Question

A 9.0V battery is connected across a 2.2kilo ohms and 6.8kilo omhs resistors connected in series. what is the potential differences across 2.2kilo ohms resistor.​

Answer 1

`V_(2.2) =2.2V`

Fast and loose: that's a classic voltage divider. the drop from the i-th resistor is
`R_i \over {\sum R}` of the drop across the whole series. In our situation, it's
`2.2\cdot 10^3 \over {2.2\cdot10^3 + 6.8\cdot 10^3` of 9 V. By plugging numbers in a calculator, it's 22/90 of 9V, or 2.2V

With Ohm's Law

The series is equivalent of a single resistor of Resistance
`2.2\cdot 10^3+6.8\cdot 10^3 \Omega = 9.0 k\Omega`. By Ohm's first Law (
`V=Ri`) the current flowing through the resistor is
`9V = 9*10^3\Omega i \rightarrow i=1mA`. At this point, the drop across the first resistor is, again by Ohm's law
`V = 2.2 \cdot 10^3 \Omega \cdot 1\cdot 10^(-3) A = 2.2V`