⚡ PHYSICS • CHAPTER 2 • TN BOARD

12th Physics Current Electricity

Prepare for the Tamil Nadu 12th Physics Public Examination with carefully selected Chapter 2 – Current Electricity Important 5 Mark Questions. These chapter-wise questions are compiled from the last six years of 1st Mid-Term, Quarterly, 2nd Mid-Term, Half-Yearly, Revision Exam 1, Revision Exam 2, and Tamil Nadu Public Examination question papers.

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1.Obtain the relation between current and drift velocity? or Microscopic model of current? (7)

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Consider a conductor with area of cross section $A$ and let an electric field $\vec{E}$ be applied to it from right to left. Suppose there are $n$ n electrons per unit volume in the conductor and assume that all the electrons move with the same drift velocity $\vec{v_d}$ ,

Current, $I = \frac{dQ}{dt}$

Total Charge

$dQ = \text{charge} \times \text{number of electrons in volume element}$

Drift Velocity

$dx = v_d \, dt$

Electrons available in volume of length $dx$

$= \text{volume} \times \text{number of electrons in unit volume}$

$= Adx \times n$

$= (A v_d \, dt)\, n$

dQ=e(Av_ddt)n

Current, $I = \frac{dQ}{dt}$

Substituting $dQ$ , $I = \frac{e \left( A v_d \, dt \right) n}{dt}$

Cancelling $dt$ ,

$I = neA v_d$ This is the relationship between current & Drift Velocity

Current density is defined as the current flowing per unit area.

$J = \frac{\text{Current}}{\text{Area}}$

$J = \frac{I}{A}$

Substituting $J = \frac{neAv_d}{A}$

$J = nev_d$

SI unit of current density: $\frac{A}{m^2} \quad \text{or} \quad Am^{-2}$

In Vector Form $\vec{J} = ne\vec{v_d}$

We know that, $\vec{v_d} = -\frac{e\tau}{m}\vec{E}$

$\vec{J} = ne \left[-\frac{e\tau}{m}\vec{E}\right]$

$\vec{J} = -\frac{ne^2\tau}{m}\vec{E}$

Let, $\sigma = \frac{ne^2\tau}{m}$

where $\sigma$ is called conductivity.

Therefore, $\vec{J} = \sigma \vec{E}$

This equation is called the microscopic form of Ohm’s law.

2.Comparison of emf of two cells with a potentiometer ?(6)

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The positive terminals of $Bt$ , $\varepsilon_1$ , and $\varepsilon_2$ are connected to the same end $C$ .
To compare the emf of two cells, a potentiometer arrangement is used.
The potentiometer wire $CD$ is connected to a battery $Bt$ and key $K$ in series. This forms the primary circuit.
The ends $M$ and $N$ are connected to a DPDT (Double Pole Double Throw) switch.
The cells with emf $\varepsilon_1$ and $\varepsilon_2$ are connected to terminals $M_1,N_1$ and $M_2,N_2$ respectively.
A galvanometer $G$ , high resistance $HR$ , and jockey are connected in the secondary circuit.

comparison of emf of a cell using potentiometer

Procedure

The DPDT switch is pressed towards $M_1,N_1$ so that the cell $\varepsilon_1$ is included in the secondary circuit.
The jockey is moved along the potentiometer wire to obtain zero deflection in the galvanometer.
The balancing length obtained is $l_1$ .
Similarly, the switch is pressed towards $M_2,N_2$ to include the second cell $\varepsilon_2$ .
The balancing length $l_2$ is obtained for zero deflection.

If $r$ is the resistance per unit length of the potentiometer wire and $I$ I is the current through the wire, then potential gradient is,

$\text{Potential gradient} = Ir$

For first cell, $\varepsilon_1 = Ir l_1$

For second cell, $\varepsilon_2 = Ir l_2$

Dividing equation (1) by equation (2), $\frac{\varepsilon_1}{\varepsilon_2} = \frac{Ir l_1}{Ir l_2}$

Result

The ratio of emfs of two cells is equal to the ratio of their balancing lengths.

$\boxed{\frac{\varepsilon_1}{\varepsilon_2} = \frac{l_1}{l_2}}$

By including a rheostat (Rh) in the primary circuit, the experiment can be repeated several times by changing the current flowing through it.

3.Obtain the condition for bridge balance in Wheatstone’s bridge (5)

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Wheatstone’s bridge is an important application of Kirchhoff’s rules.
It is used to compare resistances and in determining the unknown resistance in electrical network.
The bridge consists of four resistances $P$ , $Q$ , $R$ and $S$ connected as shown in the circuit.
A galvanometer $G$ is connected between the points $B$ and $D$ .
The battery is connected between the points $A$ and $C$ .
The current through the galvanometer is $I_G$ and its resistance is $G$ .

