0
\$\begingroup\$

I am trying to solve this current division question and I wanted to check if my work was correct.

enter image description here

I've tried applying equivalent resistance for the two parallel circuits to form 1 series circuit with (1.5 ohms) and (2 ohms) equivalent resistors. This allowed me to find the Voltage at equivalent resistor (2ohms) at 6A*2 = 12V.

As I = V/R and voltage is constant in a parallel circuit 12/6 = 2 A for the 6 ohm resistor. This however, doesn't really utilise the current division rule but the KVL rule instead. How should I approach this question instead?

\$\endgroup\$
1
  • 1
    \$\begingroup\$ The voltage across the lower resistors is the same since the bridge is shorted in the middle. From that, you can figure out the ratio between the current flowing in the lower resistors. In my head, I made the voltage across the lower resistors 6V which gives a current ratio of 1:2 between the 6 ohm and 3 ohm lower resistors. Then I scaled the currents so the sum is 6A. \$\endgroup\$
    – qrk
    Jan 11 at 4:10
2
\$\begingroup\$

Forget you ever heard that KVL and KCL were useful for solving problems like this. They're just ways to state the obvious. KVL says the sum of voltage changes around a loop is zero. Of course, it is. otherwise, the voltage at a node could have two or more values. KCL says the sum of the currents entering a node is zero. Again, of course. Otherwise, there would be charge building up at the node. The current division rule, the KVL and the KCL rules are just consequences of Ohm's law. You don't have to memorize anything more.

You were right in saying the 3//3 ohms is 1.5 ohms, and the 3//6 ohms is 2 ohms, but the two 3 ohm resisters can have any value. If one of them was a milliohm and the other a megohm, it wouldn't change the answer. There's 6 amps going through the 6//3 ohms parallel combination. The voltage across them would be 12 volts, but whatever it is, the current through the 6 ohm resistor will be 1/3 the current through the 3 ohm resistor, per ohm's law: 2 amps in this case.

\$\endgroup\$
1
\$\begingroup\$

Just think about what I've drawn below your picture: -

enter image description here

I'm not explicitly telling you how to look at this; I'm giving a hint: Is the current through the lower left hand resistor affected by either of the two resistors above it?

Here's another hint; when you see a circuit that might confuse you; redraw it.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.