A simple series circuit consists of three identical bulbs, a battery, and a wire with a switch as shown in the diagram. When the switch is open, all three bulbs glow with the same brightness. Then the switch is closed. When the switch is closed, the brightness of bulb A (increased, decreased, remained the same), and the current through the battery (increased, decreased, remained the same). Pick the right answer and explain your reasons.
Homework Answers
the bright ness of bulb A decreased
the current through the battery increased
as the switch is closed the circuit short circuited
no current flows through A
current i = V/(3R)
after switch is closed i' = v/(2R)
current increased
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