a) The figure below shows a series LCR circuit. [figure in the original paper] If the frequency of an ac source is 500 \ Hz then, i) Calculate the phase angle between current and voltage. [2] ii)…

Solution to NEB Physics 2082 (Class 12) – Question 8

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(a) Series LCR Circuit Analysis

Given:

• Frequency of AC source, $f=\frac{500}{\pi }$ Hz
• Assume the following values from the figure (since the figure is not provided, we assume standard values for a typical LCR circuit problem):
  • Inductance, $L=0.1$ H
  • Capacitance, $C=10\mu F=10\times 10^{-6}$ F
  • Resistance, $R=100\Omega$

Step 1: Calculate angular frequency ($\omega$) $$\omega =2\pi f=2\pi \left(\frac{500}{\pi }\right)=1000\text{rad/s}$$

Step 2: Calculate inductive reactance ($X_L$) and capacitive reactance ($X_C$) $$X_L=\omega L=1000\times 0.1=100\Omega$$ $$X_C=\frac{1}{\omega C}=\frac{1}{1000\times 10\times 10^{-6}}=100\Omega$$

Step 3: Calculate phase angle ($\varphi$) The phase angle between current and voltage in an LCR circuit is given by: $$\tan \varphi =\frac{X_L-X_C}{R}$$ Since $X_L=X_C$, the net reactance is zero, and the circuit behaves purely resistive. $$\tan \varphi =\frac{100-100}{100}=0⟹\varphi =0^{\circ }$$

Answer:

• (i) Phase angle = $0^{\circ }$
• (ii) Since $\varphi =0^{\circ }$, the current and voltage are in phase (neither leads nor lags).

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(b) Impedance of an AC Circuit

Impedance ($Z$) is the total opposition offered by an AC circuit to the flow of current. It is a complex quantity given by: $$Z=\sqrt{R^2+(X_L-X_C{)}^2}$$ where:

• $R$ = Resistance
• $X_L$ = Inductive reactance
• $X_C$ = Capacitive reactance

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(c) Force Between Two Parallel Current-Carrying Wires

Given:

• Two wires $P$ and $Q$ carry currents $I_1$ and $I_2$.
• Distance between wires = $r$.

Step 1: Expression for force per unit length Using Ampere’s law and the formula for magnetic force between two current-carrying conductors: $$F/L=\frac{{\mu }_0{I}_1{I}_2}{2\pi r}$$ where:

• $F/L$ = Force per unit length
• ${\mu }_0$ = Permeability of free space ($4\pi \times 10^{-7}\text{Tm/A}$)

Answer:

• (i) Expression for force per unit length: $$\boxed{\frac{F}{L}=\frac{{\mu }_0{I}_1{I}_2}{2\pi r}}$$

• (ii) Nature of force:

  • If currents are in the same direction, the force is attractive.
  • If currents are in opposite directions, the force is repulsive.

• (iii) Effect of halving the distance: If the distance $r$ is halved, the force per unit length becomes: $$\frac{F^{\prime }}{L}=\frac{{\mu }_0{I}_1{I}_2}{2\pi (r/2)}=2\times \frac{{\mu }_0{I}_1{I}_2}{2\pi r}$$ Thus, the force per unit length doubles.

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OR Section (Alternative Solution)

(a) Variation of Thermoelectric EMF with Temperature

The thermoelectric emf ($E$) in a thermocouple varies with the temperature difference between the hot ($T_h$) and cold ($T_c$) junctions. It follows the relation: $$E=\alpha (T_h-T_c)+\frac{1}{2}\beta (T_h^2-T_c^2)+\frac{1}{3}\gamma (T_h^3-T_c^3)$$ where:

• $\alpha$ = Seebeck coefficient (linear term)
• $\beta$ = Nonlinear coefficient (quadratic term)
• $\gamma$ = Higher-order coefficient (cubic term)

For small temperature differences, the emf is approximately linear: $$E\approx \alpha (T_h-T_c)$$

Graphical Representation:

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(b) Potentiometer Calculation

Given:

• Total length of potentiometer wire = $4\text{m}$
• Resistance of wire = $10\Omega$
• Balance length = $240\text{cm}=2.4\text{m}$
• Assume driver cell emf = $E_d=2\text{V}$ (standard value if not given)

Step 1: Calculate current in the circuit $$I=\frac{E_d}{R}=\frac{2}{10}=0.2\text{A}$$

Step 2: Calculate potential gradient (k) $$k=\frac{I\times R_{\text{unit length}}}{L}=\frac{0.2\times 10}{4}=0.5\text{V/m}$$ (Alternatively, $k=\frac{E_d}{L}=\frac{2}{4}=0.5\text{V/m}$)

Step 3: Calculate unknown emf (E) $$E=k\times l=0.5\times 2.4=1.2\text{V}$$

Answer: $$\boxed{E=1.2\text{V}}$$

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(c) Magnetic Field Due to a Straight Current-Carrying Conductor (Ampere’s Law)

Using Ampere’s circuital law: $$\oint \mathbf{B}\cdot d\mathbf{l}={\mu }_0{I}_{\text{enc}}$$ For a long straight wire carrying current $I$, the magnetic field at a distance $r$ is: $$B=\frac{{\mu }_{0}I}{2\pi r}$$

Answer: $$\boxed{B=\frac{{\mu }_{0}I}{2\pi r}}$$