What is the maximum emf
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What is the maximum emf

[From: ] [author: ] [Date: 11-09-11] [Hit: ]
.this is the correct answerV = 21.......

P = IV
P = V^2/Rt
P*Rt = V^2

V = sqrt( P*Rt)
V = 12.3 volts

this is your wrong answer ...

we know the current in the circuit will split at the branch nodes so

P = IV will be lower in the branch nodes if the wattage is the same for all components then the branch node items will see less power

the largest current will flow through the series resistors only

I = V/Rt for the series resistors

and Vdrop = I*R this means the largest power dissipation would be in the highest resistance and highest current.
the 40 ohm series resistor will dissipate the largest amount of power..

P = Vdrop*I
P = Vdrop * V/Rt

we know Vdrop = V*R/Rt

P = (V*R/Rt)*V/Rt
= V^2*R/Rt^2

solving for V

V^2 = Rt^2*P/R
V = sqrt( Rt^2 *P/R)

R = 40 ohms

V = 21.88 volts

its difficult to explain but I am 90 percent sure this is correct

you could do

it another way

P = I * V
P = I * I*R
P = I ^2 *R

we know I = V/Rt

so again we get ( V^2/Rt^2 )*R

where R is the resistor you are trying to measure its power ..

notice with the current equation current is more important than resistance because it gets squared ... for power

so if you double the current the power is 4 times..

same if you half the current the power is 1/4

thats why i know the power of the 50 ohm parallel resistors are lower power dissipated..

the current in the big parallel branch circuit will be about 1/3 in each branch

Ibranch = I *Rparalleltotal/R

the question was simplified because the parellel circuits had less resistance..than all the series ones.. this is the correct answer

V = 21.88
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