Current Electricity

Question 12:

Q: A 6 V battery is connected in series with two resistors, 10 $\Omega$ and 5 $\Omega$ . Find the total power dissipated in the circuit.

Solution:
First, calculate the total resistance in the circuit:

$$R_{\text{total}} = 10 + 5 = 15 \, \Omega$$

Now, calculate the total current in the circuit using Ohm’s Law:

$$I = \frac{V}{R_{\text{total}}} = \frac{6}{15} = 0.4 \, \text{A}$$

Now, calculate the total power dissipated:

$$P = I^2 R_{\text{total}} = (0.4)^2 \times 15 = 0.16 \times 15 = 2.4 \, \text{W}$$

Thus, the total power dissipated in the circuit is $2.4 \, \text{W}$ .

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Question 13:

Q: Explain the phenomenon of "current leakage" in resistors and how it affects the overall resistance in a circuit.

Solution:
Current leakage refers to the unwanted loss of current from a conductor, often due to the presence of external factors such as humidity, temperature fluctuations, or insulation defects. This leads to an increase in effective resistance because the leakage path provides an alternative route for current flow, which diverts a portion of the current and reduces the efficiency of the circuit.

The leakage current creates an additional path for current, thus increasing the total resistance. In extreme cases, it can cause the circuit to behave unpredictably, reducing the current flowing through the intended load and causing potential safety hazards. Proper insulation and shielding are critical to preventing current leakage.

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Question 14:

Q: Derive the relation for the electric field inside a uniformly charged spherical shell using Gauss's Law.

Solution: Gauss’s Law states that the electric flux through a closed surface is proportional to the charge enclosed within that surface:

$$\Phi_E = \frac{Q_{\text{enc}}}{\epsilon_0}$$

Consider a uniformly charged spherical shell with charge $Q$ and radius $R$ . Let the Gaussian surface be a sphere with radius $r$ inside the shell.

Since the charge is distributed uniformly on the shell, the electric field inside the shell at any point (where $r < R$ ) is zero. This is because no charge is enclosed by the Gaussian surface inside the shell.

Therefore, by Gauss’s Law:

$$\Phi_E = \int \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enc}}}{\epsilon_0}$$

Since there is no charge enclosed inside the spherical shell, $Q_{\text{enc}} = 0$ , so the electric flux is zero:

$$\Phi_E = 0$$

Thus, the electric field inside the shell is zero:

$$E = 0$$

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Question 15:

Q: A capacitor of capacitance $C = 4 \, \mu\text{F}$ is charged to a potential difference of $20 \, \text{V}$ . Calculate the energy stored in the capacitor.

Solution:
The energy stored in a capacitor is given by the formula:

$$U = \frac{1}{2} C V^2$$

Substitute the given values:

$$U = \frac{1}{2} \times 4 \times 10^{-6} \times (20)^2$$ $$U = \frac{1}{2} \times 4 \times 10^{-6} \times 400 = 800 \times 10^{-6} \, \text{J} = 0.8 \, \text{mJ}$$

Thus, the energy stored in the capacitor is $0.8 \, \text{mJ}$ .

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Question 16:

Q: Explain the concept of "Kirchhoff's loop rule" and apply it to calculate the current in a circuit consisting of three resistors in series.

Solution: Kirchhoff’s loop rule states that the sum of the potential differences (voltages) around any closed loop or circuit is always zero:

$$\sum V = 0$$

Consider a circuit with three resistors in series, $R_1$ , $R_2$ , and $R_3$ , connected to a battery with emf $E$ . The current $I$ flows through all resistors, so the total voltage drop across the resistors should equal the emf of the battery.

The voltage drop across each resistor is given by Ohm's Law:

$$V_1 = I R_1, \quad V_2 = I R_2, \quad V_3 = I R_3$$

Using Kirchhoff's loop rule:

$$E = I (R_1 + R_2 + R_3)$$

Solving for the current $I$ :

$$I = \frac{E}{R_1 + R_2 + R_3}$$

Thus, the current in the circuit is $\frac{E}{R_1 + R_2 + R_3}$ .

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Question 17:

Q: How does the resistance of a conductor depend on the temperature? Derive the expression for the change in resistance with temperature.

Solution: The resistance of a conductor depends on temperature according to the equation:

$$R_T = R_0 (1 + \alpha T)$$

where:

• $R_T$ is the resistance at temperature $T$ ,
• $R_0$ is the resistance at reference temperature (usually $0^\circ C$ ),
• $\alpha$ is the temperature coefficient of resistance,
• $T$ is the temperature change from $0^\circ C$ .

This equation indicates that the resistance increases linearly with temperature for most metallic conductors.

