Class 12 Physics NCERT Solutions: Important Questions on Current Electricity

Current Electricity is a crucial chapter in Class 12 Physics that covers various fundamental concepts related to the flow of electric charge, circuits, and electrical components. Understanding these concepts is essential for mastering physics and performing well in exams. This article provides a comprehensive guide to the important questions on "Current Electricity" for Class 12, focusing on NCERT solutions that will help you prepare effectively.

Overview of Current Electricity in NCERT Class 12 Physics

The chapter on Current Electricity in the NCERT Class 12 Physics textbook covers several important topics such as:

1. Electric Current
2. Ohm's Law
3. Resistance and Resistivity
4. Electromotive Force (EMF) and Potential Difference
5. Combination of Resistors (Series and Parallel)
6. Kirchhoff’s Laws
7. Wheatstone Bridge
8. Meter Bridge
9. Potentiometer
10. Electric Power and Energy

Let's dive into the important questions and solutions based on these topics that will help you prepare for your Class 12 Physics exams.

Important Questions and Solutions on Current Electricity

1. Define Electric Current and its SI Unit. Explain the difference between Conventional Current and Electron Flow.

• Solution: Electric current is defined as the rate of flow of electric charge through a conductor. The SI unit of electric current is the Ampere (A).

  • Conventional Current: This is the flow of positive charge from the positive terminal to the negative terminal of a battery.
  • Electron Flow: In reality, the current is due to the flow of electrons, which move from the negative terminal to the positive terminal. Thus, the electron flow is opposite to the direction of conventional current.

2. State Ohm's Law. Describe its limitations and the factors on which resistance depends.

• Solution: Ohm’s Law states that the current flowing through a conductor between two points is directly proportional to the potential difference across the two points, provided the temperature and other physical conditions remain constant. Mathematically, it is represented as:

  $$V = IR$$

  where:

  • $V$ is the potential difference,
  • $I$ is the current, and
  • $R$ is the resistance.

  Limitations of Ohm's Law:

  • It is not applicable to non-ohmic conductors, such as semiconductors and diodes.
  • It does not hold true if physical conditions like temperature are not constant.

  Factors Affecting Resistance:

  • Material of the Conductor: Different materials have different resistivities.
  • Length of the Conductor (L): Resistance is directly proportional to the length of the conductor, $R \propto L$.
  • Cross-sectional Area (A): Resistance is inversely proportional to the cross-sectional area, $R \propto \frac{1}{A}$.
  • Temperature: Resistance usually increases with an increase in temperature for conductors.

3. What is Resistivity? How does it vary with temperature for conductors, semiconductors, and insulators?

• Solution: Resistivity ($\rho$) is a property of a material that quantifies how strongly it resists or conducts electric current. The SI unit of resistivity is ohm-meter ($\Omega \cdot m$).

  • For Conductors: Resistivity increases with an increase in temperature. This is because, as temperature increases, atoms vibrate more vigorously, causing more collisions with electrons.
  • For Semiconductors: Resistivity decreases with an increase in temperature. As temperature increases, more electrons get enough energy to jump to the conduction band.
  • For Insulators: Generally, resistivity decreases slightly with temperature, but insulators usually have a very high resistivity that does not change significantly with temperature.

4. Derive the formula for the equivalent resistance when resistors are connected in series and parallel.

• Solution:

  • Series Connection: When resistors are connected in series, the same current flows through all resistors. The total or equivalent resistance ($R_s$) is the sum of the individual resistances:

    $$R_s = R_1 + R_2 + R_3 + \ldots + R_n$$

  • Parallel Connection: When resistors are connected in parallel, the voltage across each resistor is the same, but the current divides among them. The reciprocal of the equivalent resistance ($R_p$) is the sum of the reciprocals of the individual resistances:

    $$\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \ldots + \frac{1}{R_n}$$

5. Explain Kirchhoff's Laws of Electrical Circuits with examples.

• Solution:

