Answer any two numerical questions. a) Two resistors of resistance 1000 and 2000 are joined in series with a 100V supply. A voltmeter of internal resistance 4000 is connected to measure the…

a) Reading shown by the voltmeter

Given:

• Resistor $R_1=1000\Omega$
• Resistor $R_2=2000\Omega$
• Supply voltage $V=100\text{V}$
• Voltmeter internal resistance $R_v=4000\Omega$

Circuit Analysis: The voltmeter is connected across the $1000\Omega$ resistor, so it is in parallel with $R_1$. The equivalent resistance of the parallel combination is:

$$R_{eq}=\frac{R_1\times R_v}{R_1+R_v}=\frac{1000\times 4000}{1000+4000}=\frac{4,000,000}{5000}=800\Omega$$

Now, the total resistance of the circuit is:

$$R_{total}=R_{eq}+R_2=800+2000=2800\Omega$$

Current in the circuit:

$$I=\frac{V}{R_{total}}=\frac{100}{2800}=\frac{1}{28}\text{A}$$

Voltage across the parallel combination (which is the voltmeter reading):

$$V_{voltmeter}=I\times R_{eq}=\frac{1}{28}\times 800=\frac{800}{28}=\frac{200}{7}\approx 28.57\text{V}$$

Final Answer: The voltmeter reading is $\frac{200}{7}\text{V}$ (≈ 28.57 V).

---

b) Ratio of deflections of two galvanometers

Given:

• Galvanometer 1: $N_1=50$ turns, $R_1=10\Omega$
• Galvanometer 2: $N_2=500$ turns, $R_2=600\Omega$
• Cell emf $E=25\text{V}$
• Internal resistance $r=50\Omega$

Theory: The deflection $\theta$ of a galvanometer is proportional to the current $I$ flowing through it:

$$\theta \propto I$$

The current in each galvanometer is given by:

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

where $R$ is the resistance of the galvanometer.

Calculating currents:

For Galvanometer 1:

$$I_1=\frac{25}{10+50}=\frac{25}{60}=\frac{5}{12}\text{A}$$

For Galvanometer 2:

$$I_2=\frac{25}{600+50}=\frac{25}{650}=\frac{1}{26}\text{A}$$

Ratio of deflections:

Since deflection is proportional to current:

$$\frac{{\theta }_1}{{\theta }_2}=\frac{I_1}{I_2}=\frac{\frac{5}{12}}{\frac{1}{26}}=\frac{5}{12}\times 26=\frac{130}{12}=\frac{65}{6}$$

Final Answer: The ratio of the deflections is $\frac{65}{6}$ (≈ 10.83).

---

c) Instantaneous induced emf in the coil

Given: Magnetic flux $\varphi =4t^2+5t+2$ (in weber), where $t$ is in seconds.

Theory: The induced emf $\mathcal{E}$ is given by Faraday’s law:

$$\mathcal{E}=-\frac{d\varphi }{dt}$$

Differentiating $\varphi$ with respect to $t$:

$$\frac{d\varphi }{dt}=\frac{d}{dt}(4t^2+5t+2)=8t+5$$

At $t=2\text{sec}$:

$$\frac{d\varphi }{dt}{|}_{t=2}=8(2)+5=16+5=21\text{Wb/s}$$

Magnitude of induced emf:

$$|\mathcal{E}|=\left|-\frac{d\varphi }{dt}\right|=21\text{V}$$

Final Answer: The magnitude of the instantaneous induced emf at $t=2\text{sec}$ is 21 V.