Potentials at terminals of battery connected to ground


If we have a battery connected like so:


simulate this circuit – Schematic created using CircuitLab

Am I correct in thinking voltage on the positive terminal relative to ground is +9V and on the negative terminal is -9V?

Best Answer

Trying to build this circuit in real life can cause an electrical fire (depending on exactly how it is connected and the battery's internal state of charge). The wire and battery could get very hot, the insulation around the wire could melt or vaporize as smoke, nearby surfaces could be scorched, and it's even possible that a combustion fire could be triggered.

The ideal 9V battery makes a voltage difference of 9V between its (+) and (-) terminals, regardless of where ground is. If the (-) terminal is at ground (0V) then the (+) terminal is at +9V. If the (+) terminal is at 0V then the (-) terminal is at -9V. If the (-) terminal is at 1000V then the (+) terminal is at 1009V. But it's impossible for the (+) terminal to be +9V while the (-) terminal is at -9V, because then the total (algebraic sum) voltage would be 18V instead of 9V.

A practical battery has some internal source resistance distributed throughout the connection terminals, electrodes, and electrolyte -- this can be modeled with a single resistor in series with the ideal voltage source. The lower the battery's internal state of charge, the greater this equivalent internal source resistance becomes.

Real wires also have some non-zero resistance as well as some inductance.

If you click on the "simulate this circuit" link (underneath the schematic picture) you can see the results for yourself. The simulation shows 0V at both ends and 4.5A of current flowing through the loop. (This circuitlab simulator treats a 9V battery as a practical component rather than an ideal voltage source, so the simulator knows there is an upper limit on the current the battery can provide.)