Current Electricity
Electric Current:
PYQ2023CurrentElectricityQ11

Average current: $$I_{av} = \frac{\Delta q}{\Delta t}$$

Instantaneous current: $$i = \lim_{\Delta t \to 0} \frac{\Delta q}{\Delta t} = \frac{dq}{dt}$$
Electric Current in a Conductor:
PYQ2023CurrentElectricityQ15
$$I = nAeV$$ $$v_d = \frac{\lambda}{\tau}$$ $$v_d = \frac{\frac{1}{2}\left(\frac{eE}{m}\right)\tau^2}{\tau} = \frac{1}{2} \frac{eE}{m} \tau$$ $$I = neAV_d$$
Current Density:
$$\vec{J} = \frac{dI}{dS} \vec{n}$$
Electrical Resistance:
$$I = neAV_d = neA\left(\frac{eE}{2m}\right)\tau = \left(\frac{ne^2 \tau}{2m}\right)AE$$
$$E = \frac{V}{\ell}$$
so $$I = \left(\frac{ne^2 \tau}{2m}\right)\left(\frac{A}{\ell}\right)V = \left(\frac{A}{\rho \ell}\right)V = \frac{V}{R} \Rightarrow V = IR$$
Resistivity: $$\rho = \frac{2m}{ne^2 \tau} = \frac{1}{\sigma},$$
where $\sigma$ is conductivity.
Dependence of Resistance on Temperature: $$R = R_0(1 + \alpha \theta).$$
Electrical Power:
PYQ2023ElectrostaticsQ8, PYQ2023CurrentElectricityQ17
$$P = VI$$ $$\text{Energy}= \int P dt$$ $$P = I^2R = VI = \frac{V^2}{R}$$ $$H = VIt = I^2Rt = \frac{V^2}{R}t$$
Kirchhoff’s Laws:
PYQ2023CurrentElectricityQ12, PYQ2023CurrentElectricityQ19, PYQ2023CurrentElectricityQ23
Kirchhoff’s Current Law (Junction law): $$\Sigma I_{in} = \Sigma I_{out}$$
Kirchhoff’s Voltage Law (Loop law): $$\Sigma IR + \Sigma \text{EMF} = 0.$$
Combination of Resistances:
PYQ2023CurrentElectricityQ1, PYQ2023CurrentElectricityQ3, PYQ2023CurrentElectricityQ6, PYQ2023CurrentElectricityQ8, PYQ2023CurrentElectricityQ13, PYQ2023CurrentElectricityQ14, PYQ2023CurrentElectricityQ20, PYQ2023CurrentElectricityQ22
Resistances in Series: $$R = R_1 + R_2 + R_3 + \ldots + R_n$$
Resistances in Parallel: $$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$$
Resistance of a Conductor:
PYQ2023CurrentElectricityQ2, PYQ2023CurrentElectricityQ4, PYQ2023CurrentElectricityQ7
The resistance $R$ of a conductor can be calculated using the formula: $$R = \rho \frac{L}{A}$$
Wheatstone Network:
When current through the galvanometer is zero (null point or balance point), $$\frac{P}{Q} = \frac{R}{S}$$
Grouping of Cells:
PYQ2023CurrentElectricityQ16
 Cells in Series: Equivalent EMF $E_{eq} = E_1 + E_2 + \ldots + E_n$
 Cells in Parallel: $E_{eq} = \frac{\varepsilon_1 / r_1 + \varepsilon_2 / r_2 + \ldots + \varepsilon_n / r_n}{1 / r_1 + 1 / r_2 + \ldots + 1 / r_n}$
Ammeter:

A shunt (small resistance) is connected in parallel with a galvanometer to convert it into an ammeter.

An ideal ammeter has zero resistance.
Ammeter is represented as follows 
If maximum value of current to be measured by ammeter is I then $$I_{G} \cdot R_{G}=\left(II_{G}\right) S$$
$$S=\frac{I_{G} \cdot R_{G}}{II_{G}} \quad S=\frac{I_{G} \times R_{G}}{I} \quad when \quad I \gg I_{G.}$$
where; $I=$ Maximum current that can be measured using the given ammeter.
Voltmeter:
PYQ2023CurrentElectricityQ21

A high resistance is put in series with a galvanometer.

