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What will happen to the brightness of bulb A and bulb B in the circuit shown in the figure when we move the wiper of the rheostat from point x to point y? enter image description here When the wiper moves from point x to point y we get a smaller resistance, and hence, we get a larger intensity of current flowing in both bulb A and bulb B, so I guessed that the brightness of both bulbs will increase. However, the answer to this question in my text book was that the brightness of bulb A decreases and the brightness of bulb B doesn't change. I got confused, so what is the correct answer? And Why?

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    \$\begingroup\$ Is it the same (faulty) book you used before? \$\endgroup\$ – Andy aka Jun 10 '17 at 10:38
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    \$\begingroup\$ @Andyaka, Yes, actually it is, lol ! I think I really need to throw it away. \$\endgroup\$ – Asmaa Jun 10 '17 at 10:45
  • \$\begingroup\$ @Asmaa - Yes. You do. \$\endgroup\$ – WhatRoughBeast Jun 10 '17 at 13:12
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Whomever wrote that answer is an idiot. Clearly, bulb b must get brighter since, with the pot at x, the bulb is shorted out and must be dark.

As for bulb a, it must get brighter. At x, you have the resistance of the pot in series. At y, bulb b is in parallel with the pot, reducing the series resistance.

Edit: typo

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