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In the circuit shown in the figure, all the resistors are rated at a maximum power of 1
In the circuit shown in the figure, all the resistors are rated at a maximum power of 1.20 W.
What is the maximum emf that the battery can have without burning up any of the resistors?
-------------I tried to calculate the resistance of each component of the circuit and then use P=V2/R --> V=sqrt(PR) to find the lowest EMF that wouldn't burn out the resistor with the least resistance. This didn't seem to work.
Also had difficulty working out how the 20 Ohm resistor (just above the 40 Ohm one) factors into the circuit. i.e. is it connected to anything? if so, in series or in parallel???
Expert Solution
The current through the 40 ohm resistor equals the current through the empf and the current through each of the other resistors is less than or equal to this current. So set P(40) = 1.2W and use this to solve for the current I through the mef. If P(40) = 1.2W then P for each of the other resistors is less than 1.2W.
(I^2)*R = P gives I^2(40ohm)=1.2W, so I = .173A.
Use the equivalent resistence for series and parallel combinations to simplify the circuit. The upper parallel branch is 6.38 ohm and the lower one is 25ohm. The series sum is 25 + 30 + 6.38 + 40 + 25 = 126.38. Then E = IR = (.173)*(126.38) = 21.89
Answer: 21.89
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