(See Discussion Question Q25.14.) An ideal ammeter A is placed in a circuit with a battery and a light bulb as shown in Fig. Q25.15a, and the ammeter reading is noted. The circuit is then reconnected as in Fig. Q25.15b, so that the positions of the ammeter and light bulb are reversed. (a) How does the ammeter reading in the situation shown in Fig. Q25.15a compare to the reading in the situation shown in Fig. Q25.15b? Explain your reasoning. (b) In which situation does the light bulb glow more brightly? Explain your reasoning.

The resistance, or the ratio of voltage to current, for all or part of an electric circuit at a fixed temperature, is generally constant.

Ohm's law may be expressed mathematically as

$\frac{\mathrm{V}}{\mathrm{I}}=\mathrm{R}$

Here, V is the voltage, I is the current, and R is the resistor.

Let the resistance of the bulb is R .

We know the resistance of an ideal ammeter is 0 .

So, the current through the circuit in both cases will be

_{$\mathrm{I}=\frac{\mathrm{\epsilon }}{\mathrm{R}}$}

Here, $\epsilon$is the emf voltage.

Therefore, the ammeters in both situations have the same readings since the current is the same in both circuits.

Let the resistance of the bulb is R .

We know the resistance of an ideal ammeter is 0 .

So, the power dissipation in the resistance R is

$\mathrm{P}={\mathrm{I}}^2\mathrm{R}$.

Therefore, the light bulbs glow equally bright since the current is the same in both circuits.