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

Given, $$l = 4m,$$
$$R =$$  potentiometer wire resistance $$ = 8\,\Omega $$
Potential gradient $$ = \frac{{dV}}{{dr}} = 1\,mV/cm$$
So, for $$400\,cm,\,\Delta V = 400 \times 1 \times {10^{ - 3}} = 0.4\,V$$
Let $$a$$ resistor $${R_s}$$ connected in series, so as
$$\eqalign{ & \Delta V = \frac{V}{{R + {R_s}}} \times R \Rightarrow 0.4 = \frac{2}{{8 + R}} \times 8 \cr & \Rightarrow 8 + R = \frac{{16}}{{0.4}} = 40 \Rightarrow R = 32\,\Omega \cr} $$

According to Ohm's law
$$\frac{{dV}}{{dI}} = - r\,\,{\text{and}}\,V = \varepsilon \,\,{\text{if}}\,\,I = 0\,\,\left[ {{\text{As}}\,\,V + Ir = \varepsilon } \right]$$
So, slope of the graph $$ = -r$$  and intercept $$ = \varepsilon $$

KEY CONCEPT : We know that
$${R_t} = {R_0}\left( {1 + \alpha t} \right),$$
where $${R_t}$$ is the resistance of the wire at $${t^ \circ }C,$$
$${R_0}$$ is the resistance of the wire at $${0^ \circ }C$$
and $$\alpha $$ is the temperature coefficient of resistance.
$$\eqalign{ & \Rightarrow {R_{50}} = {R_0}\left( {1 + 50\alpha } \right)......\left( {\text{i}} \right) \cr & {R_{100}} = {R_0}\left( {1 + 100\alpha } \right)......\left( {{\text{ii}}} \right) \cr} $$
From (i), $${R_{50}} - {R_0} = 50\alpha {R_0}......\left( {{\text{iii}}} \right)$$
From (ii), $${R_{100}} - {R_0} = 100\alpha {R_0}......\left( {{\text{iv}}} \right)$$
Dividing (iii) by (iv), we get
$$\eqalign{ & \frac{{{R_{50}} - {R_0}}}{{{R_{100}} - {R_0}}} = \frac{1}{2} \cr & {\text{Here,}}\,{R_{50}} = 5\Omega {\text{ and }}{R_{100}} = 6\Omega \cr & \therefore \frac{{5 - {R_0}}}{{6 - {R_0}}} = \frac{1}{2} \cr & {\text{or,}}\,6 - {R_0} = 10 - 2{R_0}{\text{ or,}}\,{R_0} = 4\Omega . \cr} $$

After stretching, specific resistance $$\left( \rho \right)$$  will remain same.
Original resistance of wire, $$R = \frac{{\rho l}}{A}$$
Ratio of resistance before and after streching,
$$\eqalign{ & \frac{{{R_2}}}{{{R_1}}} = \frac{{{l_2}}}{{{l_1}}} \times \frac{{{A_1}}}{{{A_2}}} = {\left[ {\frac{{{l_2}}}{{{l_1}}}} \right]^2} \cr & \frac{{{R_2}}}{{{R_1}}} = \frac{{{{\left( {l + 10\% l} \right)}^2}}}{{{l^2}}} \cr & \frac{{{R_2}}}{{{R_1}}} = \frac{{{{\left( {\frac{{11}}{{10}}l} \right)}^2}}}{{{l^2}}} \cr & {R_2} = 1.21\,{R_1} \cr} $$

Using, $$I = neA{v_d}$$
$$\eqalign{ & \therefore {\text{Drift}}\,{\text{speed}}\,{v_d} = \frac{I}{{neA}} \cr & \frac{{1.5}}{{9'{{10}^{28}}'1.6'{{10}^{ - 19}}'5'{{10}^{ - 6}}}} \cr & = 0.02 \times {10^{ - 3}}m{s^{ - 1}} \cr & = 0.02mm{s^{ - 1}} \cr} $$

Let unknown resistance be $$X.$$

Then condition of Wheatstone bridge gives,
$$\frac{X}{R} = \frac{{20}}{{80}}$$
where, $$X =$$  unknown resistance
$$R =$$  known resistance
Hence, unknown resistance is
$$\eqalign{ & X = \frac{{20}}{{80}} \times R = \frac{1}{4} \times 1\,\,\left( {\because R = 1\,\Omega } \right) \cr & = 0.25\,\Omega \cr} $$

Voltage across $$2\,\Omega ,$$
$$ = 3 \times 2 = 6\,V$$
Voltage across $$4\,\Omega $$  and $$\left( {5\,\Omega + 1\,\Omega } \right)$$   resistor is same.
So, current across $$5\,\Omega = \frac{6}{{1 + 5}} = 1\,A\,\,\left[ {i = \frac{V}{R}} \right]$$
Power across $$5\,\Omega = P = {i^2}R$$
$$\eqalign{ & = {\left( 1 \right)^2} \times 5 \cr & = 5\,W \cr} $$

$${\theta _n} = \frac{{{\theta _i} + {\theta _c}}}{2}.$$

The total charge enclosed in the dotted portion when the switch $$S$$ is open is zero. When the switch is closed and steady state is reached, the current $$I$$ coming from the battery is
$$9 = I\left( {3 + 6} \right) \Rightarrow I = 1A$$

∴ Potential difference across $$3\Omega $$ resistance = $$3V$$ and
potential difference across $$6\Omega $$ resistance = $$6V$$
∴ p.d. across $$3\,\mu F$$ capacitor = $$3V$$
and p.d. across $$6\,\mu F$$ capacitor = $$6V$$
∴ Charge on $$3\,\mu F$$ capacitor $${Q_1} = 3 \times 3 = 9\mu C$$
Charge on $$6\,\mu F$$ capacitor $${Q_2} = 6 \times 6 = 36\mu C$$
∴ Charge passing the switch $$ = 36 - 9 = 27\,\mu C$$

As the voltage is same for both forty and thirty nine bulbs combination, therefore heat produced is given by $$H = \frac{{{V^2}t}}{R}$$   and $$H \propto \frac{1}{R}.$$   As equivalent resistance decreases, the combination of 39 bulbs will glow more.