It is used to measure the potential difference across a resistor in a circuit.
For maximum potential difference
$$V = I_{G}.R_{S}+ I_{G}R_{G}$$
$$R_{S}=\frac{V}{I_{G}}R_{G}$$
$$\text { if } \quad R_{G}«R_{S} \Rightarrow R_{S} \approx \frac{V}{I_{G}}$$
Potentiometer:
PYQ2023CurrentElectricityQ9, PYQ2023ExperimentalPhysicsQ2, PYQ2023ExperimentalPhysicsQ4
Used for comparing EMFs, measuring internal resistance of cells, and calibrating ammeters and voltmeters.
$$V_{A}V_{B}=\frac{\varepsilon}{R+r} \cdot R$$
Potential gradient (x): Potential difference per unit length of wire
$$x=\frac{V_{A}V_{B}}{L}=\frac{\varepsilon}{R+r} \cdot \frac{R}{L}$$
Application of potentiometer
(a)To find emf of unknown cell and compare emf of two cells.
In case I,
In figure (1) is joint to (2) then balance length $=\ell_{1} $
In case II,
$$\varepsilon_{1}=x \ell_{1} \hspace{10mm}…(i)$$
In figure (3) is joint to (2) then balance length $=\ell_{2}$
$$\varepsilon_{2}=\mathrm{x} \ell_{2} \hspace{10mm}…(ii)$$
$$\frac{\varepsilon_{1}}{\varepsilon_{2}}=\frac{\ell_{1}}{\ell_{2}}$$
If any one of $\varepsilon_{1}$ or $\varepsilon_{2}$ is known the other can be found. If $\mathrm{x}$ is known then both $\varepsilon_{1}$ and $\varepsilon_{2}$ can be found
(b) To find current if resistance is known
$$V_{A}V_{C}=x \ell_{1}$$
$$R_{1}=x\ell_{1}$$
$$I=\frac{x \ell_{1}}{R_{1}}$$
Similarly, we can find the value of $R_{2}$ also.
Potentiometer is ideal voltmeter because it does not draw any current from circuit, at the balance point.
(c) To find the internal resistance of cell.
Ist arrangement $\hspace{60mm}$ 2nd arrangement
by first arrangement: $$\varepsilon^{\prime}=\mathrm{x} \ell_{1} \hspace{10mm}…(i)$$
by second arrangement: $$\mathrm{IR}=\mathrm{x} \ell_{2}$$
$$I=\frac{\mathrm{x} \ell_{2}}{R}, \quad \text { also } I=\frac{\varepsilon^{\prime}}{r^{\prime}+R}$$
$$\therefore \quad \frac{\varepsilon^{\prime}}{r^{\prime}+R} = \frac{xl_{2}}{R} \quad \Rightarrow \frac{xl_{1}}{r^{\prime}+R} = \frac{xl_{2}}{R}$$
$$=\left[\frac{\ell_{1}\ell_{2}}{\ell_{2}}\right] R$$
(d) Ammeter and voltmeter can be graduated by potentiometer.
(e) Ammeter and voltmeter can be calibrated by potentiometer.
Metre Bridge:
PYQ2023CurrentElectricityQ10, PYQ2023ExperimentalPhysicsQ1
Used to measure unknown resistance using the principle of a balanced Wheatstone bridge.
If $A B=\ell \mathrm{cm}$, then $B C=(100\ell) \mathrm{cm}$.
Resistance of the wire between $A$ and $B, R \propto \ell$
[ $\because$ Specific resistance $\rho$ and crosssectional area A are same for whole of the wire ]
$$ \text { or } \quad R=\sigma \ell \hspace{10mm}…(i) $$
where $\sigma$ is resistance per $\mathrm{cm}$ of wire.
(a)
(b)
If $P$ is the resistance of wire between $A$ and $B$ then
$$ P \propto \ell \Rightarrow \quad P=\sigma(\ell) $$
Similarly, if $Q$ is resistance of the wire between $B$ and $C$, then
$$ \begin{array}{ll} & Q \propto 100\ell \ \therefore & Q=\sigma(100\ell)\hspace{10mm}….(2) \end{array} $$
Dividing (1) by (2), $ \frac{P}{Q}=\frac{\ell}{100\ell}$
Applying the condition for balanced Wheatstone bridge, we get $R Q=P X$
$$ \therefore \quad x=R \frac{Q}{P} \quad \text { or } \quad x=\frac{100\ell}{\ell} R $$
Since $\mathrm{R}$ and $\ell$ are known, therefore, the value of $\mathrm{X}$ can be calculated.
Ohm’s law:
PYQ2023CurrentElectricityQ5
$$ V = I R $$