# basic electronics question - series circuit Answered

hi

i have a device which normally uses a transformer rated at 2 amps outputting 5 volts.  i want to put together a battery pack to power this device.  it's easy enough to put together the batteries and get a 6 volt output, but i'm worried about burning out the device w/a 6-volt power source.  i'm thinking putting a resistor in the circuit should be enough to lower the voltage to 5.  if i remember my high school physics right, i'd need a resistor whose resistance is

r = (v1 - v0) / i = (6 - 5) / 2 = 1/2 ohm

...but this seems strange.  1/2 ohm??  my approach must be too naïve.  can anybody point me in the right direction?

thanks
doug

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You're exactly right. If you want a one-volt drop for two amps of current, you need half an ohm. However, that's the wrong way to do it, for two reasons.

First, because the voltage drop depends on the current (i.e., on the load), the more current you pull, the more the voltage will be lost across the resistor. Second, the resistor dissipates power as heat, according to W = I2R, so in your case, you'll be putting out 2*2/2 = 2 watts of heat at that 0.5 ohm resistor.

What you really want is a simple 5V regulator. This takes a high input voltage (like your 6V) and puts out 5V, no matter what the load is. You can't use the classic 7805, since that has a maximum current of 1A. A Google search for "5V regulator 2A" will give you plenty of options.