Series connection of resistors

In this first tutorial we are going to make a simple resistive circuit. The purpose of this tutorial is to get aquainted with the basc editing modes when creating a breadboard simulation.

The simulation will contain 7 resistors of 10kΩ in series connected to a 9v battery. The total current through the resistors will be \[i=\frac{9}{7 \cdot 10k} = 128.571 \mu A\] and the voltage over a single resistor will be \[v= 10k \cdot i \]

You can choose from several breadboards, depending on the size of your circuit. In this tutorial we choose the Half+ breadboard, which has smaller size than a Full breadboard.

The breadboard is selected as a component HALF+ from Components/Circuit/Breadboard. Click with he left mouse button on the HALF+, drag the mouse over your workscreen and click the left mouse button once again to place the HALF+ component. (Alternatively you could click and hold the left mouse button, drag and release the left mouse button to place the component)

Next we are going to place a battery supply on the circuit. Select this component 9v from Components/Circuit/Breadboard/Battery and place it such on the breadboard, that the two wires connect to the two upper supply rails on the breadboard.

You can always move the component on the breadboard after placing, but take care that the pins of the wires exactly match the holes on your breadboard. Otherwise there will be no connection, like on a real breadboard.

The resistors are selected from Components/Circuit/Breadboard/R. We select \(10k\Omega \) for all resistors in this simulation.

You can directly drop the resistors on the breadboard, but take care that the nodes are connected to the holes in the breadboard. If a component has to be rotated, first select the component en drop it onto the breadboard or somewhere else next to the breadboard. Second, click to component once with the left mouse button, so the component gets selected and finally press the key combination Ctrl-R on your keyboard. Alternatively you can also open the dialog box for the component by clicking the component with the right-mouse-button and change its orientation to 90 or 270 degrees.

To measure the voltage over a resistor we will use a multimeter DMM_Vdc from Components/Circuit/Breadboard/Multimeter. To measure the current we will use a multimeter DMM_Adc from Components/Circuit/Breadboard/Multimeter

Since the last resistor is connected to the lower supply rail, we need to make a connection between the upper and lower blue power rail. We could use a standard wire as provided for the schematics, but in this case we will use the wires provided for the breadboard. A wire can be selected from the Components/Circuit/Breadboard/Wire section and comes in various colours. You can change the orientation of a wire by selecting it with the left-mouse-button and pressing the Ctrl-R key combination on your keyboard. In this case we will place a blue wire of length 21.

Take care that the animation is enabled in case you directly want to see the nodes changing in colour depending on voltage. The multimeter display will always be update, even if the animation is turned off.

Start the simulation by clicking Simulation/Start Simulation from the menu

During the simulation you can observe the voltage over a resistor and the current though the whole series of resistors.

Are the simulation results correct?

A simple math reveals that 7 resistors of \(10k\Omega\) will give a voltage drop over each resistor as \[V_{Resistor}=\frac{9V}{7}= 1.286 volt\] The current through all resistors should equal \[I=\frac{9}{7 \cdot 10k} = 128,57 \mu A\]

Don't underestimate the influence of the internal resistance of the battery. Also there is no ideal connection between the minus of the battery to ground, only a high impedance pulls the minus connection of the battery towards zero.