Question

What is the rate at which thermal energy is generated in the 30-Ohm resistor shown?

Answer 1

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Answer 2

The 10 ohm resistor in series with the right hand 5 ohm

resistor is 15 ohms

this 15 ohms is in parallel with the 30 ohm resistor. is equiv

to the

sum of the conductances 1/15 mho + 1/30 mho = 1/10 mho

take the reciprocal = 10 ohms

or one can use (r1r2)/(r1+r2) (15*30)/(15+30) = 450/45= 10 ohms

This 10 ohms is in series with the bottom 5 ohm resistor

so the 30 v is being introduced into a 15 ohm load.

the current is therefore 2 amps 30v/15ohms E/R= I

2 amps is going through the bottom 5 ohm resistor,

therefore it is dropping 10 volts E=IR

the voltage being introduced to the 30 ohm resistor is

therefore 20 v --> v source - v drop in the bottom resistor

30v-10v=20v

The power dissipated by the 30 ohm resistor would be E^2/R 20

squared over 30

400/30= 13+ 1/3 joules/second = watts