The voltage drop across R_{d} is 8V. What is the current across R_{e}?

a) 2A

b) 8A

c) 14A

d) 32A

Solution

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.

I can’t really see the image clear enough to determine which one is Ra, Rb, Rc, Rd, or Re. Also, I don’t understand how the currents across Rb, Rc, and Rd were calculated.

Hi Katie! Thanks very much for the input. I’ll work on replacing this image with one in which the letters are more clearly readable. With regard to the second part of your question: remember, voltage drops across parallel branches of a circuit must always be equal. I like to think of this concept as analogous to potential energy – whichever path a particular charge happens to take, its change in potential energy must be the same – otherwise, some charges would reach the end of the circuit with more potential energy remaining than others.

Now that we know this, we know that the voltage drops across Rb and Rc (parallel branches to Rd) must also be 8 V. Ohm’s law, V = IR, works for individual resistors as well as for circuits as a whole, as long as you are very careful about which terms you plug in. For resistor B, which has a resistance of 2 ohms, 8 V = (I)(2 ohms), so I = 4 A. For resistor C, which has a resistance of 1 ohm, 8 V = (I)(1 ohm), so I = 8 A.

Hope this helps – good luck!

Hi there, I’m a little confused because I’m also thinking about the fact that currents differentially flow through different magnitude resistors. Thus wouldn’t we expect most of the current to flow through e, followed by, b followed by d?

The voltage drop is the same, but the currents are different. So I agree that the current exiting node b is 14 amps, the sum of all the currents. But the current going through resistor e is going to be some fraction of that total not te total itself. Isn’t that right?

Hi Dominic, thank you for reaching out! You’re absolutely correct in your science, but I think you misread the diagram – resistor e is the last resistor encountered before returning back to the battery. I think you confused it with resistor c, which is the resistor in the middle of the parallel 3-branch element. I understand that this diagram might be rather hard to read, so our team will look into updating it to make it more clear. Let us know if we can help with anything else!