Opening switch B will cause bulbs 2 and 3 to go out while bulb 1 will stay lit. This outcome is due to the interruption of the current path to bulbs 2 and 3. Understanding the configuration of the circuit is crucial in predicting such changes.
In an electrical circuit where three identical light bulbs are connected, the effect of opening switch B depends on the configuration of the circuit. Assuming a parallel circuit with bulbs 2 and 3 in parallel and bulb 1 in series, opening switch B would affect the circuit differently.
If switch B is opened and it only affects bulbs 2 and 3, the correct answer is A. Bulbs 2 and 3 will go out, but bulb 1 will remain lit. This is because opening the switch would remove the path for current through bulbs 2 and 3, while bulb 1 continues to have a complete circuit to the power source.
Here's a step-by-step explanation:
Initially, when switch B is closed, current flows through all three bulbs, lighting them up.
Opening switch B interrupts the current flow to bulbs 2 and 3.
Since bulb 1 is in series or has an independent path to the battery, it remains unaffected and stays lit.
This illustrates the behavior of current flow and the impact of opening a switch on certain parts of a circuit. Understanding circuit diagrams and the principle of current conservation helps in predicting these outcomes.
