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.
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