Electric Current:

PYQ-2023-Current-Electricity-Q11

• 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:

PYQ-2023-Current-Electricity-Q15

$$I = nAeV$$ $$v_d = \frac{\lambda}{\tau}$$ $$v_d = \frac{\frac{1}{2}\left(\frac{eE}{m}\right)\tau^2}{\tau} = \frac{1}{2} \frac{eE}{m} \tau$$ $$I = neAV_d$$

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:

PYQ-2023-Electrostatics-Q8, PYQ-2023-Current-Electricity-Q17

$$P = VI$$ $$\text{Energy}= \int P dt$$ $$P = I^2R = VI = \frac{V^2}{R}$$ $$H = VIt = I^2Rt = \frac{V^2}{R}t$$

Kirchhoff’s Laws:

PYQ-2023-Current-Electricity-Q12, PYQ-2023-Current-Electricity-Q19, PYQ-2023-Current-Electricity-Q23

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:

PYQ-2023-Current-Electricity-Q1, PYQ-2023-Current-Electricity-Q3, PYQ-2023-Current-Electricity-Q6, PYQ-2023-Current-Electricity-Q8, PYQ-2023-Current-Electricity-Q13, PYQ-2023-Current-Electricity-Q14, PYQ-2023-Current-Electricity-Q20, PYQ-2023-Current-Electricity-Q22

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:

PYQ-2023-Current-Electricity-Q2, PYQ-2023-Current-Electricity-Q4, PYQ-2023-Current-Electricity-Q7

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:

PYQ-2023-Current-Electricity-Q16

• 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 $$I_{G} \cdot R_{G}=\left(I-I_{G}\right) S$$

$$S=\frac{I_{G} \cdot R_{G}}{I-I_{G}} \quad S=\frac{I_{G} \times R_{G}}{I} \quad when \quad I \gg I_{G.}$$

where; $I=$ Maximum current that can be measured using the given ammeter.

Voltmeter:

PYQ-2023-Current-Electricity-Q21

• 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:

PYQ-2023-Current-Electricity-Q9, PYQ-2023-Experimental-Physics-Q2, PYQ-2023-Experimental-Physics-Q4

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:

PYQ-2023-Current-Electricity-Q10, PYQ-2023-Experimental-Physics-Q1

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 cross-sectional 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:

PYQ-2023-Current-Electricity-Q5

$$ V = I R $$