a) What do you mean by shunt? Describe its use in converting a galvanometer into an ammeter. b) State Joule's law of heating and verify it experimentally. c) State Biot and Savart law. Derive an…

(a) Shunt and its use in converting a galvanometer into an ammeter

Meaning of Shunt

A shunt is a low-resistance conductor connected in parallel with a galvanometer to convert it into an ammeter. It allows most of the current to bypass the galvanometer, protecting it from damage when measuring large currents.

Working Principle

A galvanometer is a sensitive instrument that measures small currents (typically in microamperes or milliamperes). To measure larger currents, a shunt resistor $R_s$ is connected in parallel with the galvanometer of resistance $R_g$. The current $I$ to be measured splits into two parts:

• $I_g$: Current through the galvanometer (must be within its range).
• $I_s$: Current through the shunt (major portion).

Since the shunt has very low resistance, most of the current flows through it, while only a small fraction passes through the galvanometer.

Derivation of Shunt Resistance

Let:

• $I$ = Total current to be measured.
• $I_g$ = Current through the galvanometer (full-scale deflection current).
• $R_g$ = Resistance of the galvanometer.
• $R_s$ = Resistance of the shunt.

Since the galvanometer and shunt are in parallel: $$V_g=V_s⟹I_g{R}_g=I_s{R}_s$$ But $I=I_g+I_s$. Substituting $I_s=I-I_g$: $$I_g{R}_g=(I-I_g)R_s$$ Solving for $R_s$: $$R_s=\frac{I_g{R}_g}{I-I_g}$$ This gives the required shunt resistance to convert the galvanometer into an ammeter of range $I$.

Advantages of Using a Shunt

1. Extends Range: Allows measurement of large currents without damaging the galvanometer.
2. Low Cost: Cheaper than using a high-range galvanometer.
3. Accuracy: Maintains precision if the shunt resistance is accurately calculated.

(b) Joule’s Law of Heating and Experimental Verification

Joule’s Law of Heating

Joule’s law states that:

The heat produced $H$ in a conductor is directly proportional to:

• Square of the current $I^2$ flowing through it,
• Resistance $R$ of the conductor, and
• Time $t$ for which the current flows.

Mathematically: $$>H=I^{2}Rt>$$

Experimental Verification

Aim: To verify that the heat produced in a resistor is proportional to $I^{2}Rt$.

Apparatus:

• Battery (DC supply)
• Resistor (nichrome wire)
• Ammeter
• Voltmeter
• Stopwatch
• Calorimeter (to measure heat)
• Connecting wires

Circuit Setup (Refer to Figure 2):

1. Connect a resistor $R$ in series with an ammeter and a battery.
2. Connect a voltmeter across the resistor to measure voltage $V$.
3. Immerse the resistor in a calorimeter filled with water and note the initial temperature ${\theta }_1$.
4. Pass current $I$ for time $t$ and measure the final temperature ${\theta }_2$.
5. Calculate the heat produced $H=mc({\theta }_2-{\theta }_1)$, where $m$ is the mass of water and $c$ is the specific heat capacity.

Procedure:

1. Measure resistance $R$ using a multimeter.
2. Pass different currents $I_1,I_2,I_3$ (by adjusting the battery voltage) for the same time $t$.
3. Record the temperature rise $\Delta \theta$ for each current.
4. Calculate $H$ and compare with $I^{2}Rt$.

Observations and Calculations:

Conclusion: The heat produced $H$ increases with $I^2$, verifying Joule’s law.

(c) Biot-Savart Law and Magnetic Field Due to a Long Straight Conductor

Biot-Savart Law

The Biot-Savart law gives the magnetic field $d\mathbf{B}$ at a point due to a small current element $Id\mathbf{l}$: $$d\mathbf{B}=\frac{{\mu }_0}{4\pi }\frac{Id\mathbf{l}\times \hat{\mathbf{r}}}{r^2}$$ where:

• ${\mu }_0=4\pi \times 10^{-7}\text{Tm/A}$ (permeability of free space),
• $\hat{\mathbf{r}}$ is the unit vector from the current element to the point,
• $r$ is the distance from the current element to the point.

Magnetic Field Due to a Long Straight Conductor

Consider a long straight wire carrying current $I$. We derive the magnetic field at a point $P$ at a perpendicular distance $a$ from the wire (Refer to Figure 3).

Assumptions:

• The wire is infinitely long.
• The magnetic field at $P$ is due to contributions from all current elements $Id\mathbf{l}$.

Derivation:

1. Consider a small element $dl$ at a distance $x$ from the foot of the perpendicular from $P$.
2. The distance from $dl$ to $P$ is $r=\sqrt{a^2+x^2}$.
3. The magnetic field contribution $dB$ is perpendicular to the plane containing $dl$ and $P$.
4. Using symmetry, the net field is tangential to a circle of radius $a$.

From Biot-Savart law: $$dB=\frac{{\mu }_0}{4\pi }\frac{Idl\sin (90^{\circ })}{r^2}=\frac{{\mu }_0}{4\pi }\frac{Idl}{a^2+x^2}$$ The component along the tangent (perpendicular to $r$) is: $$d{B}_{\text{net}}=dB\cos \theta =\frac{{\mu }_0}{4\pi }\frac{Idla}{(a^2+x^2{)}^{3/2}}$$ Integrate over the entire wire (from $x=-\infty$ to $x=+\infty$): $$B={\int }_{-\infty }^{\infty }\frac{{\mu }_{0}Ia}{4\pi (a^2+x^2{)}^{3/2}}dx$$ Using the substitution $x=a\tan \theta$, the integral evaluates to: $$B=\frac{{\mu }_{0}I}{2\pi a}$$ This is the magnetic field at a distance $a$ from a long straight conductor.

(d) Alternating Current in an L-R Circuit

Circuit Description

An AC circuit contains:

• A resistor $R$,
• An inductor $L$,
• An AC voltage source $V=V_0\sin (\omega t)$.

The current $I$ lags the voltage by a phase angle $\varphi$.

Impedance of the Circuit

The impedance $Z$ is given by: $$Z=\sqrt{R^2+X_L^2}$$ where $X_L=\omega L$ is the inductive reactance.

Current in the Circuit

Using Ohm’s law for AC circuits: $$I=\frac{V}{Z}=\frac{V_0\sin (\omega t)}{\sqrt{R^2+(\omega L{)}^2}}$$ The current can be written as: $$I=I_0\sin (\omega t-\varphi )$$ where:

• $I_0=\frac{V_0}{Z}=\frac{V_0}{\sqrt{R^2+(\omega L{)}^2}}$ is the peak current,
• $\varphi ={\tan }^{-1}\left(\frac{\omega L}{R}\right)$ is the phase angle.

Phase Relation

The voltage leads the current by $\varphi$ (Refer to Figure 4). This is because the inductor opposes changes in current, causing a lag.

Power in the Circuit

The average power $P_{\text{avg}}$ dissipated is: $$P_{\text{avg}}=\frac{1}{2}{V}_0{I}_0\cos \varphi =V_{\text{rms}}{I}_{\text{rms}}\cos \varphi$$ where $\cos \varphi$ is the power factor.

flowchart TD
    A["AC Voltage Source\nV = V₀ sin(ωt)"] --> B["Resistor R"]
    A --> C["Inductor L"]
    B --> D["Current I = I₀ sin(ωt - φ)"]
    C --> D
    D --> E["Phase Lag φ = tan⁻¹(ωL/R)"]
    D --> F["Impedance Z = √(R² + (ωL)²)"]