Physics is as good a place as any, I guess.
And, before we roll up our sleeves, may I also note that assigning 48 fill-ins
(4 each for 12 problems) is exceeded in its excess only by expecting all of
the answers in return for 5 points. I'll give you the tips and hints you need
in order to solve these, and then I'll wish you well.
I see from the label in the upper corner that all of this refers to series circuits.
That's important to know, and I was about to ask. So we're going to assume
that for the whole bunch on the page, R1 and R2 are in series, connected
across VS, and there are no other components involved besides the two
resistors.
OK. Here are the tools you need:
On a resistor . . .
' K ' = 'thousand'
' Meg ' = 'million'
Total resistance . . . RT = R1 + R2 ohms
Voltage across each resistor:
VR1 = VS [ R1 / (R1+R2) ] volts
VR2 = VS [ R2 / (R1+R2) ] volts
Total current in the series circuit:
IT = VS / RT = VS / (R1+R2) Amps
And that's it. Easier than I thought. There are some other things that
it could have asked for, but it didn't.
Example:
Line #4:
VS = 5 volts
R1 = 470 ohms
R2 = 1,000 ohms
RT = R1 + R2 = 1,470 ohms
VR1 = VS ( R1/RT ) = 5 (470 / 1,470) = 1.598... volts
VR2 = VS ( R2/RT ) = 5 (1,000 / 1,470) = 3.401 volts
(Notice that VR1 + VR2 always = VS.)
IT = VS / RT = 5 / 1,470 = 0.0034 Amperes = 3.4 mA (milliamps)
The power supplied by the battery is (VS)²/RT = 0.017 watt ,
but it doesn't ask for that.
And that's it ! That's everything you need. Go get 'em, champ !
