## anonymous one year ago In a series circuit with 9V battery and two resistors, one 100 Ohms and one at 350 Ohms, this is the voltage across each one respectively. **What is that voltage?

1. Michele_Laino

the current in your circuit is: $\Large I = \frac{{\Delta V}}{{{R_1} + {R_2}}} = \frac{9}{{100 + 350}}$ since your resistors are connected in series, and the equivalent resistance is: R_1+R_2 |dw:1433611823658:dw|

2. anonymous

ok! and solving we get 0.02?

3. Michele_Laino

correct!

4. anonymous

is that the voltage across each one? :/ 0.02 is our solution?

5. Michele_Laino

now using the Ohm's law, we can compute the requested voltage drop as below: |dw:1433612018865:dw| $\Large \begin{gathered} \Delta {V_1} = I \times {R_1} = 0.02 \times 100 \hfill \\ \Delta {V_2} = I \times {R_2} = 0.02 \times 350 \hfill \\ \end{gathered}$

6. anonymous

2 & 7? so those are our solution? 2 and 7 ?

7. anonymous

is it ohms as well? :/

8. Michele_Laino

yes! 2 volts, and 7 volts

9. anonymous

oh oops volts haha :P

10. anonymous

yay so this problem is complete? :O

11. Michele_Laino

yes!

12. anonymous

yay!! thank you!:D

13. Michele_Laino

:)

14. Michele_Laino

please, as you can check, we have: 2 +7 = 9 volts which is the voltage of our battery

15. anonymous

ohh yes:) thank you!!:D