A 16 $$\Omega$$ wire is bent to form a square loop. A 9 V supply having an internal resistance of 1 $$\Omega$$ is connected across one of its sides. The potential drop across the diagonals of the square loop is _________ $$\times 10^{-1}$$ V.

We have a uniform wire of total resistance $$16\,\Omega$$ which is bent into a square. Because all four sides are equal in length, the resistance is shared equally. Hence the resistance of each side is obtained simply by dividing the total resistance by four:

$$R_{\text{side}} = \dfrac{16\,\Omega}{4}=4\,\Omega.$$

Let us name the corners of the square $$A,\,B,\,C,\,D$$ in clockwise order. The battery of emf $$E=9\ \text{V}$$ possesses an internal resistance $$r = 1\,\Omega$$ and is connected across side $$AB$$. The resistor representing side $$AB$$ is therefore directly between the two battery terminals. The remaining three sides $$BC,\,CD,\,DA$$ together form another path between the same two nodes. We can now view the external portion of the circuit between points $$A$$ and $$B$$ as two resistors in parallel:

• direct path $$AB$$  ⇒  $$R_1 = 4\,\Omega$$
• indirect path $$A \to D \to C \to B$$  ⇒  $$R_2 = 4\,\Omega + 4\,\Omega + 4\,\Omega = 12\,\Omega$$

For two resistors in parallel, the equivalent resistance is found from the formula

$$\dfrac1{R_{\text{eq}}} = \dfrac1{R_1} + \dfrac1{R_2}.$$

Substituting $$R_1 = 4\,\Omega$$ and $$R_2 = 12\,\Omega$$ gives

$$\dfrac1{R_{\text{eq}}} = \dfrac1{4} + \dfrac1{12} = \dfrac3{12} + \dfrac1{12} = \dfrac4{12},$$

so

$$R_{\text{eq}} = \dfrac{12}{4} = 3\,\Omega.$$

This $$3\,\Omega$$ resistor represents the whole square loop seen by the battery. It is in series with the internal resistance $$1\,\Omega$$, therefore the total resistance in the circuit is

$$R_{\text{total}} = r + R_{\text{eq}} = 1\,\Omega + 3\,\Omega = 4\,\Omega.$$

Using Ohm’s law $$I = \dfrac{E}{R_{\text{total}}}$$, the current delivered by the battery is

$$I = \dfrac{9\ \text{V}}{4\,\Omega} = 2.25\ \text{A}.$$

The potential drop across the internal resistance is

$$V_r = I\,r = 2.25\ \text{A}\times 1\,\Omega = 2.25\ \text{V}.$$

Therefore the potential difference that actually appears across the external network (points $$A$$ and $$B$$) is

$$V_{AB} = E - V_r = 9\ \text{V} - 2.25\ \text{V} = 6.75\ \text{V}.$$

Because $$R_1$$ and $$R_2$$ are in parallel, the same voltage $$V_{AB}=6.75\ \text{V}$$ is applied to each path. The individual currents are hence

$$I_1 = \dfrac{V_{AB}}{R_1} = \dfrac{6.75}{4} = 1.6875\ \text{A},$$

$$I_2 = \dfrac{V_{AB}}{R_2} = \dfrac{6.75}{12} = 0.5625\ \text{A}.$$

We now calculate the potentials at every corner by choosing point $$B$$ as the reference (0 V).

Side $$AB$$ (direct path) carries current $$I_1$$, so the potential at $$A$$ is

$$V_A = V_{AB} = 6.75\ \text{V}.$$

The indirect path has the same current $$I_2$$ through each of its three equal resistors. The drop across one side in that path is

$$\Delta V = I_2 \times 4\,\Omega = 0.5625 \times 4 = 2.25\ \text{V}.$$

Starting from $$A$$ and moving to $$D$$, $$C$$, $$B$$ we obtain

$$V_D = V_A - 2.25 = 6.75 - 2.25 = 4.5\ \text{V},$$

$$V_C = V_D - 2.25 = 4.5 - 2.25 = 2.25\ \text{V},$$

$$V_B = V_C - 2.25 = 2.25 - 2.25 = 0\ \text{V},$$

just as expected.

The potential difference across diagonal $$AC$$ is therefore

$$V_{AC} = V_A - V_C = 6.75\ \text{V} - 2.25\ \text{V} = 4.5\ \text{V}.$$

In exactly the same way, the potential difference across the other diagonal $$BD$$ is

$$V_{BD} = V_D - V_B = 4.5\ \text{V} - 0\ \text{V} = 4.5\ \text{V}.$$

Thus the potential drop across a diagonal of the square loop is

$$4.5\ \text{V} = 45 \times 10^{-1}\ \text{V}.$$

So, the answer is $$45$$.