Electric Current MCQ Questions & Answers in Electrostatics and Magnetism | Physics

Initial energy of capacitor, $${E_1} = \frac{{q_1^2}}{{2C}}$$
Final energy of capacitor, $${E_2} = \frac{1}{2}{E_1} = \frac{{q_1^2}}{{4C}} = {\left( {\frac{{\frac{{{q_1}}}{{\sqrt 2 }}}}{{2C}}} \right)^2}$$
$$\therefore {t_1}$$  = time for the charge to reduce to $$\frac{1}{{\sqrt 2 }}$$ of its initial value and $${t_2}$$ = time for the charge to reduce to $$\frac{1}{4}$$ of its initial value
$$\eqalign{ & {\text{We}}\,{\text{have,}}\,{q_2} = {q_1}{e^{ - \frac{t}{{CR}}}} \cr & \Rightarrow \ln \left( {\frac{{{q_2}}}{{{q_1}}}} \right) = - \frac{t}{{CR}}\quad \therefore \ln \left( {\frac{1}{{\sqrt 2 }}} \right) = \frac{{ - {t_1}}}{{CR}}\,......\left( 1 \right) \cr & {\text{and}}\,\ln \left( {\frac{1}{4}} \right) = \frac{{ - {t_2}}}{{CR}}\,......\left( 2 \right) \cr} $$
By (1) and (2),
$$\frac{{{t_1}}}{{{t_2}}} = \frac{{\ln \left( {\frac{1}{{\sqrt 2 }}} \right)}}{{\ln \left( {\frac{1}{4}} \right)}} = \frac{1}{2}\frac{{\ln \left( {\frac{1}{2}} \right)}}{{2\ln \left( {\frac{1}{2}} \right)}} = \frac{1}{4}$$

At steady state, there is no current in capacitor.
$$2\Omega $$ and $$10\Omega $$ are in series. There equivalent resistance is $$12\Omega .$$ This $$12\Omega $$ is in parallel with $$4\Omega $$ and there combined resistance is $$\frac{{12 \times 4}}{{\left( {12 + 4} \right)}}.$$   This resistance is in series with $$6\Omega .$$ Therefore, current drawn from battery
$$i = \frac{V}{R} = \left( {\frac{{72}}{{6 + \frac{{12 \times 4}}{{12 + 4}}}}} \right) = 8A$$

Current in $$10\Omega $$ resistor
$$i' = \left( {\frac{4}{{4 + 12}}} \right)8 = 2A$$
$$Pd$$  across capacitor, $$V = i'R = 2 \times 10 = 20V$$
∴ Charge on the capacitor, $$q = CV$$
$$ = 10 \times 20 = 200\mu C.$$

III < II < IV < I.

Power dissipation, $${P_1} = {I^2}\left( {3R} \right) = 3{I^2}R$$

Power dissipation, $${P_2} = {I^2}\left( {\frac{{2R}}{3}} \right) = 0.67{I^2}R$$

Power dissipation, $${P_3} = {I^2}\left( {\frac{R}{3}} \right) = 0.33{I^2}R$$

Power dissipation, $${P_4} = {I^2}\left( {\frac{3}{2}R} \right) = 1.5{I^2}R$$

The equivalent circuit is shown in figure.
$$R = \frac{{6 \times 4}}{{6 + 4}} + \frac{{7 \times 3}}{{7 + 3}} = 4.5\,\Omega $$

Resistance of wire $$\left( R \right) = \rho \frac{l}{A}$$
If wire is bent in the middle then
$$l' = \frac{l}{2},A' = 2A$$
$$\therefore $$ New resistance, $$R' = \rho \frac{{l'}}{{A'}} = \frac{{\rho \frac{l}{A}}}{{2A}}$$
$$ = \frac{{\rho l}}{{4A}} = \frac{R}{4}.$$

Given ; $$J = a{r^2}.$$
$$\eqalign{ & i = \int_2^2 {J \times 2\pi \,r\,dr} = \int_{\frac{R}{3}}^{\frac{R}{2}} {a{r^2} \times 2\pi \,r\,dr} \cr & = 2\pi a\int_{\frac{R}{3}}^{\frac{R}{2}} {{r^3}dr} = 2\pi \,a\left| {\frac{{{r^4}}}{4}} \right|_{\frac{R}{3}}^{\frac{R}{2}} \cr & = \frac{{\pi a}}{2}\left[ {{{\left( {\frac{R}{2}} \right)}^4} - {{\left( {\frac{R}{3}} \right)}^4}} \right] \cr & = \frac{{\pi a{R^4}}}{2} \times \frac{{65}}{{81 \times 16}} = \frac{{65\pi \,a{R^4}}}{{2592}} \cr} $$

When resistors are connected in series, then effective resistance of series combination
$$\eqalign{ & {R_S} = R + R + \ldots + n\,{\text{terms}} \cr & = nR\,.......\left( {\text{i}} \right) \cr} $$
When resistors are connected in parallel, then effective resistance
$$\eqalign{ & \frac{1}{{{R_P}}} = \frac{1}{R} + \frac{1}{R} + \ldots + n\,{\text{terms}} \cr & {\text{or}}\,\,\frac{1}{{{R_P}}} = \frac{n}{R}\,.......\left( {{\text{ii}}} \right) \cr & {R_P} = \frac{R}{n} \cr} $$
From Eqs. (i) and (ii), we get
$$\frac{{{R_S}}}{{{R_P}}} = \frac{{{n^2}}}{1}$$

As, $$I = \frac{E}{{R + r}}$$
$$E =$$  emf of cell
$$R =$$  external resistance
$$r =$$  internal resistance
$$I =$$  current in circuit
$$\eqalign{ & {\text{or}}\,\,E = I\left( {R + r} \right),2.1 = 0.2\left( {10 + r} \right) \cr & 10 + r = \frac{{2.1}}{2} \times 10 \cr & \therefore r = 10.5 - 10 = 0.5\,\Omega \cr} $$