Statement: An Electrical network contains two voltage sources of 10 V and 20 V, and three resistors 5 k, 25 k and 15 k powering the load resistance. Find the Norton equivalent circuit.

Note: Before solving this problem you should have a basic knowledge about:

### Solution:

Step 1: Remove the load resistor and mention the terminals. Here label 'ab' is used.

Step 2: Find the equivalent resistance at 'ab' terminals:

Step 3: Now apply the superposition principle to find the current passing through ab.

Step 3a: For 10 V source.

Remove the 20 V source and convert 10 V source to an equivalent circuit. The final circuit which we achieve is a 2 mA current source in series with three resistors, By applying current divider we achieve:

Step 3 b: Now for 20 V source, remove the 10 V input and apply the Ohm's law. Finally, a current source of 1.33 mA joins the three resistors. (Same circuit as above)

By applying KCL we get the current 0.173 mA.

Step 4: Final Norton Equivalent

Finally, add 0.173 mA to 0.260 mA and we achieve a 433 µA current source which connects parallel to the 28.75 kΩ resistance.

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Tell me how (5k || 25k) + 15k = 28.75k

ReplyDeleteIts 19.2k