The instantaneous current I is defined as the limit of the average current \(\Delta t\rightarrow 0\) 
\(I=\underset { \Delta t-0 }{ lim } \cfrac { \Delta Q }{ \Delta t } =\cfrac { dQ }{ dt } \)

If a net charge Q passes through any cross section of a conductor in time t, then the current is defined as \(I=\cfrac { Q }{ t } \)

The relation between drift velocity of electrons and current in a wire of cross- sectional area A is
\({ v }_{ d }=\frac { I }{ neA } =\frac { 0.2 }{ 8.4\times { 10 }^{ 28 }\times 1.6\times { 10 }^{ -19 }\times 0.5\times { 10 }^{ -6 } }\)
v_d = 0.03 x 10^-3 m s^-1

E = 570 N C^-1, e = 1.6 x 10^-19 C,
m = 9.11 x 10^-31 kg and a = ?
F = ma = eE
\(a=\frac { eE }{ m } =\frac { 570\times 1.6 \times { 10 }^{ -19 } }{ 9.11\times { 10 }^{ -31 } } \)
\(=\frac { 912\times { 10 }^{- 19 }\times { 10 }^{ 31 } }{ 9.11 } \)
= 1.001 x 10¹⁴ ms^-2

Temperature coefficient of resistance is defined as the ratio of increase in resistivity per degree rise in temperature to its resistivity at T_o.
\(\alpha=\frac{\rho_{T}-\rho_{0}}{\rho_{0}\left(T-T_{0}\right)}\)

Electrical resistivity of a material is defined as the resistance offered to current flow by a conductor of unit length having unit area of cross section.

Current is defined as the ratio of the net (i.e. amount of) charge (Q) passing through any cross section of a conductor to time.
\(I=\frac{Q}{t}\)
Since current is the ratio of two scalar quantities, it is a scalar. In addition current I is defined as the scalar product of the current density and area vector at which the charges cross.
I = \(\vec{J}.\vec{A}\)

The current (rate of flow of charge) in the wire is 
\(I=\frac { Q }{ t } =\frac { 120 }{ 60 } =2A\)

(i) For conductors a is positive. If the temperature of a conductor increases, the average kinetic energy of electrons in the conductor increases. This results in more frequent collisions and hence the resistivity increases.
(ii) The graph of the Even though, the resistivity of conductors like metals varies linearly for a wide range of temperatures, there also exists a nonlinear region at very low temperatures.
(iii) The resistivity approaches some finite value as the temperature approaches absolute zero.
(iv) As the resistance is directly proportional to the resistivity of the material, we can also write the resistance of a conductor at temperature T °C as
\({ R }_{ T }=R\left[ 1+\alpha \left( T-{ T }_{ 0 } \right) \right] \)
\(\alpha =\cfrac { { R }_{ T }-{ R }_{ 0 } }{ { R }_{ 0 }\left( T-{ T }_{ 0 } \right) } =\cfrac { 1\Delta R }{ { R }_{ 0 }\Delta T } \)
\(\alpha =\cfrac { 1 }{ { R }_{ 0 } } \cfrac { \Delta R }{ \Delta T } \)
where \(\Delta R={ R }_{ 1 }-{ R }_{ 0 }\) is a change in resistance during the changing temperature \(\Delta T=T-{ T }_{ 0 }\)

The drift velocity is the average velocity acquired by the electrons inside the conductor when it is subjected to an electric field. The average time between successive collisions is called the mean free time denoted by \(\tau \). The acceleration \(\vec { a } \) experienced by the electron in an electric field \(\vec { E } \)  is given by
\(\vec { a } =\cfrac { -e\vec { E } }{ m } \left( since\vec { F } =-e\vec { E } \right) \)
The drift velocity is given by
\({ \vec { V } }_{ d }=\vec { a } \tau \)
\({ \vec { V } }_{ d }=\cfrac { e\tau }{ m } \vec { E } \)
\({ \vec { V } }_{ d }=-\mu \vec { E } \)
Here \(\mu =\cfrac { e\tau }{ m } \) is the mobility of the electron and it is defined as the magnitude of the drift velocity per unit electric field \(\mu =\cfrac { \left| { \vec { v } }_{ d } \right| }{ \left| \vec { E } \right| } \)

