Question

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 configuration? not enough information. the resistors are in series. the resistors are in parallel. r1 is in series with the pair, r2 and r3 in parallel with each other.

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.