91Ó°ÊÓ

1. Resistance of the resistor,$\mathrm{R}=3.00×{10}^6\text{Ω}$
2. Capacitance of the capacitor,$\mathrm{C}=1.00×{10}^{−6}\text{F}$
3. Emf of the ideal battery,$\mathrm{Î\mu }=4.0\text{V}$
4. Time at which the connection is made,$\mathrm{t}=1.0\text{s}$

The rate of charge increasing is the value of the current flowing in the circuit. Now, the energy stored in the capacitor upon differentiation will give the next required value. The rate at which energy is used by the resistor is the power generated in the process. Now, the power delivered by the battery is given the emf of the battery using the current value.

Formulae:

The energy stored in the capacitor, $\mathbf{U}\mathbf{=}\frac{{\mathbf{q}}^{\mathbf{2}}}{\mathbf{2}\mathbf{C}}$ (i)

The charge on one plate of the capacitor,$\mathbf{q}\mathbf{=}{\mathbf{q}}_{\mathbf{0}}\mathbf{(}\mathbf{1}\mathbf{-}{\mathbf{e}}^{\mathbf{-}\mathbf{t}\mathbf{/}\mathbf{Ï„}}\mathbf{)}$ (ii)

The time constant of the RC circuit, $\mathbf{Ï„}\mathbf{=}\mathbf{RC}$ (iii)

The current flowing within the given time, $\mathbf{i}\mathbf{=}\frac{\mathbf{dq}}{\mathbf{dt}}$ (iv)

The charge stored within the plates of a capacitor, $\mathbf{q}\mathbf{=}\mathbf{CV}$ (v)

The power of dissipation in the resistor, $\mathbf{P}\mathbf{=}{\mathbf{i}}^{\mathbf{2}}\mathbf{R}\mathbf{}\mathbf{or}\mathbf{}\mathbf{iV}$ (vi)

(a)

The value of the time constant can be given using the given data in equation (iii) as follows:

$\begin{array}{c}\mathrm{τ}=(3.00×{10}^6\text{Ω})(1.00×{10}^{−6}\text{F})\\ =3\text{s}\end{array}$

The rate at which the charge of the capacitor is increasing can be given using equation (ii) in equation (iv) as follows:

$\begin{array}{c}\mathrm{i}=\frac{\mathrm{dq}}{\mathrm{dt}}\\ =\frac{\mathrm{d}\left[{\mathrm{q}}_0(1−{\mathrm{e}}^{−\mathrm{t}/\mathrm{Ï„}})\right]}{\mathrm{dt}}\\ =\frac{\mathrm{°äÎ\mu }}{\mathrm{Ï„}}{\mathrm{e}}^{−\mathrm{t}/\mathrm{Ï„}}(âˆ\mu \text{Fromequation(v),}{\mathrm{q}}_0=\mathrm{°äÎ\mu })\\ =\frac{(1.00×{10}^{−6}\text{F})\left(4.0\text{V}\right)}{3.00\text{s}}{\mathrm{e}}^{−1.0\text{s}/3.0\text{s}}\end{array}$

$=9.55×{10}^{−7}\text{C/s}$

Hence, the rate at which the charge is increasing is $=9.55×{10}^{−7}\text{C/s}$.

(b)

The rate at which the energy is stored within the capacitor can be given using equation (i) as follows:

$\frac{\mathrm{dU}}{\mathrm{dt}}=\frac{\mathrm{q}}{\mathrm{C}}\frac{\mathrm{dq}}{\mathrm{dt}}\mathrm{...........}\left(\mathrm{a}\right)$

Now, the charge value can be given using the given data in equation (ii) as follows:

$\begin{array}{c}\mathrm{q}=\mathrm{°äÎ\mu }(1−{\mathrm{e}}^{−\mathrm{t}/\mathrm{Ï„}})(âˆ\mu {\mathrm{q}}_0=\mathrm{°äÎ\mu },\text{fromequation(v)})\\ =(1.00×{10}^{−6}\text{F})\left(4.0\text{V}\right)(1−{\mathrm{e}}^{−1/.0\text{s}/3.0\text{s}})\\ =1.13×{10}^{−6}\text{C}\end{array}$

The rate at which the energy is being stored in the capacitor can be given using the given data in equation (a) as follows:

$\begin{array}{c}\frac{\mathrm{dU}}{\mathrm{dt}}=\frac{(1.13×{10}^{−6}\text{C})}{(1.00×{10}^{−6}\text{F})}(9.55×{10}^{−7}\text{C/s})\\ =1.08×{10}^{−6}\text{W}\end{array}$

Hence, the value of the rate of energy is $1.08×{10}^{−6}\text{W}$.

(c)

The rate at which the thermal energy is appearing in the resistor of the circuit can be given using the calculated data in equation (vi) as follows:

$\begin{array}{c}\mathrm{P}={(9.55×{10}^{−7}\text{A})}^2(3.00×{10}^6\text{Ω})\\ =2.74×{10}^{−6}\text{W}\end{array}$

Hence, the value of the rate of energy is $2.74×{10}^{−6}\text{W}$.

(d)

The rate at which the energy is being delivered by the battery can be given using the given data in equation (vi) as follows:

$\begin{array}{c}\mathrm{P}=\mathrm{¾\pm Î\mu }\\ =(9.55×{10}^{−7}\text{A})\left(4.0\text{V}\right)\\ =3.82×{10}^{−6}\text{W}\end{array}$

Hence, the value of the rate of the energy is $3.82×{10}^{−6}\text{W}$.