Electricity problem from Jan. 1959 issue of QST;
Using Ohm's law find.

Question

Electricity problem from Jan. 1959 issue of QST;
Using Ohm's law find.

radar · Answer

"There are three (3) resistors in series, but you only know three things: R1 has a value of 2 Ohms, R2 dissipates 2 Watts, and R3 has a voltage drop of 2 volts across it.  Given that the power supply is a 10.5 Volt battery, how can you puzzle out the value of R2 and R3, as well as the current, I?

radar · Answer

Hint:  Don't be surprised that there may be more than one solution.

anonymous · Answer

$$I ^{2}*R2=P2=2wats,so I=\sqrt{P2/R2}$$ ,where I is current,P2 is power dissipation of R2 resistor,R2 resistane of R2 resistor,Ubattery=U3+I*R1+I*R2,where Ubattery is 10.5 U3 is 2V,R1 is 2 Ohms,and from there its quadratic journey to unknown yet still reachable positive solution of R2

radar · Answer

Yes it results in a quadratic thus the two solutions.

radar · Answer

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