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Creating a voltage divider using resistors

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, VB, 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 = Vsource / Rtotal, and Vres = I∙R. So:

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

  1. The total resistance of R2 and R3 is 2k, so VA = 2k∙3mA = 6V.

  2. The voltage across each resistor is 3V, so VA = 6V.

  3. Using the equation above: VA = 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 VB 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!