t}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{500}\mathbf{s}$  find (a) the charge on the capacitor, (b) the current into the capacitor, and (c) the power output from the power supply.

Step-by-Step Solution

Answer

1. The charge on the capacitor is 225$\mathrm{\mu C}$ .
2. The current into the capacitor is $60\mathrm{\mu A}$ .
3. The power output from the power supply is 0.450 mW.

Step 1: The given data

1. The capacity of the conductor is $\mathrm{C}=30\mathrm{\mu F} \mathrm{or} 30\times {10}^{-6}\mathrm{F}$
2. The function of the voltage is $V\left(t\right)=6.00+4.00t-2.00t^2$
3. The time for voltage supply is $t=0.500s$

Step 2: Understanding the concept of the capacitance and charge

We can use the concept of capacity in the conductor and rate of electric energy transfer. The voltage between two plates of capacitors is directly proportional to the current in the capacitor.

Formulae:

The charge contained within the plates of capacitor, $Q=CV$                                  (i)

The current equation due to the rate of the potential difference, $I=C\frac{dV}{dt}$              (ii)

The electric power due to the potential difference, $P=IV$                                      (iii)

Step 3: a) Calculation of the charge on the capacitor

The expression of capacitance of the programmed power supply is calculated using the given values substituted in equation (i) as follows:

$$\begin{gathered} \mathrm{Q}=30\times {10}^{-6}\mathrm{F}\times {\left|6.00+4.00\mathrm{t}-2.00{\mathrm{t}}^2\right|}_{\mathrm{t}=0.500\mathrm{s}} \\ =30\times {10}^{-6}\mathrm{F}\times \left(\left(6.00\right)+\left(4.00\right)\left(0.500\right)-\left(2.00\right){\left(0.500\right)}^2\right) \\ =225\times {10}^{-6}\mathrm{C} \\ =225\mathrm{\mu C} \end{gathered}$$

Hence, the charge of the capacitor is $225\mathrm{\mu C}$ .

Step 4: b) Calculation of the current in the capacitor

The current in the capacitor can be given using the data in equation (ii) as follows:

 $$\begin{gathered} \mathrm{I}=\mathrm{C}\frac{\mathrm{d}\left(6.00+4.00\mathrm{t}-2.00{\mathrm{t}}^2\right)}{\mathrm{dt}} \\ =\mathrm{C}\left(4.00-4.00\mathrm{t}\right) \\ =30\times {10}^{-6}\mathrm{F}\times \left(4.00-4.00\times 0.500\right) \\ =60\times {10}^{-6}\mathrm{A} \\ =60\mathrm{\mu A} \end{gathered}$$

Hence, the value of the current is $60\mathrm{\mu A}$.

Step 5: c) Calculation of the output power from the power supply

The output power from the power supply is calculated using equation (iii) as follows:

$$\begin{gathered} \mathrm{P}=60\times {10}^{-6}\mathrm{A}\times {\left|6.00+4.00\mathrm{t}-2.00{\mathrm{t}}^2\right|}_{\mathrm{t}=0.500 \mathrm{s}} \\ =60\times {10}^{-6}\mathrm{A}\times \left(6.00+(4.00)-(2.00){(0.500)}^2\right) \\ =450\times {10}^{-6} \\ =0.450\times {10}^{-3} \\ =0.450\mathrm{mW} \end{gathered}$$ 

Hence, the value of the output power is 0.450mW.