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Question 18:

Q: Two capacitors of capacitance $5 \, \mu\text{F}$ and $10 \, \mu\text{F}$ are connected in series across a $12 \, \text{V}$ battery. Calculate the equivalent capacitance of the combination.

Solution: For capacitors in series, the reciprocal of the equivalent capacitance $C_{\text{eq}}$ is the sum of the reciprocals of the individual capacitances:

$$\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2}$$

Substitute the given values:

$$\frac{1}{C_{\text{eq}}} = \frac{1}{5} + \frac{1}{10} = \frac{2}{10} + \frac{1}{10} = \frac{3}{10}$$ $$C_{\text{eq}} = \frac{10}{3} = 3.33 \, \mu\text{F}$$

Thus, the equivalent capacitance is $3.33 \, \mu\text{F}$ .

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Question 19:

Q: A wire of resistance $10 \, \Omega$ is stretched to double its length. What is the new resistance of the wire?

Solution: When a wire is stretched, its length $L$ becomes twice the original length. The resistance $R$ of a wire is given by:

$$R = \rho \frac{L}{A}$$

where:

• $\rho$ is the resistivity of the material,
• $L$ is the length,
• $A$ is the cross-sectional area.

If the length is doubled, the cross-sectional area $A$ becomes halved (because the volume of the wire remains constant). Thus, the new resistance is:

$$R' = \rho \frac{2L}{A/2} = 4R$$

Therefore, the new resistance is $4 \times 10 = 40 \, \Omega$ .

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Question 20:

Q: A current of 3 A flows through a conductor of length 2 m and cross-sectional area $1 \, \text{mm}^2$ . If the resistivity of the material is $2 \times 10^{-7} \, \Omega \, \text{m}$ , calculate the potential difference across the conductor.

Solution: The resistance $R$ of the conductor is given by:

$$R = \frac{\rho L}{A}$$

Substitute the given values:

$$R = \frac{(2 \times 10^{-7}) (2)}{1 \times 10^{-6}} = 0.4 \, \Omega$$

Now, using Ohm’s Law, the potential difference $V$ across the conductor is:

$$V = IR = 3 \times 0.4 = 1.2 \, \text{V}$$

Thus, the potential difference is $1.2 \, \text{V}$ .

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Question 21:

Q: Explain the concept of "superconductivity" and its significance in practical applications.

Solution: Superconductivity is a phenomenon in which certain materials, at very low temperatures, exhibit zero electrical resistance. This means that once an electrical current is established in a superconducting material, it can flow indefinitely without any energy loss.

The significance of superconductivity lies in its potential applications, such as in the development of highly efficient power transmission lines, powerful magnets for MRI machines, and advanced particle accelerators. However, the requirement for extremely low temperatures (near absolute zero) currently limits its widespread use.

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Question 22:

Q: A wire of resistance $R$ is bent into a circular loop. Find the effective resistance between two points on the wire when the loop is connected to a battery.

Solution: When a wire is bent into a circular loop, the resistance between two points is equivalent to two resistors (half the length each) connected in parallel. Each half of the loop has resistance $R/2$ .

Thus, the effective resistance between the two points is:

$$R_{\text{eff}} = \frac{R/2 \times R/2}{R/2 + R/2} = \frac{R}{4}$$

Thus, the effective resistance between the two points is $R/4$ .

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Question 23:

Q: Derive the expression for the time constant of an RC circuit and explain its significance.

Solution: The time constant $\tau$ of an RC circuit is given by:

$$\tau = R C$$

This represents the time required for the charge on the capacitor to either charge up to 63% of its maximum value or discharge to 37% of its initial value.

Significance: The time constant determines the speed of charging and discharging of a capacitor in an RC circuit. A larger time constant means slower charging/discharging.

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Question 24:

Q: A battery of emf 12 V is connected in series with two resistors $4 \, \Omega$ and $8 \, \Omega$ . Find the current in the circuit and the potential difference across each resistor.

Solution: The total resistance in the circuit is:

$$R_{\text{total}} = 4 + 8 = 12 \, \Omega$$

Now, calculate the total current using Ohm’s Law:

$$I = \frac{V}{R_{\text{total}}} = \frac{12}{12} = 1 \, \text{A}$$

Now, calculate the potential difference across each resistor:

$$V_1 = IR_1 = 1 \times 4 = 4 \, \text{V}, \quad V_2 = IR_2 = 1 \times 8 = 8 \, \text{V}$$

Thus, the current is $1 \, \text{A}$ , and the potential differences are $4 \, \text{V}$ across $4 \, \Omega$ and $8 \, \text{V}$ across $8 \, \Omega$ .

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