carla draws two circuit diagrams that connect the same components in different ways, as shown. which…

Explanation:

Step1: Analyze series and parallel circuits

Circuit A is a series circuit. In a series circuit, the total resistance (R_{A}=R_1 + R_2+R_3) (assuming each bulb has resistance (R_1 = R_2=R_3 = R)). Circuit B is a parallel circuit. In a parallel circuit, (\frac{1}{R_{B}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}), so (R_{B}=\frac{R}{3}). Since (R_{A}=3R) and (R_{B}=\frac{R}{3}), (R_{A}>R_{B}).

Step2: Analyze bulb - out situation

In a series circuit (Circuit A), if one bulb goes out (opens the circuit), the current stops flowing and all bulbs go out. In a parallel circuit (Circuit B), each bulb has its own independent path for current. If one bulb goes out, the current through the other bulbs remains unchanged.

Step3: Analyze bulb brightness

The power of a bulb is (P = \frac{V^{2}}{R}). In Circuit A, the voltage across each bulb (V_{A - bulb}=\frac{V}{3}) (by voltage division in series, (V) is the source voltage). In Circuit B, the voltage across each bulb (V_{B - bulb}=V). Using (P=\frac{V^{2}}{R}), (P_{A - bulb}=\frac{V^{2}}{9R}) and (P_{B - bulb}=\frac{V^{2}}{R}), so bulbs in Circuit B are brighter.

Answer:

The total resistance in Circuit A is greater than that in Circuit B.