A10.0 v battery is connected across two resistors in series. one resistor has resistance of 840.0 ω and the other has resistance of 590.0 ω. what is the voltage drop across the 590.0 ω resistor?

Explanation:

There are a couple of ways you could do this.

The easiest is to use E*R1/(R1 + R2)

So the result would be

E_590 = 10 * 590/(590 + 840)

E_590 = 10 * 590/ (1430)

E_590 = 4.13 volts rounded.

You could do this a slightly longer way.

R = 1430 (total ohms in series.

E = 10 volts

I = ???

I = E/R

I = 10 / 1430

I = 0.00699

Now use this current to figure out the voltage drop.

E = I * R

I = 0.00699 amps

R = 590 ohms

E = 0.00699 * 590

E = 4.13 volts

Pick the way of doing it you like best.