resistors
Other then touching the leads and getting shocked, if I plug a 1M, 0.5W resistor into the 120V outlet, would it be bad?
I calculated that \$\frac{120 V}{1 M\Omega} = 0.12 mA\$, which shouldn't do anything bad if only 0.12 mA is going through, but I don't know if AC is different or something.
They are simulating a (more or less) real 9V battery. They've modeled the battery as an ideal voltage source of 9V with a series resistance of ~1.5 ohms.
simulate this circuit – Schematic created using CircuitLab
If you work this out, the current is 9V/(11.5 ohms) = 0.783A, so the voltage across the 10 ohm resistor must be 7.83V.
Supposedly the operation of this circuit is v0= 3v1 + 1.5v2, but I don't know what that means.
This means that you should be able to calculate the output voltage from the two input voltages using algebra, according to the given formula.
If you work out the circuit, you will find a formula that gives \$v_o\$ in terms of \$v_1\$, \$v_2\$, \$R_1\$, \$R_2\$, \$R_{f1}\$, and \$R_{f2}\$. Your job is to find values of the resistors that makes the constants in the formula come out to 3 and 1.5.
Best Answer
Yes, you've basically got the right idea except for the units in your calculation being off. 120V / 1MΩ = 120µA. That's very little. The more relevant calculation is how much power the resistor will dissipate. That is the voltage accross it times the current thru it. By Ohm's law you can rearrange those equations to realize that is also the square of the voltage accross it divided by the resistance:
(120V)2 / 1MΩ = 14.4 mW
That's again very little. You probably wouldn't even notice that getting warm if you touched it, although in this case touching it would be a bad idea due to the high voltage accross the two leads.
Note that since 120 V is the RMS voltage, that is the equivalent DC voltage that will deliver the same power, so the above is correct and would also work identically with 120 V DC.
However, there is one additional wrinkle with AC, which is the peak voltage the resistor can withstand. That does NOT average out since insulation breaks down with instantaneous voltage regardless of what the average over some time might be. Since the 120 V RMS is a sine, the peaks are sqrt(2) higher, which is 170 V. The resistor needs to be rated for at least that much voltage, else its insulation might break or it might arc between the leads.