Current Electricity

Let the emf of the cell be $E$ and its internal resistance be $r$.
The power delivered to external resistance $R$ is:

$$P = I^2 R = \left(\frac{E}{R + r}\right)^2 R$$

Given $P_1 = P_2$ for resistances $R_1$ and $R_2$:

$$\frac{E^2 R_1}{(R_1 + r)^2} = \frac{E^2 R_2}{(R_2 + r)^2}$$

Taking the square root on both sides:

$$\frac{\sqrt{R_1}}{R_1 + r} = \frac{\sqrt{R_2}}{R_2 + r}$$ $$\sqrt{R_1}(R_2 + r) = \sqrt{R_2}(R_1 + r)$$ $$R_2\sqrt{R_1} + r\sqrt{R_1} = R_1\sqrt{R_2} + r\sqrt{R_2}$$ $$r(\sqrt{R_2} - \sqrt{R_1}) = R_2\sqrt{R_1} - R_1\sqrt{R_2} = \sqrt{R_1 R_2}(\sqrt{R_2} - \sqrt{R_1})$$

Since $R_1 \neq R_2$:

$$r = \sqrt{R_1 R_2}$$

Analysis of the circuit with the voltmeter connected:

• Two resistors of $100\,\Omega$ each are in series across a $40\,\text{V}$ DC supply.
• The voltmeter of resistance $R_V = 150\,\Omega$ is connected in parallel with the first $100\,\Omega$ resistor (across points A and B).

• Step 1: Calculate the equivalent resistance between A and B ($R_{AB}$):
  $$R_{AB} = \frac{100 \times 150}{100 + 150} = \frac{15000}{250} = 60\,\Omega$$
• Step 2: Total resistance of the circuit:
  $$R_{\text{total}} = R_{AB} + R_{BC} = 60\,\Omega + 100\,\Omega = 160\,\Omega$$
• Step 3: Total current in the circuit:
  $$I = \frac{V_{\text{total}}}{R_{\text{total}}} = \frac{40\,\text{V}}{160\,\Omega} = 0.25\,\text{A}$$
• Step 4: Reading of the voltmeter (potential difference across AB):
  $$V_{AB} = I \times R_{AB} = 0.25\,\text{A} \times 60\,\Omega = 15\,\text{V}$$

• Step 1: Calculate current in the primary circuit:
  $$I = \frac{E_1}{R_{AB} + r_1} = \frac{2.4\,\text{V}}{10\,\Omega + 2\,\Omega} = \frac{2.4}{12} = 0.2\,\text{A}$$
• Step 2: Potential drop across the potentiometer wire AB ($L = 2\,\text{m}$):
  $$V_{AB} = I \times R_{AB} = 0.2\,\text{A} \times 10\,\Omega = 2.0\,\text{V}$$
• Step 3: Potential gradient along the wire ($k$):
  $$k = \frac{V_{AB}}{L} = \frac{2.0\,\text{V}}{2.0\,\text{m}} = 1.0\,\text{V/m}$$
• Step 4: At balance point ($l = 1.6\,\text{m}$), no current is drawn from the standard cell, so its internal resistance ($0.2\,\Omega$) produces zero voltage drop:
  $$V = k \times l = 1.0\,\text{V/m} \times 1.6\,\text{m} = 1.6\,\text{V}$$