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I've tried running following circuit in simulator, It gives value of current across top right most resistor(of 2k)=0.2mA(200microA), why it isn't 0.3mA as provided by current source? enter image description here

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  • \$\begingroup\$ FYI, 0.2mA != 0.2μA \$\endgroup\$ – Phil N DeBlanc Dec 8 '17 at 20:45
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The equivalent resistance of the left half of the circuit is 10k. That of the right half is 5k.

Total resistance of the circuit is 3.33k, thus the voltage at the top node is i*r = 0.3m*3.33k = 1V.

1 volt dropped across 5k is i = v/r = 1/5k = 0.2mA 1 volt dropped across 10k is i = 1/10k = 0.1mA

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The current will be split between the right and left sides of the circuit proportional to the resistance. It may help to visualize this as water flowing through tubes with flow restrictors where the resistors are. It's easy to see in this way that SOME water will flow through both sides of the circuit.

Also note: you can use parallel and series resistor equations to so simplify both sides of the circuit into single resistors to ground.

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The current source provides 300uA to both sides of the circuit. Equivalent resistance for right circuit is (2+3)k = 5k. Equivalent resistance for left circuit is (30k || 10k) + 2.5k = 10k. Hence, current flowing into right circuit is twice the current flowing into the left circuit. This results in

\begin{equation} I_{total} = I_{left} + I_{right} = 100 \mu A + 200 \mu A = 300 \mu A \end{equation}

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Because of Kirchhoff’s Current Law:

You can read more here for example. http://www.electronics-tutorials.ws/dccircuits/kirchhoffs-current-law.html

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