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# Electrical Current Study Guide (page 3)

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Updated on Sep 27, 2011

#### Example

How does the increase in the distance between the two parallel plates of a capacitor affect capacitance?

#### Solution

If the distance between the plates increases two times, then we have an initial distance d0 and a final distance 2 · d0, and then we can compare the two capacitances:

Hence, the capacitance decreases by 2.

Table 14.2 lists some of most common insulators and shows the value of their dielectric constant.

Considering the definition of the capacitance and the few examples in the table, you can say that the capacitance will increase when a dielectric other than a vacuum is placed between the plates, and it increases elecby the value of the dielectric constant.

## Kirchoff's Laws

Conservation of charge is expressed in the Kirchoff's laws. In order to introduce these laws, we must define an electrical circuit. An electrical circuit is a set of electrical consumers and electrical sources that form a closed path in which an electrical current can be established. A closed electrical circuit will have batteries or other electrical sources, resistors and capacitors, switches, and conductors. The point where two or more conductors meet is called a junction, and a closed path is called a loop.

The junction law expresses the fact that the total current entering a junction has to be equal to the current leaving the junction. In other words, the charge arriving at the junction should be the same as the amount of charge leaving the junction. In a closed circuit, the voltage difference at the ends of the circuit provides the energy for charge circulation through the circuit.

In a closed circuit, the total potential drop on the consumers is equal to the potential supplied from the sources. And again we have a rule very similar to the conservation of energy we worked with in mechanics and heat. The closed circuit is the analog of the previous mechanical and thermal isolated systems.

The two laws provide us with sufficient equations to be able to solve different characteristics of a circuit.

#### Example

In a simple circuit, we have two identical light bulbs and two identical batteries connected to each other as in Figure 14.5. Considering the electrical resistance is 100 Ω, find out the emf of the batteries and the currents through all the branches of the circuit. The current through the top resistor is 2 mA.

#### Solutions

First, determine the given quantities and then set up the equation to solve for the unknowns. The currents are:

R = 100 Ω

I1 = 2 mA

E = ?

I2 and I3 = ?

According to the convention of positive-charge motion determining the direction of the current, we will choose a convenient direction for the three currents in this circuit. When we analyze the results regarding the currents, we are able to see if our choice was valid for this circuit by looking at the sign. If the result for a current is negative, the current flows in the opposite direction to the one considered. As an example: If the result in the current below will be I1 = –2A, then the direction of the real current is toward the left, not the right as we considered.

Using Kirchoff's laws, we consider first the junctions A and B. They are equivalent in this case, as they connect the same branches of the circuit. Let's look at the currents entering junction A: I2 and I3. And exiting junction A: I1. Then the first law is:

1. I2 + I3 = I1
2. And it is the same for the junction B.

Now we can see there are three loops in the circuit, and we will consider a direction for looping (which is completely arbitrary, and therefore, I have decided to use clockwise). First, one will be considering the upper half of the circuit:

1. The second loop is the lower part of the circuit:

2. And lastly, we can disregard the middle part and have a circuit composed only of the outside conductors and electrical components.

Out of the total number of loops N, only N – 1 are independent (as a condition, we need to be able to write independent equations), in this case 3 loops – 1 = 2 loops.

We will look at Figure 14.6 and 14.7.

For Figure 14.6, there is only a resistor (batteries have no resistance considered in this example, and they run at maximum energy, emf).

3. I1 · R + I3 · 0 = E
4. For Figure 14.7, there is again one resistor but two batteries. Around the circuit, the two batteries are supplying opposite voltages to the loop.

5. I2 · 0 + I3 · R = EE
6. And so, we have three equations and three unknowns, and we can solve for each of the unknowns. From the third of these equations (Equation 3), we can solve for I3.

I2 · 0 + I3 · R = EE = 0

I3 · R = 0

Because R is a nonzero resistor, the only possibility is for the current to be zero:

I3 = 0 A

Next, we use Equation 2, and solve for E:

I1 · R + I3 · 0 = E

I1 · R + 0 = E

I1 · R = E

2 mA · 100 Ω = E

E = 200 mV

And lastly, we can use the junction law to figure the current through the middle branch of the circuit.

I2 + I3 = I1

I2 + 0 = 2 mA

I2 = 2 mA

All currents are positive, so the direction considered in the diagram was correct. As you can see, no current runs through the lower branch of the battery.

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