Answer, in brief, any four questions: a) You are given n wires, each of resistance R. What is the ratio of maximum to minimum resistance obtainable from these wires? b) Why do we prefer a…

a) Ratio of maximum to minimum resistance from n wires

Maximum resistance: To obtain the maximum resistance, connect all the wires in series. The total resistance $R_{\text{max}}$ is: $$R_{\text{max}}=R+R+\cdots +R=nR$$

Minimum resistance: To obtain the minimum resistance, connect all the wires in parallel. The total resistance $R_{\text{min}}$ is: $$\frac{1}{R_{\text{min}}}=\frac{1}{R}+\frac{1}{R}+\cdots +\frac{1}{R}=\frac{n}{R}$$ $$R_{\text{min}}=\frac{R}{n}$$

Ratio of maximum to minimum resistance: $$\frac{R_{\text{max}}}{R_{\text{min}}}=\frac{nR}{\frac{R}{n}}=n^2$$

Answer: The ratio of maximum to minimum resistance is $n^2:1$.

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b) Preference of potentiometer over voltmeter for measuring emf

Conclusion: A potentiometer is preferred because it measures the true emf of a cell without any loss due to internal resistance, whereas a voltmeter measures the terminal voltage, which is less than the emf.

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c) Angle of dip and its relation to Earth’s magnetic field

Angle of dip (δ): The angle of dip is the angle made by the Earth’s magnetic field (B) with the horizontal plane at a given location. It is also called the inclination angle.

Relation to Earth’s magnetic field components: The Earth’s magnetic field at any point can be resolved into two components:

1. Horizontal component ($B_H$): Parallel to the Earth’s surface.
2. Vertical component ($B_V$): Perpendicular to the Earth’s surface (downward in the Northern Hemisphere).

The angle of dip is related to these components by the tangent function: $$\tan \delta =\frac{B_V}{B_H}$$ where:

• $\delta$ = angle of dip,
• $B_V$ = vertical component of Earth’s magnetic field,
• $B_H$ = horizontal component of Earth’s magnetic field.

Key points:

• At the magnetic equator, the angle of dip is 0° (only horizontal component).
• At the magnetic poles, the angle of dip is 90° (only vertical component).
• The angle of dip varies with latitude.

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d) Use of soft iron in transformer cores

Why soft iron?

1. High magnetic permeability: Ensures most of the magnetic flux is confined to the core, improving coupling between primary and secondary windings.
2. Low hysteresis and eddy current losses: Reduces heat generation and improves efficiency.
3. Quick magnetization/demagnetization: Follows AC changes efficiently, crucial for transformer operation.

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e) Effect of doubling turns on self-inductance of a solenoid

Self-inductance ($L$) of a solenoid is given by: $$L=\frac{{\mu }_0{N}^{2}A}{l}$$ where:

• $N$ = number of turns,
• $A$ = cross-sectional area,
• $l$ = length of the solenoid,
• ${\mu }_0$ = permeability of free space.

If the number of turns is doubled ($N^{\prime }=2N$): $$L^{\prime }=\frac{{\mu }_0(2N{)}^{2}A}{l}=\frac{4{\mu }_0{N}^{2}A}{l}=4L$$

Answer: The self-inductance increases by a factor of 4.

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f) Peak voltage and frequency from $E=300\sin 314t$

Given equation: $$E=300\sin 314t$$ This is in the standard form: $$E=E_0\sin \omega t$$ where:

• $E_0$ = peak voltage (amplitude),
• $\omega$ = angular frequency in rad/s.

Step 1: Identify peak voltage ($E_0$) $$E_0=300\text{ V}$$

Step 2: Calculate frequency ($f$) The angular frequency $\omega$ is related to frequency by: $$\omega =2\pi f$$ Given $\omega =314$ rad/s: $$f=\frac{\omega }{2\pi }=\frac{314}{2\times 3.1416}\approx 50\text{ Hz}$$

Answer:

• Peak voltage = 300 V
• Frequency = 50 Hz