A simulation will not run if it has no ground. When you probe voltages in a simulation, all voltages are measured with respect to ground. Therefore before it can show you any meaningful voltage values, the simulator needs to know were that ground is in your circuit. So, when you create an active simulation schematic (i.e. a schematic that you wish to run in simulation) then there must be at least one point or node in the circuit that is at ground (0 or 0V). To establish a ground node, you must place one of the two available ground symbols from the Wiring Tools pallete onto one of the wires in your circuit diagram. Please see: <https://easyeda.com/file_view_All-simulation-schematics-MUST-have-a-ground-01_y20YWvRPp.htm> and: <https://easyeda.com/file_view_All-simulation-schematics-MUST-have-a-ground-02_4WUSQNLJ5.htm> in: <https://easyeda.com/project_view_Spice-tutorials_AysKEWevN.htm> What is not quite so clear is that every point in a simulation schematic must have a DC path to ground (0). So, for example, a circuit that has two capacitors in series (a favourite of elementary electrical engineering classes and a very tricky problem to solve using a simulator!) must have a DC path to ground from the common point between the two caps. Similarly, a transformer coupled circuit where the primary side is driven from mains live and neutral and one side of the secondary circuit is connected to earth (which is probably where you would place the ground symbol in your circuit diagram), must have a DC path back to this ground from one or both ens of the transfromer primary pins. That does not mean that you have to connect such nodes directly to ground. The necessary DC path to ground can be provided by a resistor. Depending on your application, the resistance can be anywhere between a milliOhm right the way up to 1G Ohms (1e9 Ohms). An example of where very high value resistors would be added is in transformer coupled datacomms networks such a 100BaseTx or 1000BaseT Ethernet. Here, the connections between the two transceivers are floating and so have to have DC paths to ground added to keep the simulator happy. Using resistances that are high in comparison to the network cable characteristic impedance and hence termination resistance is important in order not to introduce an impedance mismatch and so disrupt the signal integrity. In practice it is simplest therefore to add two resistors: one from each of the two signal wires to ground. This preserves the symmetry of the signalling on the wires and doubles the effective resistance between them. Note that using resistors beyond these extremes is perfectly possible but can sometimes give simulation convergence problems if you have resistances or impedances at both ends of the extremes in the same circuit. Basically the simulator may run out of dynamic range trying to handle both the very small and very large currents that may be flowing in the two extreme parts of the circuit. Note also that in ngspice (the simulation engine used by EasyEDA) you cannot have a resistor of zero: it will throw an error. This is basically because any voltage difference across a zero resistance (such as may occur normally in a circuit or even just as a consequence of numerical "noise" in the simulation calculations) will generate an infinite current, which will obviously crash out of the top end of the dynamic range. See: <https://easyeda.com/project_view_Transformers-and-coupled-inductors_LWewOI0ic.htm> for some examples of this use of the ground return path resistor. Don't forget that all voltages probed in a schematic are with respect to ground. This is particularly important to remember when you want to probe signals that are floating, such as the transformer coupled examples discussed above. This is when B or E Sources can be used to probe two floating voltages and subtract them to simply generate the difference between them. Alternatively, to probe the difference between two voltages, V(a) and V(b), you can add a spice directive to your schematic to do the subtraction for you. Like this: >probe V(a)-V(b) or >probe V(a,b) A DC path to ground is often provided by the rest of the circuit but here are some cases to watch out for: i) All current sources (independent I and dependent B and F sources) are ideal: they have an infinite DC resistance (and AC impedance) and so do not have a DC path through them. Connecting a current source across a capacitor with one side grounded will throw an error even if the current source is set to zero (the capacitor voltage would ramp to infinity if a non-zero current were set and that would throw a different error!). So a resistor must be connected across the current source, to provide the DC path to ground to the other side of the current source / capacitor. A similar problem arises if two current sources are connected in series even if the two currents are identical (if they're not then the common node flies off to infinity again); ii) Capacitors (unless using the more advanced options in ngspice) are ideal: they have no parasistic (leakage) DC resistance in parallel with them. This is why both ends of a capacitor must have a DC path to ground either through the external circuit or explicitly by adding a resistor; iii) switches do not have an infinite open circuit resistance and so they always have their own internal DC path between their terminals. However, this can sometimes cause confusion if you are expecting the switch to actually have an infinite OFF resistance! See this example for more information: <https://easyeda.com/file_view_Effects-of-finite-switch-resistances_Ba4mEWewq.htm> iv) Voltage sources (including independent V and dependent B and E voltage sources) have the opposite problem. They are ideal so they have zero resistance. The same is true for simple inductors. If you connect voltage sources directly in parallel then ngspice will throw an error even if the voltage sources are set to the same value (if they're not then an infinite current would flow and that would then throw a different error). The same problem arises if you connect a voltage source directly in parallel with a simple, ideal, inductor because the voltage source tries to drive an infinite current through the inductor. This problem often occurs when driving a transformer from a voltage source. Adding a small series resistance fixes this little gotcha (in practice there will always be a finite winding resistance anyway!).