02/23 - Lab 1: Solder-less Breadboards, Open and Short Circuits

02/23

Lab 1 - Solderless Breadboards, Open-circuits and Short-circuits

In this lab, we acquainted ourselves with the solder-less bread boards using a multimeter and various connection setups.

1) Check the resistance between two holes in the same row on the breadboard:

Multimeter indicates a resistance of 1.4 Ohms between the two holes. 

This means the circuit is closed.

2) Check the resistance between two rows of holes on opposite sides of the central channel:

The multimeter reads a resistance of infinity, which means there is no connection between ports on opposite sides of the central channel of the breadboard. This means the circuit is open.

3) Check the resistance between two arbitrary holes of the breadboard:

Again, the multimeter is showing a resistance of infinity, which indicates that there is no connection and therefore the circuit is open.

4) Use a jumper wire to connect the the two rows of the arbitrary holes:

The multimeter displays a resistance of 3.1 Ohms, which makes sense because there is a small resistance added due to the extra wire. This means the circuit is closed.

After running this series of tests, we were instructed to draw a schematic of the breadboards and indicate how each of the nodes are connected internally.

As shown in the diagram, each row is connected horizontally accross all 5 nodes of a column, but the two colums are separated by a central channel. The + and - columns on the outside of the breadboard are connected vertically along the entire length of the breadboard.

In-Class Examples

1) In the first exercise, we were shown a simple circuit with two power sources and three lightbulbs. With the switch off, both light bulbs on the outside of the circuit are equally bright, and the lightbulb in the middle along the switch wire is off. We were asked to determine what would happen to the lights as soon as the switch is turned on, connecting the middle wire to the rest of the circuit.

We took a guess that the two lit bulbs would dim, and the middle bulb would light up to the same brightness as the other two.

In reality, when the switch is flipped, nothing happens. This is because there is no difference in potential between the left and right junctions. As a result, the flow of the current does not change when the switch is turned on, and the lights stay the same.

2) The charge transferred in time is given as shown below where the peak value is 10 C and the time at the maxima is 0.785s. Determine the current, I(t), flowing through the wire.

3) Determine the total charge flowing through an element for 0 < t < 2 seconds when the current entering its positive terminal is i(t) = e^(-2t)

Note: The numbers are good, but we forgot to convert our answer to the appropirate magnitude (mC).

4) Given v = 10V and i(t) as shown below, sketch the power and energy as a function of time

5) The figure shows the current through and voltage across a device. Find the total energy absorbed by the device for the period 0 < t < 4 seconds.

6) Given the circuit below, find V0. Hint: Be sure to use conservation of energy or that the sum of the power in all the elements is 0.