Thanks for the help!! Kirchhoff\'s Loop Rule Goal: Investigate Kirchhoff\'s Loop

Question

Thanks for the help!!

Kirchhoff's Loop Rule

Goal: Investigate Kirchhoff's Loop Rule: S ?V = 0.

Kirchhoff''s loop rule states that the sum of all the potential differences around a closed loop equals zero. In other words, S ?V = 0 for a complete loop.

Click or cut and paste the link below for virtual lab animation and then answer the questions. Please wait for the animation to completely load.  Thanks so much, you are a lifesaver!!

http://lpa.feri.uni-mb.si/Pedagosko_delo/PhysletPhisics/contents/circuits/dc_circuits/illustration30_7.html

In the circuit shown, current from the battery flows through the resistors before returning to the battery. This illustration follows a hypothetical charge as it flows through the upper of the parallel resistors.

A) In the Lab window click on [Restart] to initialize the lab. Then click on [play] to observe the charge as it moves in the circuit.

Follow the energy of the unit charge as it passes through each circuit element. Each voltage drop represents the amount of energy that is lost or gained when the charge passes through a circuit element. This demonstrates that, as charge flows around a complete loop, the gains in energy are always offset by the losses. The total change in energy is zero.

This is, of course, just a simulation. Current also flows through the lower resistor, and the current through these two resistors is not the same. (In fact, an accurate microscopic simulation would need ~1020 electrons moving counterclockwise around the circuit.) Because an electron-flow representation of current would be awkward, we use the standard definition of current and show a hypothetical unit of positive charge flowing out of the positive battery terminal, through the resistors, and into the negative terminal.

In a circuit, there are charges moving through the potential differences from the battery and across the resistor. So another way to state the loop rule is that, when a charge goes around a complete loop and returns to its starting point, its potential energy must be the same. Positive charges gain energy when they go through batteries from the - terminal to the + terminal, and they give up that energy to resistors as they pass through them.

Question (1): Use the loop rule to determine the current through the battery in a circuit consisting of a 16-volt battery connected to a set of three resistors: a 2-O resistor in series with a parallel combination of a 2-O resistor and a 3-O resistor.



Now use this link for the last animation and to answer questions 2 thru 8:

Http://pages.towson.edu/jrsimpson/PhysletsCD/contents/circuits/dc_circuits/ex30_1.html

This Exploration begins with four identical light bulbs connected to a battery (voltage is given in volts and current is given in amperes). Often, you will be asked to find the current through, the voltage across, the resistance, and/or the power consumed by a given bulb (or group of bulbs). To solve these types of problems you will use Ohm's law, V = IR, and an equation for power, P = VI = I2R = V2/R, where V is the voltage, I is the current, R is the resistance, and P is the power. You will also need to use two rules that are based on conservation laws:

B.) current in = current out. Since charge is not created or destroyed (conservation of charge), charge flowing into some point must flow back out unless there is a circuit element that can store charge (a capacitor).
C.) ?V through a complete loop = 0. The electric force is a conservative force (which is why we can define an electrostatic potential: V). This means that if you start at one point in a circuit and add up all the potential increases and subtract all the decreases in potential as you trace out a loop, then when you get back to the place you started, the potential must be the same.
Let's use these rules for the initial circuit. The brightness of the bulbs is an indication of the current through the bulbs (brightness actually goes as I2).

D.) In the Lab window click on Restart to initialize the lab.

Question (2): Rank the bulbs in order of brightness (and therefore in order of current through them), from highest to lowest.



E.) In the Lab window click on Show the currents (in the data table) through the bulbs to check your answer. The arrows indicate the direction of the current through the circuit.

Question (3): Find the current through bulb D by determining how much current must be coming into the node (the dot where the wires come together) above bulb D. The current is coming from bulbs A and B. Check your answer by verifying that the current coming into the node below bulb D (from bulb C and bulb D) is equal to the current going out of the node and into the battery.

Now consider the voltage across various elements.


F.) In the Lab window click on Show the voltages (in the data table) across the bulbs..

Question (4): Why is the voltage across bulb C the same as the voltage across the battery (think about tracing a path around the outside "loop" of the circuit)?

Question (5): Why is the voltage across bulb A equal to the voltage across bulb B?

Question (6): Find the voltage across D by picking a complete loop to trace around  (battery ? bulb A ? bulb D ? battery) OR (battery ? bulb B ? bulb D ? battery) OR (bulb C ? bulb D ? bulb A) OR (bulb C ? bulb D ? bulb B) and finding the value of the voltage for bulb D that makes the change in potential equal to zero.

Question (7): Using V = IR, what is the resistance of bulb D? (Check that it is the same as bulbs A, B , and C).

Question (8): What is the power dissipated by bulb D?

Explanation / Answer

Can Not get the link to work but answering question 1


Question (1): Use the loop rule to determine the current through the battery in a circuit consisting of a 16-volt battery connected to a set of three resistors: a 2-O resistor in series with a parallel combination of a 2-O resistor and a 3-O resistor

So you have the parallel combination of the 2 & 3 ohm => ((1/3)+(1/2))^-1 + 2 = 3.2 Ohms
Then by source transformation
I=V/R =16/3.2 = 5amps

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.