Kirchhoff’s law states that the algebraic sum of currents meeting at a point (say at a junction) is equals to zero i.e.,
$$\sum i = 0$$

or $$i = {i_1} + {i_2} + {i_3}$$
$$\eqalign{ & \frac{{dQ}}{{dt}} = \frac{{d{Q_1}}}{{dt}} + \frac{{d{Q_2}}}{{dt}} + \frac{{d{Q_3}}}{{dt}} \cr & {\text{or}}\,\,dQ = d{Q_1} + d{Q_2} + d{Q_3} \cr} $$
So, we can say that when a steady current flows through circuit, then neither accumulation of charges takes place at the junction in the circuit nor any charge is removed from there. That’s why Kirchhoff's first law deals with the conservation of charge.

The given circuit can be shown as,

From figure, $$\frac{P}{Q} = \frac{{10}}{{10}} = 1$$
$$\eqalign{ & \frac{R}{S} = \frac{{10}}{{10}} = 1 \cr & \therefore \frac{P}{Q} = \frac{R}{S} \cr} $$
Hence, Wheatstone bridge is balanced. Therefore, the galvanometer will be ineffective. The above Wheatstone bridge can be redrawn as

Resistances $$P$$ and $$Q$$ are in series, so
$$R' = 10 + 10 = 20\,\Omega $$
Resistances $$R$$ and $$S$$ are in series, so
$$R'' = 10 + 10 = 20\,\Omega $$
Now, $${R'}$$ and $${R''}$$ are in parallel hence, net resistance of the circuit
$$ = \frac{{R' \times R''}}{{R' + R''}} = \frac{{20 \times 20}}{{20 + 20}} = 10\,\Omega $$

Resistance of wire
$$R = \frac{{\rho l}}{A} = \frac{{\rho {l^2}}}{C}\left( {{\text{where}}\,Al = C} \right)$$
∴ Fractional change in resistance
$$\frac{{\Delta R}}{R} = 2\frac{{\Delta l}}{l}$$
∴ Resistance will increase by 0.2%

Applying kirchoff's loop law in $$AB{P_2}{P_1}A$$   we get
$$ - 2i + 5 - 10\,{i_1} = 0\,......\left( {\text{i}} \right)$$

Again applying kirchoff’s loop law in $${P_2}CD{P_1}{P_2}$$   we get, $$10\,{i_1} + 2 - i + {i_1} = 0\,......\left( {{\text{ii}}} \right)$$
From (i) and (ii) $$11{i_1} + 2 - \left[ {\frac{{5 - 10{i_1}}}{2}} \right] = 0$$
$$ \Rightarrow {i_1} = \frac{1}{{32}}\,A\,{\text{from}}\,{P_2}\,{\text{to}}\,{P_1}$$

$$\frac{{{R_{AB}}}}{{{R_{BC}}}} = \frac{{{R_{AD}}}}{{{R_{DC}}}}$$

As bridge is in balanced condition, no current will flow through $$BD.$$
$$\eqalign{ & {R_1} = {R_{AB}} + {R_{BC}} \cr & = R + R \cr & = 2\,R \cr & {R_2} = {R_{AD}} + {R_{CD}} \cr & = R + R \cr & = 2\,R \cr} $$
$${R_1}$$ and $${R_2}$$ are in parallel combination.
Hence, equivalent resistance between $$A$$ and $$C$$ will be
$$\therefore {R_{{\text{eq}}}} = \frac{{{R_1}{R_2}}}{{{R_1} + {R_2}}} = \frac{{4{R^2}}}{{4R}} = R$$

KEY CONCEPT : The current in $$RC$$ circuit is given by $$I = {I_0}{e^{ - \frac{t}{{RC}}}}$$
$$\eqalign{ & {\text{or}}\,lnI = ln{I_0} - \frac{t}{{RC}}{\text{ or }}lnI = \left( {\frac{{ - t}}{{RC}}} \right) + ln{I_0} \cr & lnI = \left( {\frac{{ - t}}{{RC}}} \right) + ln\left( {\frac{{{E_0}}}{R}} \right) \cr} $$
On comparing with $${\text{ }}y = mx + C$$
$${\text{Intercept }} = ln\left( {\frac{{{E_0}}}{R}} \right){\text{ and slope }} = - \frac{1}{{RC}}$$
When $$R$$ is changed to $$2R$$  then slope increases and current becomes less. New graph is $$Q.$$

In discharge tube, the current is due to flow of positive ions and electrons. Moreover, secondary emission of electrons is also possible. So, $$V-I$$  curve is non-linear, hence its resistance is non-ohmic.

KEY CONCEPT :
$$\eqalign{ & {I_g}G = \left( {I - {I_g}} \right)S \cr & {\text{Here,}}\,{I_g} = 100 \times {10^{ - 6}}A;G = 100\,\Omega ;S = 0.1\,\Omega \cr} $$

$$\eqalign{ & \therefore I = {I_g}\left( {\frac{G}{S} + 1} \right) = 100 \times {10^{ - 6}}\left( {\frac{{100}}{{0.1}} + 1} \right) \cr & = 100 \times {10^{ - 6}} \times 1000.1 = 100.01\,mA \cr} $$

Thermistors are usually made of metaloxides with high temperature coefficient of resistivity.

Internal resistance of the cell is given by
$$r = \left( {\frac{{E - V}}{V}} \right)R$$
Given, $$E = 1.5\,V,V = 1.0\,V,R = 2\,\Omega $$
$$\therefore r = \left( {\frac{{1.5 - 1.0}}{{1.0}}} \right) \times 2 = \frac{{0.5}}{{1.0}} \times 2 = 1\,\Omega $$