a circuit has three 10 ω resistors and a 9 v source. if an ammeter reads 0.3 a next to r2, what is the…

Explanation:

Step1: Analyze the current - reading and voltage - source information

We know the voltage of the source $V = 9V$ and the current through the ammeter next to $R_2$ is $I=0.3A$. Using Ohm's law $V = IR$, if the resistors were in series, the total resistance $R_{total}$ for a series - connected circuit with three $10\Omega$ resistors would be $R_{total}=10 + 10+10=30\Omega$, and the current in the circuit $I=\frac{V}{R_{total}}=\frac{9}{30}=0.3A$. But this is the current through $R_2$ only. If they were in parallel, the equivalent resistance $R_{eq}$ of three $10\Omega$ resistors in parallel is $\frac{1}{R_{eq}}=\frac{1}{10}+\frac{1}{10}+\frac{1}{10}=\frac{3}{10}$, so $R_{eq}=\frac{10}{3}\Omega$, and the total current in the circuit would be $I_{total}=\frac{V}{R_{eq}}=\frac{9}{\frac{10}{3}} = 2.7A$. Since the ammeter reads $0.3A$ next to $R_2$, it means the resistors are not in parallel.

Step2: Determine the circuit configuration

Since the current through one of the resistors ($R_2$) is the same as the current that would flow through a single - resistor in a series - connected circuit of three equal resistors with the given voltage source, the resistors are in series.

Answer:

The resistors are in series.