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

Using Kirchhoff’s law at $$P$$ we get
$$\frac{{V - 12}}{1} + \frac{{V - 13}}{2} + \frac{{V - 0}}{{10}} = 0$$
[Let potential at $$P,\,Q,\,U = 0$$   and at $$R = V$$  ]
$$\eqalign{ & \Rightarrow \frac{V}{1} + \frac{V}{2} + \frac{V}{{10}} = \frac{{12}}{1} + \frac{{13}}{2} + \frac{0}{{10}} \cr & \Rightarrow \frac{{10 + 5 + 1}}{{10}}V = \frac{{24 + 13}}{2} \cr & \Rightarrow V\left( {\frac{{16}}{{10}}} \right) = \frac{{37}}{2} \cr & \Rightarrow V = \frac{{37 \times 10}}{{16 \times 2}} = \frac{{370}}{{32}} = 11.56\,volt \cr} $$

The energy stored in the capacitor $$ = \frac{1}{2}C{V^2}.$$
This energy will be converted into heat in the resistor.
$$\therefore H = \frac{1}{2}C{V^2}$$
where, $$C =$$  capacitance of capacitor
$$V =$$  voltage across the plate of capacitor
$$ \Rightarrow H = \frac{1}{2} \times 4 \times {10^{ - 6}} \times {\left( {400} \right)^2} = 0.32\,J$$

Since current $$I$$ is independent of $${R_6},$$  it follows that the resistance $${R_1},{R_2},{R_3}$$  and $${R_4}$$ must form the balanced
Wheatstone bridge. ∴ $${R_1}{R_4} = {R_2}{R_3}$$

KEY CONCEPT : To convert a galvanometer into a voltmeter we connect a high resistance in series with the galvanometer.
The same procedure needs to be done if ammeter is to be used as a voltmeter.

When a cell is balanced against potential drop across a certain length of potentiometer wire, no current flows through the cell
∴ emf of cell = potential drop across balance length of potentiometer wire.
So, potentiometer is a more accurate device for measuring emf of a cell or no current flows through the cell during measurement of emf.

In an open circuit, emf of cell
$$E = 2.2\,V$$
In a closed circuit, terminal potential difference
$$V = 1.8\,V$$
External resistance, $$R = 5\,\Omega $$
Thus, internal resistance of cell is
$$\eqalign{ & r = \left( {\frac{E}{V} - 1} \right)R = \left( {\frac{{2.2}}{{1.8}} - 1} \right)5 \cr & = \left( {\frac{{11}}{9} - 1} \right)5 = \frac{2}{9} \times 5 = \frac{{10}}{9}\,\Omega \cr} $$

Let each half side has resistance $$r\left( { = \frac{{\rho d}}{2}} \right)$$

$$\eqalign{ & \therefore R = \frac{1}{2}\left[ {2r + \frac{{\left( {2r} \right)\left( {r\sqrt 2 } \right)}}{{\left( {2 + \sqrt 2 } \right)r}}} \right] = r\sqrt 2 \,\,\left( {{\text{on}}\,{\text{solving}}} \right) \cr & \therefore R = \frac{{\rho d}}{{\sqrt 2 }} \cr} $$

$$BC$$ and $$AC$$ are in series
$$\therefore {R_{BCA}} = 30 + 30 = 60\Omega $$
Now $$BA$$ and $$DC$$ are in parallel.
$$\eqalign{ & \frac{1}{{{R_{eq}}}} = \frac{1}{{30}} + \frac{1}{{60}} = \frac{{90}}{{30 \times 60}} \cr & {R_{eq}} = 20\Omega ;\,\,\,\,\,\,\, \cr & V = IR \cr & \Rightarrow I = \frac{2}{{20}} = 0.1\,Amp \cr} $$

From $$KVL$$
$$ - 6 + {3I_1} + 1\left( {{I_i} - {I_2}} \right) = 0$$

$$\eqalign{ & 6 = 3{I_1} + {I_1} - {I_2} \cr & 4{I_1} - {I_2} = 6\,......\left( 1 \right) \cr & - 9 + 2{I_2} - \left( {{I_1} - {I_2}} \right) + 3{I_2} = 0 \cr & - {I_1} + 6{I_2} = 9\,......\left( 2 \right) \cr} $$
On solving (1) and (2)
$$I_1 = 0.13A$$
Direction $$Q$$ to $$P,$$ since $${I_1} > {I_2}.$$
Alternatively

$$\eqalign{ & Eq = \frac{{\frac{{{E_1}}}{{{r_1}}} + \frac{{{E_2}}}{{{r_2}}}}}{{\frac{1}{{{r_1}}} + \frac{1}{{{r_2}}}}} = \frac{{\frac{6}{3} - \frac{9}{5}}}{{\frac{1}{3} + \frac{1}{5}}} = \frac{3}{{8V}} \cr & \therefore I = \frac{{\frac{3}{8}}}{{\frac{{15}}{8} + 1}} = \frac{3}{{23}} = 0.13A \cr} $$
Considering potential at $$P$$ as $$0V$$ and at $$Q$$ as $$x$$ volt, then
$$\eqalign{ & \frac{{x - 6}}{3} + \frac{{x - 0}}{1} + \frac{{x + 9}}{5} = 0 \cr & \therefore x = \frac{2}{{23}} \cr & \therefore i = \frac{{x - 0}}{1} = \frac{2}{{23}} = 0.13A \cr} $$
From $$Q$$ to $$P$$

Energy $$= 2eV$$
$$\eqalign{ & e{V_0} = 2eV \cr & \Rightarrow {V_0} = 2 \cr} $$
So, electric field, $$E = \frac{2}{{4 \times {{10}^{ - 8}}}}$$
$$\eqalign{ & = 0.5 \times {10^8} \cr & = 5 \times {10^7}\,V/m \cr} $$