To check which bulb will be fused, the voltage drop across each bulb has to be calculated.
The resistance of a bulb,
\(R=\frac { V^{ 2 } }{ P } =\frac { { (Ratedvoltage) }^{ 2 } }{ Ratedpower } \)
For 20W - 220V bulb,
\({ R }_{ 1 }=\frac { { \left( 220 \right) }^{ 2 } }{ 20 } \Omega =2420\Omega \)
For 100W - 220V bulb,
\({ R }_{ 2 }=\frac { { \left( 220 \right) }^{ 2 } }{ 20 } \Omega =484\Omega \)
Both the bulbs are connected in series. So same current will pass through both the bulbs. The current that passes through the circuit, \(I=\frac { V }{ { R }_{ tot } } \)
R_tot (R₁ + R₂)
R_tot = (484 + 2420) \(\Omega\) = 2904 \(\Omega\)
\(I=\frac { 440V }{ 2904\Omega } \approx 0.151A\)
The voltage drop across the 20W bulb is
\(V_1=IR_1=\frac { 440V }{ 2904 }\times2420 \approx 366.6V\)
The voltage drop across the 100W bulb is
\({ V }_{ 2 }=I{ R }_{ 2 }=\frac { 440 }{ 2904 } 484\approx 73.3A\)
The 20 W bulb will get fused because the voltage across it is more than the voltage rating.

(a) The power delivered by the battery P = VI. Since the bulbs are connected in parallel, the voltage drop across each bulb is the same. If the voltage is kept fixed, then the power is directly proportional to current (P ∝ I). So 60 W bulb draws twice as much as current as 30 W and it will glow brighter than 30 W bulb.
(b) To calculate the resistance of the bulbs, we use the relation \(P=\frac { { v }^{ 2 } }{ R } \) In both the bulbs, the voltage drop is the same, so the power is inversely proportional to the resistance or resistance is inversely proportional to the power \(\left( R∝ \frac { 1 }{ P } \right) \). It implies that, the 30W has twice as much as resistance as 60 W bulb.
(c) When these two bulbs are connected in series, the current passing through each bulb is the same. It is equivalent to two resistors connected in series. The bulb which has higher resistance has higher voltage drop. So 30W bulb will glow brighter than 60W bulb. So the higher power rating does not always imply more brightness and it depends whether bulbs are connected in series or parallel.

\(\mathrm{V}=230 \mathrm{~V}  \)
\(P=\frac{V^2}{R} \text { since } \mathrm{V} \text { is same } \mathrm{P} \propto \frac{1}{R}  \)
\(\frac{1}{R_2}=\frac{1}{\frac{R_1}{2}}+\frac{1}{\frac{R_1}{2}}=\frac{2}{R_1}+\frac{2}{R_1}=\frac{2+2}{R_1}  \)
\(\frac{1}{R_2}=\frac{4}{R_1}  \)
\(\therefore R_2=\frac{R_1}{4}  \)
\(R_1=4 R_2  \)
\(\therefore \frac{P_2}{P_1}=\frac{R_1}{R_2}=\frac{4 R_2}{R_2}=4\)

\(R \propto \frac{1}{A}, R \propto \frac{1}{r^2}  \)
\(R_A \propto \frac{1}{r_A^2}, R_B \propto \frac{1}{r_B^2} \)
\(\frac{r_A}{r_B}=\left(\frac{R_B}{R_A}\right)^{1 / 2}=\left(\frac{R_B}{3 R_B}\right)^{1 / 2}=\frac{1}{3^{\frac{1}{2}}}=\frac{1}{\sqrt{3}}\)

Yellow - 4
Violet - 7
Orange - 10³
Silver - Tolerance - 10%

\(P=\frac{V^2}{R}=\frac{240 \times 240}{120}=480 \ W\)

R = 37 X 102 ± 10 %
The colour of bands corresponding to
3 - orange
7 - violet
102 - red
10% - silver.