Free MCAT practice question
 by
 Nov 25, 2013
 MCAT Blog, MCAT Long Form, MCAT Physics, MCAT Prep
Item 16
The voltage drop across R_{d} is 8V. What is the current across R_{e}?
a) 2A
b) 8A
c) 14A
d) 32A
Explanation:
This problem requires 3 electrical circuit concepts:
1) V=IR relating voltage, current, and resistance
2) The voltage drop across resistors in parallel is equal in each resistor
3) Kirchoff’s current law states that the sum of currents entering a node equals the sum of currents leaving a node
The voltage drop is equivalent for each resistor between red nodes A and B. The voltage drop across R_{d} is given as 8V, and therefore the voltage drop across resistors R_{b }and R_{c} is also 8V. Using V=IR for each resistor, the current in resistor R_{b}=4A, R_{c}=8A, and R_{d}=2A. Using Kirchoff’s current law at node B: 4A + 8A + 2A = 14A. The current leaving node B equals the current across R_{e} because there are no other nodes in between.
a) 2A, incorrect, this implies that the current across R_{d}=R_{e}.
b) 8A, incorrect, this implies that the current across R_{d} equals the voltage and the current across R_{e}.
c) 14A, correct.
d) 32A, incorrect, this answer incorrectly assumes a voltage of 8V across R_{e} and then uses a false relationship VR=I.
Search the Blog
Free Consultation
Interested in our Online MCAT Course or one of our OneonOne MCAT Tutoring packages? Set up a free consultation with one of our experienced Academic Managers!
Schedule NowPopular Posts

MCAT Blog How To Prep For The MCAT During COVID19

MCAT Blog How is the MCAT Scored?

MCAT Blog Should You Retake the MCAT?
Free MCAT Practice Account
Need great MCAT practice?Get the most representative MCAT practice possible when you sign up for our free MCAT Account, which includes a halflength diagnostic exam and one of our fulllength MCAT practice exams.
Learn More
Submit a Comment