Cracking the SAT Physics Subject Test

(Marvins-Underground-K-12) #1

this current splits. Since R 1 is twice R 2 , half as much current will flow through R 1


as through R 2 ; the current through R 1 is I 1 = 2 A, and the current through R 2 is I 2 = 4


A. The voltage drop across each of these resistors is I 1 R 1 = I 2 R 2 = 1,200 V


(matching voltages verify the values of currents I 1 and I 2 ). Since the ammeter is in


the branch that contains R 2 , it will read I 2 = 4 A.


The voltmeter will read the voltage drop across R 3 , which is V 3 = IR 3 = (6 A)(150


Ω) = 900 V. So the potential at point b is 900 V lower than at point a.



  1. The following diagram shows a point a at potential V = 20 V
    connected by a combination of resistors to a point (denoted G) that
    is grounded. The ground is considered to be at potential zero. If
    the potential at point a is maintained at 20 V, what is the current
    through R 3?


Here’s How to Crack It


R 1 and R 2 are in parallel; their equivalent resistance is RP, where

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