Question

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

circuit a (series) and circuit b (parallel) diagrams

which statement about the circuits is correct?

• the total resistance in circuit a is greater than that in circuit b.
• when a bulb in circuit a goes out, the other bulbs keep shining.
• the bulbs in circuit a shine brighter than those in circuit b.
• when a bulb in circuit b goes out, the other bulbs go out.

Explanation:

Step1: Analyze Circuit Types

Circuit A is a series circuit (components in a single path), Circuit B is a parallel circuit (components in multiple paths).

Step2: Evaluate Resistance

In series, total resistance \( R_{total} = R_1 + R_2 + R_3 \). In parallel, \( \frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \), so parallel total resistance is less than any individual \( R \). Thus, Circuit A (series) has greater total resistance than Circuit B (parallel).

Step3: Evaluate Bulb Behavior in Series

In series (Circuit A), if one bulb fails (open circuit), the whole circuit is broken—other bulbs go out. So the statement about Circuit A’s bulbs staying on is wrong.

Step4: Evaluate Bulb Brightness

Brightness depends on power \( P = \frac{V^2}{R} \) (for parallel) or \( P = I^2 R \) (for series). In series, current \( I \) is same, but total voltage is split. In parallel, each bulb gets full voltage. For same bulbs, parallel bulbs have higher power (brighter) than series. So Circuit B’s bulbs are brighter—statement about Circuit A’s bulbs being brighter is wrong.

Step5: Evaluate Bulb Behavior in Parallel

In parallel (Circuit B), each bulb has its own path. If one bulb fails, others still have current—they stay on. So the statement about Circuit B’s bulbs going out is wrong.

Answer:

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