Question

(See Discussion Question Q25.14.) Will a light bulb glow more brightly when it is connected to a battery as shown in Fig. Q25.16a, in which an ideal ammeter is placed in the circuit, or when it is connected as shown in Fig. 25.16b, in which an ideal voltmeter V is placed in the circuit? Explain your reasoning.

Short answer

The bulb in fig 25.16a glows normallycausingcurrentto passnormally since there is no extra resistance due tothe ammeter. And in fig 25.16b current doesn’t pass at all.

The bulb in fig 25.16a glows more brightly.

Step-by-step solution

Definition of power dissipation

Power dissipation is a measure of how much power $\left(\begin{array}{c}P=I\times V\end{array}\right)$ in a circuit is converted into heat.

The formula of power$\left(\mathrm{P}\right)$ dissipation is written as current $\left(\mathrm{I}\right)$multiply by voltage$\left(\mathrm{V}\right)$

Determine the power dissipation

The resistance of an ideal ammeter is 0 and the resistance of an ideal voltmeter is infinite.

So, inthe first case in figure 25.16a current passes normally since there is no extra resistance due tothe ammeter.

$\mathrm{P}=\mathrm{IV}$

But in 25.16b current doesn’t pass at all since the resistance ofthe voltmeter is infinite

$\mathrm{P}=\mathrm{IV}$

Since $\mathrm{I}=0$the equation will be

$\mathrm{P}=0$

Therefore, in fig 25.16a the bulb glows normally causingthe current to pass normally since there is no extra resistance due to the ammeter. And in fig 25.16b current doesn’t pass at all.

So, the bulb in fig 25.16a glows more brightly.