Take a look at the picture on the right. We see three series
connected resistors. We've already learned that the total resistance is
3k. So the current I will be 9V / 3k = 3mA. The voltage at point B,
V_{B}, equals 1k∙3mA = 3V. (Do you still remember what
is meant by 'voltage at point B'? It means: connect the red wire of the
volt meter to point B and the black wire to ground.)

The general way of calculating the voltage across a resistor in a series connection is:

I = V_{source} / R_{total},
and V_{res} = I∙R. So:

There are three ways to calculate the voltage at point A:

The total resistance of R2 and R3 is 2k, so V

_{A}= 2k∙3mA = 6V.The voltage across each resistor is 3V, so V

_{A}= 6V.Using the equation above: V

_{A}= 9V∙(2k/3k) = 6V.

Does this mean that you can connect your 3V portable cassette player
to point B? Well, of course you could, but don't expect it to work! The
player acts like a resistor of, say, 50 ohms. That resistor is parallel
connected with R3, resulting in a resistance of 47.6 ohms. So
V_{B} will drop to 9V∙(47.6/2047.6) = 0.2V. And that
will never be enough for your player.

Conclusion: If you design a voltage divider, don't forget to take the load into account!