Power In Electric Circuits

In this circuit, we have a combination of resistors. First, we need to find the total resistance in the circuit.
The two resistors,

10%20%5C%2C%20%5COmega

and

20%20%5C%2C%20%5COmega

, are in series, so the total resistance for this part of the circuit is:

R_%7B%5Ctext%7Bseries%7D%7D%20%3D%2010%20%5C%2C%20%5COmega%20%2B%2020%20%5C%2C%20%5COmega%20%3D%2030%20%5C%2C%20%5COmega

Now, this combined resistance is in parallel with the

10%20%5C%2C%20%5COmega

resistor. To find the total resistance of these parallel resistors, we use the formula for parallel resistance:

%5Cfrac%7B1%7D%7BR_%7B%5Ctext%7Bparallel%7D%7D%7D%20%3D%20%5Cfrac%7B1%7D%7BR_%7B%5Ctext%7Bseries%7D%7D%7D%20%2B%20%5Cfrac%7B1%7D%7BR_%7B%5Ctext%7B10%7D%7D%7D

%5Cfrac%7B1%7D%7BR_%7B%5Ctext%7Bparallel%7D%7D%7D%20%3D%20%5Cfrac%7B1%7D%7B30%7D%20%2B%20%5Cfrac%7B1%7D%7B10%7D

%5Cfrac%7B1%7D%7BR_%7B%5Ctext%7Bparallel%7D%7D%7D%20%3D%20%5Cfrac%7B1%20%2B%203%7D%7B30%7D%20%3D%20%5Cfrac%7B4%7D%7B30%7D

R_%7B%5Ctext%7Bparallel%7D%7D%20%3D%20%5Cfrac%7B30%7D%7B4%7D%20%3D%207.5%20%5C%2C%20%5COmega

Now, the total resistance in the circuit is

R_%7B%5Ctext%7Bparallel%7D%7D%20%3D%207.5%20%5C%2C%20%5COmega

. Using Ohm’s Law, the total current

I

drawn from the 5 volt source is given by:

I%20%3D%20%5Cfrac%7BV%7D%7BR%7D

I%20%3D%20%5Cfrac%7B5%20%5C%2C%20%5Ctext%7BV%7D%7D%7B7.5%20%5C%2C%20%5COmega%7D%20%3D%200.67%20%5C%2C%20%5Ctext%7BA%7D

Thus, the current drawn from the source is 0.67 A.

About Power In Electric Circuits - ASSAM-CEE

Power In Electric Circuits is a vital chapter for ASSAM-CEE aspirants. Mastering the concepts covered in this chapter is essential for securing a top rank.

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About Power In Electric Circuits - ASSAM-CEE

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Power In Electric Circuits

High-Yield Trend

Chapter Questions 1 MCQs

Official Solution

About Power In Electric Circuits - ASSAM-CEE

Frequently Asked Questions

Why focus on Power In Electric Circuits PYQs?

How to best use this analysis?

Chapter Questions
1 MCQs