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

Given : Number of cells, $$n = 5,$$  emf of each cell $$= E$$
Internal resistance of each cell $$ = r$$
In series, current through resistance $$R$$
$$I = \frac{{nE}}{{nr + R}} = \frac{{5E}}{{5r + R}}$$
In parallel, current through resistance $$R$$
$$I' = \frac{E}{{\frac{r}{n} + R}} = \frac{{nE}}{{r + nR}} = \frac{{5E}}{{r + 5R}}$$
According to question, $$I = I'$$
$$\eqalign{ & \therefore \frac{{5E}}{{5r + 5R}} = \frac{{5E}}{{r + 5R}} \Rightarrow 5r + R = r + 5R \cr & {\text{or}}\,R = r \cr & \therefore \frac{R}{r} = 1 \cr} $$

$$\eqalign{ & R = \frac{{\rho l}}{{\pi {r^2}}}.\,{\text{But}}\,m = \pi {r^2}ld \cr & \therefore \pi {r^2} = \frac{m}{{ld}} \cr & \therefore {R_1} = \frac{{\rho l_1^2d}}{{{m_1}}},{R_2} = \frac{{\rho l_2^2d}}{{{m_2}}};{R_3} = \frac{{\rho l_3^2d}}{{{m_3}}} \cr & {R_1}:{R_2}:{R_3} = \frac{{l_1^2}}{{{m_1}}}:\frac{{l_2^2}}{{{m_2}}}:\frac{{l_3^2}}{{{m_3}}} = 125:45:3 \cr} $$

Electric power, $$P = {i^2}R$$
∴ Current, $$i = \sqrt {\frac{P}{R}} $$
For resistance of $$9\,\Omega ,$$
$$\eqalign{ & {i_1} = \sqrt {\frac{{36}}{9}} = \sqrt 4 = 2\,A \cr & {i_2} = \frac{{{i_1} \times R}}{6} = \frac{{2 \times 9}}{6} = 3\,A \cr} $$
$$I = {i_1} + {i_2} = 2 + 3 = 5\,A$$
So, $${V_2} = I{R_2} = 5 \times 2 = 10\,V$$

Copper sulphate in water dissociates as follows
$$CuS{O_4} \to C{u^{ + + }} + SO_4^{ - - }$$
At cathode,
$$2\,C{u^{ + + }} + 4{e^ - } \to 2Cu$$
At anode,
$$2{H_2}O \to 4{H^ + } + {O_2} + 4{e^ - }$$
Hence, copper is liberated at cathode and oxygen is liberated at anode.

Equivalent resistance of $$n$$ resistances each of $$r$$ ohm in parallel is given by
$$\frac{1}{R} = \frac{1}{r} + \frac{1}{r} + ... + n\,{\text{times}} = \frac{n}{r}$$
So, $$r = nR$$
When these resistances are connected in series, effective resistance is
$$\eqalign{ & R' = r + r + ... + n\,{\text{times}} = nr \cr & \therefore R' = n\left( {nR} \right) = {n^2}R \cr} $$

At null point
$$\frac{{{R_1}}}{{{R_2}}} = \frac{{{R_3}}}{{{R_4}}} = \frac{x}{{100 - x}}$$
If radius of the wire is doubled, then the resistance of $$AC$$ will change and also the resistance of $$CB$$  will change. But since $$\frac{{{R_1}}}{{{R_2}}}$$ does not change so, $$\frac{{{R_3}}}{{{R_4}}}$$ should also not change at null point. Therefore the point $$C$$ does not change.

As voltage of appliance remains constant, the amount of heat produced is given by,
$$H = \frac{{{V^2}}}{R}t\,.......\left( {\text{i}} \right)$$
when resistance is reduced by $$20\% ,$$  new resistance is $$R' = R - 0.2\,R = 0.8\,R$$
$$H' = \frac{{{V^2}}}{{0.8R}}t'\,.......\left( {{\text{ii}}} \right)$$
Equating Eqs. (i) and (ii), we get
$$\eqalign{ & \frac{{{V^2}}}{R}t = \frac{{{V^2}}}{{0.8R}}t' \cr & \Rightarrow t' = 0.8t = 0.8 \times 12 = 9.6\,\min \cr} $$

$$\eqalign{ & Q = at - b{t^2} \cr & i = \frac{{dQ}}{{dt}} = a - 2bt \cr & i = 0\,{\text{for}}\,t = {t_0} = \frac{a}{{2b}}, \cr} $$
i.e., current flow from $$t = 0\,{\text{to}}\,t = {t_0}.$$
The heat produced $$ = \int\limits_0^{{t_0}} {{i^2}R\,dt} .$$   Putting the value of $$i$$ we get heat produced $$ = \frac{{{a^3}R}}{{6b}}.$$

Let $$\theta $$ be the smallest temperature difference that can be detected by the thermocouple, then
$$I \times R = \left( {25 \times {{10}^{ - 6}}} \right)\theta $$
where $$I$$ is the smallest current which can be detected by the galvanometer of resistance $$R.$$
$$\eqalign{ & \therefore {10^{ - 5}} \times 40 = 25 \times {10^{ - 6}} \times \theta \cr & \therefore \theta = {16^ \circ }C. \cr} $$

Faraday's laws are based on the conversion of electrical energy into mechanical energy, which is in accordance with the law of conservation of energy.