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

Applying $$KVL,\,{V_A} + \Sigma V = {V_B} + 2 \times 2 + 2 \times 1$$
$${V_A} - {V_B} - 3 = 4 + 2;\,{V_A} - {V_B} = 9\,V$$

When heating coil is cut into two equal parts and these parts are joined in parallel, then the resistance of the coil is reduced to $$\frac{1}{4}$$ of the previous value. As $$\left( {H \propto \frac{1}{R}} \right)$$   for constant voltage so, energy liberated per second becomes 4 times, i.e. $$4 \times 100 = 400\,J.$$

Resistance of the wire of a semicircle $$ = \frac{{12}}{2} = 6\,\Omega $$
For equivalent resistance between two points on any diameter, $$6\,\Omega $$  and $$6\,\Omega $$  are in parallel.

Mass deposited
$$\eqalign{ & m = Zq \Rightarrow Z \propto \frac{1}{q} \Rightarrow \frac{{{Z_1}}}{{{Z_2}}} = \frac{{{q_2}}}{{{q_1}}}\,......\left( {\text{i}} \right) \cr & {\text{Also }}q = {q_1} + {q_2}\,......\left( {{\text{ii}}} \right) \cr & \Rightarrow \frac{q}{{{q_2}}} = \frac{{{q_1}}}{{{q_2}}} + 1\,\,\,\,\,\,\,\,\left( {{\text{Dividing (ii) by}}\,{q_2}} \right) \cr & \Rightarrow {q_2} = \frac{q}{{1 + \frac{{{q_1}}}{{{q_2}}}}}\,......\left( {{\text{iii}}} \right) \cr} $$
From equations (i) and (iii),
$${q_2} = \frac{q}{{1 + \frac{{{Z_2}}}{{{Z_1}}}}}$$

The distribution of currents in the circuit is shown in the Fig. (a). Due to symmetry, current in arm $$AE$$  is equal to current in the arm $$EB.$$  Since, current in the arm $$CE$$  is equal to the current in the arm $$ED,$$  therefore the resistance of network will not change, if the wires $$CED$$  and $$AEB$$  are disconnected at $$E,$$ as shown in Fig. (b).


Now, resistance of path $$AEB = r + r = 2r$$
Resistance of path $$ACDB = r + \frac{{\left( {2r} \right) \times r}}{{\left( {2r} \right) + r}} + r$$
$$ = \frac{{8r}}{3}$$
The paths $$AEB$$  and $$ACDB$$   are in parallel, therefore the effective resistance between $$A$$ and $$B$$ will be
$$\eqalign{ & \frac{1}{R} = \frac{1}{{2r}} + \frac{3}{{8r}} = \frac{{4 + 3}}{{8r}} = \frac{7}{{8r}} \cr & {\text{or}}\,\,R = \frac{{8r}}{7} \cr & {\text{But}}\,\,r = 1\,\Omega , \cr & {\text{Therefore,}}\,\,R = \frac{{8 \times 1}}{7} = \frac{8}{7}\,\Omega \cr} $$

The potential difference across $$4\,\Omega $$  resistance given by Ohm's law
$$\eqalign{ & = 4 \times {i_1} \cr & = 4 \times 1.2 = 4.8\,V\,\,\left( {{\text{as}}\,{i_1} = 1.2\,A} \right) \cr} $$
As resistances $$4\,\Omega $$  and $$8\,\Omega $$  are in parallel, so potential difference across $$8\,\Omega $$  resistance will also be $$4.8\,V.$$
$$\therefore $$ Current through $$8\,\Omega $$  resistance
$$\eqalign{ & {i_2} = \frac{V}{R} = \frac{{4.8}}{8} = 0.6\,A \cr & \therefore i = {i_1} + {i_2} = 1.2 + 0.6 = 1.8\,A \cr} $$
$$\therefore $$ Potential difference across $$2\,\Omega $$  resistance
$$\eqalign{ & {V_{BC}} = i \times 2 \cr & = 1.8 \times 2 = 3.6\,V \cr} $$

Kirchhoff's first law supports law of conservation of charge. This is because a point in a circuit cannot act as a source or sink of charge.

As $$R \propto \frac{{{V^2}}}{P}$$   or $$R \propto \frac{1}{P},$$   so resistance of heater is less than that of fan.

Effective resistance,

$$\eqalign{ & \frac{1}{{{R_{{\text{eff}}}}}} = \frac{1}{{{R_1}}} + \frac{1}{{{R_2}}}\,.......\left( {\text{i}} \right) \cr & {\text{then,}}\,\,{R_1} = 10 + 30 \cr & {R_1} = 40 \cr & {\text{Now,}}\,\,{R_2} = 90 + 30 = 120 \cr & {R_2} = 120 \cr} $$
By Eq. (i),
$$\eqalign{ & \frac{1}{{{R_{{\text{eff}}}}}} = \frac{1}{{40}} + \frac{1}{{120}} \cr & {R_{{\text{eff}}}} = \frac{{40 \times 120}}{{120 + 40}} = \frac{{4800}}{{160}} = 30\,\Omega \cr} $$
In the balancing condition,
$$\therefore {\text{Current,}}\,\,I = \frac{7}{{\left( {30 + 5} \right)}} = \frac{7}{{35}} = 0.2\,A\,\,\left[ {\because I = \frac{E}{{R + r}}} \right]$$

If each resistor is $$r,$$ then
$${R_0} = 3r\,\,{\text{or}}\,\,r = \frac{{{R_0}}}{3}$$

When keys are closed, then
$${R_{{\text{eq}}}} = 7\frac{r}{3} = \frac{{7\,{R_0}}}{9}$$