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curcuit

I hope this is not a wrong place to ask. I've just started studying electrical engineering and I have (a newbie) question and would really appreciate it if you guys could help me. So I have this curcuit and I have to calculate all the currents in this curcuit.
R1 = 400 Ω R4 = 400 Ω
R2 = 0.8 kΩ R5 = 1.2 kΩ
R3 = 1.8 kΩ R6 = 2 kΩ
V = 36 V
I already calculated the equivalent resistance which is 1600 Ω and current in the curcuit which is 22.5 mA. I don't know how would I calculate current through R1, R2, R3, R5, R6.

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    \$\begingroup\$ The circuit is just drawn in a confusing way, first step is to simplify things. Like R1 and R4 are in series, so can be replaced by a single resistor. Do this wherever possible. Resistors in parallel can also be combined if you're not interested in the individual currents flowing through them. If I only wanted to know the total current then I would reduce the whole circuit to one resistor. \$\endgroup\$ – Bimpelrekkie Oct 11 '17 at 13:15
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Simplifying then labelling this circuit will make everything easier:

schematic

simulate this circuit – Schematic created using CircuitLab

This is the beginning stage to simplifying it. From here you can turn this into 2 resistors by solving the top and bottom halves of the circuit. Once you have done this, you can find V2 and use that information to solve the currents through the different resistors.

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  • \$\begingroup\$ I calculated that V2 is 17V, but I don't know how to calculate the current through R1 and R4 \$\endgroup\$ – Venoox Oct 11 '17 at 16:02
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    \$\begingroup\$ How did you come to 17V? If you simplify this further and end up with 2 resistors, what 2 values do you get? You can then use Ohm's Law to work out V2, and then the rest of the circuit \$\endgroup\$ – MCG Oct 11 '17 at 16:16
  • \$\begingroup\$ Sorry, I meant 27V. (R124=400Ω, U=0.0225A*400Ω=9, 36V-9V=27V). I tried calcuting the current through R1 and R4: I=36V/800Ω and I got 0,045 but I think that's not right. \$\endgroup\$ – Venoox Oct 11 '17 at 19:58
  • \$\begingroup\$ Yeah that's not right. There was a reason for calculating V2. Btw, 27V is right. That means there is 27V across the second half of the circuit. If you take 27 away from 36 then you are left with the voltage across that top resistor set \$\endgroup\$ – MCG Oct 11 '17 at 22:04
  • \$\begingroup\$ You were also able to solve the current (0.0225A) so you can know that the current through R1,R4 and the current through R2 will be equal to 0.0225A \$\endgroup\$ – MCG Oct 11 '17 at 23:15
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The reason this problem appears difficult at first glance is because the circuit was deliberately drawn to be confusing.

The first thing to do is to draw the circuit with high to low voltages decending down the page. In your case, that would mean all resistors would be drawn vertically.

Once you do that, some simplifications will be more obvious. For example, R1 and R4 are in series, and so are R5 and R3.

Once you simplify those, you will see pairs of resistors in parallel. You can collapse parallel pairs for the purpose of their effect on the remaining circuit. Once you know the voltage across the parallel pair, you can go back and find the current thru the two separate branches.

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Make all the series/parallel combinations you can to simplify the circuit, and figure out the voltage at your nodes. Once you know the voltage at your nodes, go back to the original circuit and figure out how much current goes through each resistor using Ohm's Law.

For example, R1 and R4 are in series, so they just add to form one bigger resistor. That bigger resistor is in parallel to R2, so you can represent R1, R4, and R2 as one resistor. You can do the same with the lower three resistors. Use this to figure out the voltage where R2, R3 and R4 meet.

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