look at two circuit diagrams that connect the same components in different ways, as shown. which statement…

Explanation:

Step1: Analyze series - circuit resistance

In a series - circuit (Circuit A), the total resistance $R_{totalA}=R_1 + R_2+R_3$ (assuming three bulbs with resistances $R_1$, $R_2$, $R_3$). In a parallel - circuit (Circuit B), the total resistance $R_{totalB}=\frac{1}{\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}}$, so $R_{totalA}>R_{totalB}$.

Step2: Analyze series - circuit failure

In a series - circuit (Circuit A), if one bulb burns out, the circuit is broken and all bulbs go out.

Step3: Analyze parallel - circuit failure

In a parallel - circuit (Circuit B), each bulb is on a separate branch. If one bulb burns out, the other bulbs remain lit because the current has alternative paths.

Step4: Analyze brightness

The brightness of a bulb is related to the power $P = \frac{V^{2}}{R}$. In a parallel - circuit, the voltage across each bulb is the same as the source voltage. In a series - circuit, the voltage is divided among the bulbs. So, bulbs in parallel (Circuit B) are brighter than bulbs in series (Circuit A) when the bulbs are identical.

Answer:

The total resistance in Circuit A is greater than that in Circuit B. When a bulb in Circuit A goes out, all other bulbs go out. The bulbs in Circuit B are brighter than those in Circuit A. When a bulb in Circuit B goes out, the other bulbs keep shining. So the correct statement is: The total resistance in Circuit A is greater than that in Circuit B.