Learn the basics of Electrical Engineering.

Series-Parallel Circuit Example # 1: A network containing 4k, 8k and 12k [1bee]

Statement: An electrical network contains three resistors of 4k, 8k and 12k in series-parallel configuration as shown in the figure below.

  1. Find the original voltage across the 8 k resistor.
  2. A resistor of 15 k is added in parallel to the 8k resistor. Now find the new voltage across 8 k resistor.

Part 1: 8 k ohm and 12 k ohm resistors are in parallel configuration. both will have the same voltage across them. First solving them in parallel will yield a 4.8 k resistor.

Now 4.8 k and 4 k resistor are in series. Apply the voltage divider to solve them. The voltage across 4.8 k is 5.45 V. This is the voltage across both 8k and 12k resistors.

So originally the voltage across 8k resistor is 5.45 V.

Part 2: Now add the 15k resistor in parallel to 8k. Since both 8k and 12k resistors are in parallel, the new 12k resistor will be in parallel to both of these.

Now again solve 8k, 15k and 12k resistors in parallel configuration this gives us an equivalent resistance of 3.636k.

Again apply the voltage divider we get the output voltage 4.76 V.

The new voltage across 8k resistor is 4.76 V.
So that was all about this problem. To strengthen your basics check out more examples and problems on Basics of electrical engineering here.

Important points from this example:
  1. Adding a new resistance in parallel decreases the overall resistance.
  2. Adding a new resistance in parallel to Series-parallel resistance in this configuration also decreases the voltage.
More important topics to learn:

  1. Voltage divider
  2. Current divider
  3. Branch current analysis
  4. Superposition principle
  5. Thevenin theorem
  6. Norton theorem

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