  • Kirchhoff’s Current Law (KCL): It states that the algebraic sum of currents entering a junction is equal to the algebraic sum of currents leaving the junction. Mathematically,

    $$\sum I_{\text{in}} = \sum I_{\text{out}}$$

  • Kirchhoff’s Voltage Law (KVL): It states that the algebraic sum of all the potential differences in any closed loop of a circuit is equal to zero. Mathematically,

    $$\sum V = 0$$

  Example: Consider a circuit with two resistors $R_1$ and $R_2$ connected in series with a battery of EMF $E$. By applying KVL, we get:

  $$E - I R_1 - I R_2 = 0$$

  This can be used to find the current $I$ flowing through the circuit.

6. What is a Wheatstone Bridge? Derive the condition for its balance.

• Solution: A Wheatstone bridge is a circuit used to measure an unknown resistance by balancing two legs of a bridge circuit. It consists of four resistors arranged in a diamond shape.

  Condition for Balance:

  For the Wheatstone bridge to be balanced, the ratio of the two resistances in one branch must be equal to the ratio of the two resistances in the other branch:

  $$\frac{R_1}{R_2} = \frac{R_3}{R_4}$$

  When this condition is satisfied, there is no current through the galvanometer connected between the two branches.

7. Describe the principle and working of a Potentiometer. How is it used to compare the EMF of two cells?

• Solution: A potentiometer is an instrument used for measuring the EMF of a cell and comparing the EMFs of two cells. It works on the principle that the potential drop across a segment of a wire of uniform cross-sectional area and composition is directly proportional to its length.

  Comparison of EMF of Two Cells:

  • Connect the positive terminals of both cells to the same end of the potentiometer wire.

  • The other terminals are connected to a galvanometer.

  • Adjust the sliding contact for the first cell and note the balancing length $L_1$.

  • Repeat the same for the second cell and note the balancing length $L_2$.

  • The EMFs $E_1$ and $E_2$ of the two cells are related by:

    $$\frac{E_1}{E_2} = \frac{L_1}{L_2}$$

8. Calculate the power dissipated in a resistor of resistance $R$ when connected to a battery of EMF $E$ and internal resistance $r$.

• Solution: The power dissipated in a resistor is given by:

  $$P = I^2 R$$

  where $I$ is the current flowing through the circuit. Using Ohm's Law and considering the internal resistance $r$ of the battery:

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

  Therefore, the power dissipated in the resistor $R$ is:

  $$P = \left( \frac{E}{R + r} \right)^2 R$$

9. Explain the working of a Meter Bridge. How is it used to measure unknown resistance?

• Solution: A meter bridge is based on the principle of the Wheatstone bridge and is used to measure an unknown resistance by balancing two legs of the bridge. It consists of a 1-meter long wire of uniform cross-sectional area, with a known resistance and an unknown resistance placed in the two gaps.

  Working:

  • Adjust the sliding contact until the galvanometer shows zero deflection.

  • Note the lengths $l_1$ and $l_2$ on either side of the bridge.

  • The unknown resistance $R_x$ is given by:

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

  where $R$ is the known resistance.

10. What is Electrical Power? Derive the expressions for electrical power in terms of $I$, $V$, and $R$.

• Solution: Electrical power is defined as the rate at which electrical energy is consumed or converted into other forms of energy. The SI unit of electrical power is the watt (W).

  Expressions for Electrical Power:

  • In terms of current and voltage: $P = IV$

  • Using Ohm's Law, $V = IR$, power can also be expressed as:

    $$P = I^2 R$$

  • Similarly, using $I = \frac{V}{R}$, power can be expressed as:

    $$P = \frac{V^2}{R}$$

Conclusion

The "Current Electricity" chapter in Class 12 Physics is vital for understanding the behavior of electric circuits, components, and various electrical phenomena. Mastering the important questions and concepts in this chapter is essential for scoring well in the exams. The NCERT solutions provided here will help you comprehend the key topics, solve complex problems, and gain confidence in tackling the exam questions effectively. Practice regularly and use these solutions as a guide to excel in your studies.