12th Physics Chapter 2 – 3 Mark Questions (Important & Repeated – TN Board)

Prepare for exams with 12th Physics Chapter 2 (Current Electricity) important 3 mark questions. This collection includes most repeated and previously asked questions based on TN Board exams, helping students revise key concepts quickly and score better.

Current Electricity – 3 Mark Questions

12th physics ln2 current electricity important 3 mark questions

Most Repeated 3 Marks

1.Explain the equivalent resistance of a series resistor network?(4)

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When two or more resistors are connected end to end, they are called series connection. In a series circuit, the same current flows through all the resistors because charge cannot accumulate anywhere in the circuit. Let the resistors be $R_1, R_2,$ and $R_3$ , and the current flowing be $I$ .

According to Ohm’s law: $V_1 = IR_1,\quad V_2 = IR_2,\quad V_3 = IR_3$

The total supply voltage is equal to the sum of the potential differences across each resistor:

$V = V_1 + V_2 + V_3$

= IR₁ + IR₂ + IR₃

= I(R₁ + R₂ + R₃)

V = IR_s [R_s is equivalent resistance]

When several resistors are connected in series, the total or equivalent resistance is the sum of the individual resistances

2.Explain the equivalent resistance of a parallel resistor network?(3)

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Definition

Resistors are said to be connected in parallel when they are connected across the same potential difference.

Current in Parallel Connection

The total current $I$ from the battery splits into different branches.

$I = I_1 + I_2 + I_3$

Applying Ohm’s Law

Since voltage across each resistor is same, $I_1 = \frac{V}{R_1}, \quad I_2 = \frac{V}{R_2}, \quad I_3 = \frac{V}{R_3}$

Substituting in the current equation, $I = \frac{V}{R_1} + \frac{V}{R_2} + \frac{V}{R_3}$

$I = V \left(\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}\right)$

If $R_p$ is the equivalent resistance, $I = \frac{V}{R_p}$

Therefore, $\frac{1}{R_p}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$

Important Point

The equivalent resistance in parallel combination is always less than the smallest individual resistance.

3.Internal resistance of a cell using voltmeter?(3)

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The emf $\varepsilon$ of a cell is measured using a high resistance voltmeter without connecting external resistance $R$ .
Since the voltmeter draws very little current, the circuit is considered open. Therefore, the voltmeter reading gives the emf of the cell.
When external resistance $R$ is connected, current $I$ flows through the circuit and the terminal voltage becomes less than emf due to internal resistance $r$ .

Internal resistance of a cell using voltmeter

Using Ohm’s law:

V = IR (1)

Also,

V=ε−Ir

$Ir = \varepsilon – V \qquad (2)$

Dividing equation (2) by equation (1): $\frac{Ir}{IR} = \frac{\varepsilon – V}{V}$

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

Therefore, $r = \left(\frac{\varepsilon – V}{V}\right) R$

Since ε, V and R are known, internal resistance r can be determined. We can also find the total current that flows in the circuit.

4.What is Seebeck Effect? State the application of Seebeck Effect?(3)