## Power distribution in series circuits Assignment Help

Assignment Help: >> Electrical Engineering - Power distribution in series circuits

Power distribution in series circuits

When calculating the power in a circuit containing resistors in series, all you are required to do is to find out the current, I that the circuit is carrying. Then it' is easy to calculate the power Pn, based on formula Pn   =  I 2Rn.

Problem:1

Suppose we have a series circuit with the supply of 150 V and 3 resistors: R1 = 330O, R2 = 680O, and R3 = 910O. What is power which is dissipated by R2?

You should find out the current in circuit. To do this, compute total resistance first. As the resistors are in series, the total is R= 330 + 680 + 910 = 1920O.Then the current is I =150/1920 = 0. 07813 A = 78.1 mA. The power in R2 is P2 = I2R2 = 0.07813 *0.07813*680 = 4.151 W. Round off to 2 significant digits, because that is all we have in the data, to obtain 4.2 W.

The total power dissipated in the series circuit is equal to sum of the wattages dissipated in each resistor. In this manner, the distribution of power in a series circuit is like distribution of the voltage.

#### Assured A++ Grade

Get guaranteed satisfaction & time on delivery in every assignment order you paid with us! We ensure premium quality solution document along with free turntin report!

All rights reserved! Copyrights ©2019-2020 ExpertsMind IT Educational Pvt Ltd