# Circuit, maximum emf without burning up any of the resistors?

In the circuit (shown below) all the resistors are rated at a maximum power of 2.90 W.

10+20/2=20Ω ,50/2=25Ω
R2515=(25*15)/(25+15)=9.375Ω
R=(20*9.375)/(20+9.375)=6.383Ω
Ro= 25+30+40+25+6.383= 126.383Ω

If it is based [ that a current value is large and resistance is large ] on 40 ohms from P=RI^2.
RI^2=2.9
I=√(2.9/40) = 0.269A
E=RI =126.383*0.269= 33.997= 34[V] max voltage.

if my English is weird I’m sorry.

Source(s): My sixth sense

I got 34volts

You are correct in your equivalent resistance computation.

What you get is a series circuit of several resistors.
6.36 30 25 25 and 40. Now consider kirchoff’s law for a series circuit, the current is the same for allor these equivalent resistances. Next consider that power is current squared times resistance. With these two concepts in mind we can understand that in this equivalent series circuit the largest resistor will have the largest power . At this time we don’t need to know the exact value of current whetere we use the squared value or not current is the same for all resistances in a series circuit so we can make a simple deduction that the largest resistance in the series will have the most power

This mean that we can now concentrate on the largest resistance which is 40 ohms. We a know it has 2.9 watts which equals the square of the current time resistance. Rearrange the equation and you get 2.9/ 40 = current squared = 0.0725. Take the square root of this number and you get 0.26925 amps

You have just figured out the the current in the series circuit. To compute voltage add up all of your series resistance which is126.38 ohms and multiply this by the current of 0.26925 and you will get 34volts