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

According to Faraday’s first law of electrolysis $$m = Z \times q$$
$$\eqalign{ & {\text{For}}\,{\text{same}}\,q,\,\,\,\,\,\,\,m \propto Z \cr & \therefore \frac{{{m_{Cu}}}}{{{m_{Zn}}}} = \frac{{{Z_{Cu}}}}{{{Z_{Zn}}}} \cr & \Rightarrow {m_{Cu}} = \frac{{{Z_{Cu}}}}{{{Z_{Zn}}}} \times {m_{Zn}} = \frac{{31.5}}{{32.5}} \times 0.13 = 0.126g \cr} $$

$$\eqalign{ & R = \frac{{\rho {\ell _1}}}{{{A_1}}},{\text{now}}\,{\ell _2} = 2{\ell _1} \cr & {A_2} = \pi {\left( {{r_2}} \right)^2} = \pi {\left( {2{r_1}} \right)^2} = 4\pi r_1^2 = 4{A_1} \cr & \therefore {R_2} = \frac{{\rho \left( {2{\ell _1}} \right)}}{{4{A_1}}} = \frac{{\rho \ell }}{{2A}} = \frac{R}{2} \cr} $$
$$\therefore $$ Resistance is halved, but specific resistance remains the same.

Current in circuit connected with battery of emf $$E$$ with internal resistance $$r$$ is given by
Current, $$i = \frac{E}{{R + r}}$$
I
$$2 = \frac{E}{{2 + r}}\,.......\left( {\text{i}} \right)$$
II
$$0.5 = \frac{E}{{9 + r}}\,.......\left( {{\text{ii}}} \right)$$
From Eqs. (i) and (ii), we have, $$\frac{2}{{0.5}} = \frac{{9 + r}}{{2 + r}}$$
$$4 = \frac{{9 + r}}{{2 + r}} \Rightarrow 3r = 1 \Rightarrow r = \frac{1}{3}\Omega $$

$$\eqalign{ & I \propto \frac{1}{{{r^2}}}; \cr & V \propto \frac{1}{r}; \cr & V \propto {r^0} \cr} $$

The current upto which bulb of marked $$25 W - 220 V,$$   will not fuse $${I_1} = \frac{{{W_1}}}{{{V_1}}} = \frac{{25}}{{220}}Amp$$
$${\text{Similarly,}}\,{I_2} = \frac{{{W_2}}}{{{V_2}}} = \frac{{100}}{{220}}Amp$$
The current flowing through the circuit

$$\eqalign{ & I = \frac{{440}}{{{R_{eff}}}},{R_{eff}} = {R_1} + {R_2} \cr & {R_1} = \frac{{V_1^2}}{{{P_1}}} = \frac{{{{\left( {220} \right)}^2}}}{{25}};\,\,{R_2} = \frac{{V_2^2}}{P} = \frac{{{{\left( {220} \right)}^2}}}{{100}} \cr & I = \frac{{440}}{{\frac{{{{\left( {220} \right)}^2}}}{{25}} + \frac{{{{\left( {220} \right)}^2}}}{{100}}}} = \frac{{440}}{{{{\left( {220} \right)}^2}\left[ {\frac{1}{{25}} + \frac{1}{{100}}} \right]}} \cr & I = \frac{{40}}{{220}}Amp \cr & \because {I_1}\left( { = \frac{{25}}{{220}}A} \right) < I\left( { = \frac{{40}}{{220}}A} \right) < {I_2}\left( { = \frac{{100}}{{200}}A} \right) \cr} $$
Thus the bulb marked $$25 W - 220 W$$   will fuse.

$$\frac{P}{Q} = \frac{R}{S}\,{\text{where}}\,S = \frac{{{S_1}{S_2}}}{{{S_1} + {S_2}}}$$

Let $${A_1},{A_2},{A_3}$$   be the area of cross-section of three wires of copper of masses $${m_1},{m_2},{m_3}$$   and length $${l_1},{l_2},{l_3}$$   respectively. Given $${m_1} = m,{m_2} = 3m,{m_3} = 5m,{l_1} = 5l,{l_2} = 3l,{l_3} = l$$
$${\text{Mass}} = {\text{volume}} \times {\text{density}}$$
So, $$m = {A_1} \times 5l \times \rho \,......\left( {\text{i}} \right)$$
$$\eqalign{ & 3m = {A_2} \times 3l \times \rho \,\,......\left( {{\text{ii}}} \right) \cr & 5m = {A_3} \times l \times \rho \,......\left( {{\text{iii}}} \right) \cr} $$
From Eqs. (i) and (ii), we get
$${A_2} = 5{A_1}$$
From Eqs. (i) and (iii),
$$\eqalign{ & {A_3} = 25{A_1} \cr & \therefore {R_1} = \rho \frac{{{l_1}}}{{{A_1}}} = \rho \frac{{5l}}{{{A_1}}} \cr & {R_2} = \rho \frac{{{l_2}}}{{{A_2}}} = \rho \times \frac{{3l}}{{5{A_1}}} = \frac{3}{{25}}{R_1} \cr & {R_3} = \rho \frac{{{l_3}}}{{{A_3}}} = \frac{{\rho \times l}}{{25{A_1}}} = \frac{{{R_1}}}{{125}} \cr & \therefore {R_1}:{R_2}:{R_3} = {R_1}:\frac{3}{{25}}{R_1}:\frac{{{R_1}}}{{125}} \cr & = 125:15:1 \cr} $$

We know that $$P = \frac{{{V^2}}}{R}$$
For constant value of potential difference $$\left( V \right)$$  we have $$P \propto \frac{1}{R}$$
Case (i)

This is a case of balanced Wheatstone bridge $${R_1} = 1\Omega $$
Case (ii)

Clearly the equivalent resistance $$\left( {{R_2}} \right)$$ will be less than $$1\Omega .$$
Case (iii)

$$\eqalign{ & {\text{Thus}}\,{R_3} = 2\Omega \cr & {\text{Since,}}\,{R_2} < {R_1} < {R_3} \cr & \therefore {P_2} > {P_1} > {P_3} \cr} $$

Since due to wrong connection of each cell the total emf reduced to $$2\varepsilon $$ then for wrong connection of three cells the total emf will reduced to $$\left( {n\varepsilon - 6\varepsilon } \right)$$   whereas the total or equivalent resistance of cell combination will be $$nr.$$

$$\eqalign{ & {R_1} = {R_0}\left( {1 + \alpha {t_1}} \right) \Rightarrow 1 = {R_0}\left[ {1 + 0.00125 \times 27} \right] \cr & {R_2} = {R_0}\left( {1 + \alpha {t_2}} \right) \Rightarrow 2 = {R_0}\left[ {1 + 0.00125 \times {t_2}} \right] \cr} $$
On solving we get
$${T_2} = {854^ \circ }C = 1127\,K$$