Draw the circuit using ideal circuit elements:

Now, the series current is:

$$I = \dfrac{\mathcal{E}}{R_{internal}+ R_{load}}$$

The voltage across the internal resistance is:

$$V = \mathcal{E} \dfrac{R_{internal}}{R_{internal}+ R_{load}} $$

The power dissipated by the internal resistance can by found three equivalent ways:

$$P = VI = \dfrac{V^2}{R_{internal}} = I^2 R_{internal} = \mathcal{E}^2 \dfrac{R_{internal}}{(R_{internal}+ R_{load})^2}$$

Clearly, setting $R_{load} = 0$ yields:

$$I = \dfrac{\mathcal{E}}{R_{internal} + 0} = \dfrac{\mathcal{E}}{R_{internal}}$$

$$V = \mathcal{E} \dfrac{R_{internal}}{R_{internal + 0}} = \mathcal{E}$$

$$P = \dfrac{\mathcal{E}^2}{R_{internal}}$$

Simply put, the entire emf appears across the internal resistance. Zero volts appears across the *source plus internal resistance* due to the short circuit.

My answer is r. 2 resistances on the vertical line are redundant as no current flows via that route due to the symmetry of the problem.

Consider this circuit joined to the source with A joined to the positive terminal, then current will equally split along two possible routes from A. Due to symmetry of the problem, the electric potential energy above the vertical line is same as the point below it. So no current flows via that part of the circuit. So that section of wire can be removed, yet giving same equivalent resistance. In our simplified circuit, 2 resistors of resistance 2r are in parallel giving equivalent resistance r.

## Best Answer

look at the final equivalent circuit and solve you will get c)39/7