Embedded Files
Lab 7: Series circuit.pdfNew terms
New terms
dangling wires are loose/unconnected wires freely hanging
a break is an interruption in electrical current flow in a circuit
TBA
TBA
Lab
Lab
Overview
Overview
Build this circuit on a protoboard to show that Ohm's law and all series rules matter.
It's powered by a 120-10 V transformer.
Have 2.2k Ω, 3.3k Ω, 4.7k Ω, 5.6k Ω, and 1k Ω resistors.
We use a 10 V doorbelt transformer on the circuit.
It's actually a voltage of 13-14 V.
10 V is done by drawing 1 amp of current.
Circuit built on a protoboard
Circuit built on a protoboard
This is one way to build a 5-resistors series on a protobard.
R1 on left side. R5 on right side.
Measure each resistor's voltage: Here we measure voltage across R1, placed across the resistor or in parallel with it.
After measuring each resistors, measure the voltage across the supply voltage or across all resistors (from left side R1 to right side R5).
total voltage drop = supply voltage's voltage drop--called the Kirchoff's Voltage Law
Use an anmeter to measure current through a component: break the circuit and put it in series.
Easiest way is to move a resistor's end being measured, like the image.
Put a new wire where the original component was from, now with 2 dangling wires: 1 of the resistor, 1 from the new wire.
The anmeter connects to the 2 dangling wires. Here it's set to miliamps as current is low. The lead is moved from volt ohm terrminal to the 400 miliamp terminal.
With this method, we already have the miliammeter in series, a method that also measures all 5 resistors' current, which all have the same currents (IT = IR1 + IR2 + IR3 + IR4 + IR5).
Measure total current pull a transformer's lead out (one dangling wires), replaced by another wire.
What if a series has an open or break?
It wouldn't have current flow, so no voltage drop in any resistors, as V = IR, I = 0, so V = 0.
Here the break was where R2 is.
Voltage across R3 is 0 V.
Each components' voltage drop of a series circuit with a break is 0 V.
Voltage across the break is the full supply voltage (here ~13-14).
As this circuit has a break without R2, current is 0, proven by measuring R3's current, the same previous method: R3's one end pulled out, replaced by a wire and put the miliammeter between the 2 dangling wires.
What happens if between a component in a serise circuit.
Here, we place a jumper across R5, with 0 resistance.
Now RT went down to, so VT went up, so does each remaining components' voltages.
The voltage in each components should be 0 V.
With the short in R5, the IT goes up with the previous method, we should measure each components' currents to see it went up.
In series, we only see 3 resistors in series. 40 - largest, 100 - medium, 200 - smallest.
So the light bulb won't dissipate the rated power.
40 W is the brighest, it's a series circuit, they all have the same current, and the bulb has the highest resistance since P = I^2 * R, so it has the highest power
Light the light bulbs
Light the light bulbs
(Lab photo 1#)
Hook 3 light bulbs in series (40, 100, 200 W).
40 W light bulb has the highest and 200 W light bulb, the lowest resistance.
A 200 W light bulb is brighter than a 40 W bulb if both are hooked up to 120 V.
Here, they're hooked in series, so they aren't 120 V in each light bulb.
Each voltage drop's sum here = supply voltage of the disconnect.
Now without the 1st light bulb, there shouldnt be current flow anywhere. Voltage in other 2 light bulbs is zero and voltage drop in the 1st light bulb's socket should be 120 V supply voltage.
With power off, insert a jump across the 1st light bulb, causing a short circuit around the 40 W light bulb (RT goes down).
More current flows in the 100 and 200 W light bulb, so they're brighter.
Voltage drop in the 200 and 100 W light bulb adds up to 120 V (supply voltage). No voltage drop across the 40 W light bulb and is 0.
References
References
Lab photos
Lab photos
1#
Page updated
Google Sites
Report abuse