Kirchhoff Current Rule at B & D

$I_1 – I_G – I_3 = 0 \qquad (1)$

$I_2 + I_G – I_4 = 0 \qquad (2)$

Kirchhoff’s Voltage Rule

$I_1P + I_GG – I_2R = 0 \qquad [ABDA] \qquad (3)$

$I_1P + I_3Q – I_4S – I_2R = 0 \qquad [ABCDA] \qquad (4)$

Substituting $I_G = 0$ ,

From (1), $I_1 = I_3$

From (2), $I_2 = I_4$

From (3), $I_1P = I_2R \qquad (5)$

Substituting $I_1P = I_2R$ in (4), $I_3Q – I_4S = 0$

$I_3Q = I_4S \qquad (6)$

Dividing equation (5) by equation (6),

$\frac{I_1P}{I_3Q} = \frac{I_2R}{I_4S}$

Since, $I_1 = I_3 \quad \text{and} \quad I_2 = I_4$

$\frac{P}{Q} = \frac{R}{S}$

This is the condition for bridge balance. Only under this condition, galvanometer shows null deflection. Suppose we know the values of two adjacent resistances, the other two resistances can be compared. If three of the resistances are known, the value of unknown resistance (fourth one) can be determined.

4.Explain the determination of unknown resistance using meter bridge (2)

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The meter bridge is another form of Wheatstone’s bridge.
It consists of a uniform wire of manganin $AB$ of one meter length.
This wire is stretched along a metre scale on a wooden board between two copper strips $C$ and $D$ .
Between these two copper strips another copper strip $E$ is mounted to enclose two gaps $G_1$ and $G_2$ .
An unknown resistance $P$ is connected in $G_1$ and a standard resistance $Q$ is connected in $G_2$ .
A jockey (conducting wire-contact maker) is connected to the terminal $E$ on the central copper strip through a galvanometer $G$ and a high resistance $HR$ .
A Leclanche cell and a key $K$ are connected between the ends of the bridge wire.
The position of the jockey on the wire is adjusted so that the galvanometer shows zero deflection.
Let the position of jockey on the wire be at $J$ .
The resistances corresponding to $AJ$ and $JB$ of the bridge wire form the resistances $R$ and $S$ of the Wheatstone’s bridge.
Then, for bridge balance,

$\frac{P}{Q} = \frac{AJ}{JB} = \frac{l_1}{l_2}$

$\frac{P}{Q} = \frac{l_1}{l_2}$

$P = Q \frac{l_1}{l_2}$

To find the specific resistance of the material of the wire in the coil $P$ P, the radius $a$ a and length $l$ l of the wire are measured.

The specific resistance or resistivity $\rho$ can be calculated using the relation,

$\text{Resistance} = \rho \frac{l}{A}$

By rearranging the above equation,

$\rho = \text{Resistance} \times \frac{A}{l}$

If $P$ is the unknown resistance, equation becomes,

$\rho = P \frac{\pi a^2}{l}$

5.Internal resistance of a cell using Potentiometer ? (1)

▶️ Watch Vide0

To measure the internal resistance of a cell, the circuit connections are made using a potentiometer arrangement.
The end $C$ of the potentiometer wire is connected to the positive terminal of the battery $Bt$ .
The negative terminal of the battery is connected to the end $D$ through a key $K_1$ . This forms the primary circuit.
The positive terminal of the cell of emf $\varepsilon$ , whose internal resistance is to be determined, is also connected to the end $C$ of the wire.
The negative terminal of the cell $\varepsilon$ ε is connected to a jockey through a galvanometer and a high resistance.
A resistance box $R$ and key $K_2$ are connected across the cell $\varepsilon$ . With $K_2$ open, the balancing point $J$ is obtained and the balancing length $CJ = l_1$ is measured.
Since the cell is in open circuit, its emf is measured.

ε=l₁

The current passing through the cell and resistor $R$ is

$I = \frac{\varepsilon}{R+r}$

Potential difference across $R$

$V = \frac{\varepsilon R}{R+r}$

When potentiometer is balanced

$\frac{\varepsilon R}{R+r} = l_2$

Substituting $\varepsilon = l_1$ , $\frac{l_1R}{R+r} = l_2$

$\frac{R}{R+r} = \frac{l_2}{l_1}$

$\frac{R+r}{R} = \frac{l_1}{l_2}$

$1+\frac{r}{R} = \frac{l_1}{l_2}$

$\frac{r}{R} = \frac{l_1}{l_2} – 1$

$\frac{r}{R} = \frac{l_1-l_2}{l_2}$

$r = R \left(\frac{l_1-l_2}{l_2}\right)$

Substituting the values of the R, l₁ and l₂, the internal resistance of the cell is determined. The experiment can be repeated for different values of R. It is found that the internal resistance of the cell is not constant but increases with increase of external resistance connected across its terminals.

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12th Physics Current Electricity

02

Five Marks

TN Board

Most Repeated 5 Mark Questions

1.Obtain the relation between current and drift velocity? or Microscopic model of current? (7)

Total Charge

Drift Velocity

Electrons available in volume of length dxdx

2.Comparison of emf of two cells with a potentiometer ?(6)

Procedure

Result

3.Obtain the condition for bridge balance in Wheatstone’s bridge (5)

Kirchhoff Current Rule at B & D

Kirchhoff’s Voltage Rule

4.Explain the determination of unknown resistance using meter bridge (2)

5.Internal resistance of a cell using Potentiometer ? (1)

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Electrons available in volume of length $dx$