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\prob{32}
9277_32

In the circuit shown above, the resistance are given in ohms and the battery is assumed ideal with emf equal to 3.0 volts.

The resistor that dissipates the most power is


  1. $R_1$
  2. $R_2$
  3. $R_3$
  4. $R_4$
  5. $R_5$

Electromagnetism}Circuits

Power is related to current and resistance by P=I^2R. The resistor that has the most current would be R_1 and R_{eq} (the equivalent resistance of all the resistors except for R_1), since all the other resistors share a current that is split from the main current running from the battery to R_1. Since R_{eq}<R_1, the most power is thus dissipated through R_1, as in choice (A).

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Comments
NoPhysicist3
2017-03-23 10:13:45
Notice that the total resistance of the smallest circuit is 20 ohms. Therefore total resistance of the right vertical bar is 50 ohms, which is the same as R2. Since the current distributes between R2 end effective R (combined from R3, R4, R5) equally, we imply that the maximum current is at the R1, and therefore the maximum power is.NEC
tatitechno
2011-09-24 03:50:22
Hope we get one the these in the test !!
scottsaw
2013-08-24 11:05:35
Does anybody know if ETS ever re-uses old questions?
mrshroom29
2013-09-19 22:25:32
Ive seen some similar questions between test. Then again I've seen test that are completely different from one another. If there is a possibility of being repeated or similar questions, they will be off the most recent tests that have been released.
NoPhysicist3
2017-03-23 10:14:23
They stopped doing this after the cheating in china was revealed.
NEC

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Notice that the total resistance of the smallest circuit is 20 ohms. Therefore total resistance of the right vertical bar is 50 ohms, which is the same as R2. Since the current distributes between R2 end effective R (combined from R3, R4, R5) equally, we imply that the maximum current is at the R1, and therefore the maximum power is.

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