Thevenin Equivalent Resistance

This sample essay on Thevenin Equivalent Resistance offers an extensive list of facts and arguments related to it. The essay’s introduction, body paragraphs, and the conclusion are provided below.

When using a voltage or current that varies, certain components that would not work with a direct current become functional. The capacitor is this type of component. It is made up of two conducting pieces of material that are a small distance apart and are separated by an insulator (or dielectric).

The following experiments will show the way in which the capacitor works when placed in a working circuit in different combinations with a resistor. They will show how the time constant can be calculated experimentally as well as theoretically.

One of the most underlying laws when dealing with electronics, which was introduced by Georg Ohm in 1821 [Gough], is Ohm’s Law. These experiments will validate this law. It will also investigate how the characteristics of a circuit may change when introducing resistors in parallel or series and as a result, verify the voltage divider rule.

They will look at voltage drops around complete loops and how by manipulating a circuit can be used to make complex networks simpler, thereby validating Kirchoff’s Voltage and Current Laws, and Thevenin’s Theorem.

“The Wheatstone bridge circuit was developed in 1843 by Charles Wheatstone in order to determine the values of unknown resistances” – [Gough], This will be investigated to check the validity of the Wheatstone bridge theory and prove it’s usefulness. Theory Experiment P-IE-R-1 (Ohm’s Law) Ohm’s Law indicates that the current through a conductor is proportional to the difference in potential between its ends.

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This, in equation form, is shown by V=I R (V is potential difference, I is the current and R is the constant of proportionality, or resistance).

Wheatstone Bridge Thevenin Equivalent

So if a current is passed through a circuit with an unknown resistance, this resistance can be calculated by plotting a graph of voltage against current. This should produce a straight line with a slope equalling the value of R. Experiment P-IE-R-2 (Resistor Networks) If a number of components are connected so that the current through each of them is equal then they are connected in series. So if you have two resistors connected in series, as shown below in Figure 1, then V1 = R1 I and V2 = R2 I.

If you total all the separate potential differences around the circuit in Figure 1, then the sum will be 0, this is true for any complete loop in a circuit. It is known as Kirchoff’s Voltage Law. As a result of this, each value of resistance can be combined to give an equivalent resistance, referred to as Req, this has no effect on the characteristics of the circuit. However, the components within a circuit can be connected up so that the potential difference across each of them is identical. This is a parallel connection. The two resistors in Figure 2 show components in parallel.

The current of each is given as I1 = V/R1 and I2 = V/R2. As charge is conserved, it can be said the amount of current going into a node is equal to the total amount that leaves it, i. e. the sum of the currents is 0 (This is known as Kirchoff’s current law). Therefore, the amount of current that passes through the two resistors in Figure 2 has to be equal to the current that is generated by the supply. It can be expressed as I = I1 + I2. By manipulating this equation and applying Ohm’s law, the equivalent resistance of the circuit can be calculated using the following equation 1/Req = 1/R1 + 1/R2.

But when there are only two resistors it can be written as Req = R1 R2 / (R1 + R2), this is known as the product over the sum rule. Experiment P-IE-R-3 (Kirchoff’s Laws and Thevenin Resistor Networks) Kirchoff’s Laws: As mentioned above, Kirchoff’s Voltage Law is defined as ‘The algebraic sum of the potential differences around any complete loop of a circuit is zero’ – [Gough]. Therefore if you refer to Figure 1, V = V1 + V2. But as Figure 2 indicates, current flows into the positive side of a resistor but at the same time out of the positive terminal of an emf source.

As a result potential difference can be called the ‘Voltage Drop’. Also mentioned above, Kirchoff’s current law can be defined as ‘ the algebraic sum of the currents into any node is zero’ – [Gough]. So where three or more conductors connect the total current through the node will equal the current from the supply. Referring back to Figure 2, this can be shown by writing I = I1 + I2. Thevenin’s Theorem Thevenin’s Theorem can be defined as ‘ any network of resistors and batteries having two terminals is equivalent, as far as its terminal behaviour is concerned, to the series combination of a resistor and a DC voltage supply’.

– [Gough] With a Voltage divider (Figure 3), by moving the switch to certain possible connections, different fractions of the supply can be created at the output. So if the switch is connected to the point B as shown, the output voltage can be obtained using the formula Vout = I R3. The current can be calculated by first working out the total resistance of the circuit and then by using Ohm’s law. If a load resistance is put across the output terminals as shown in Figure 3, then the current in the circuit will no longer be the same.

The new value for the current will now be obtainable by using the formula I = V (R3 + RL)/ RL (R1 + R2 + R3) + (R1 + R2) R3. If a load is connected across Vout, then the current through the load resistance will be given by IL = Vout/RL. This shows that by using a combination of Ohm’s law and Kirchoff’s Current and Voltage Laws, more complex circuits can be analysed faster and more easily. Experiment P-IE-R-4 (The Wheatstone Bridge) As Figure 4 shows the unknown resistor is R4, the other resistances, apart from R5, are known and can be a combination of different values.

This circuit works by varying the resistance of R1, R2 and R3 so that the current through R5 is equal to zero. When the circuit is in this situation the bridge is known to be ‘balanced’. The value of the unknown resistor can then be worked out by using the values of the now known resistors. By using Thevenin’s theorem the current through R5 can be found by changing the rest of the circuit to its Thevenin’s equivalent, this gives the circuit shown in Figure 5. The Thevenin equivalent resistance (RT) across DB is ascertained by connecting these two parallel resistor combinations across R5, giving the Formula:

RT = (R1 R3/ R1 + R3) + (R2 R4 / R2 + R4) The Thevenin equivalent voltage is determined by measuring the potential difference between the points D and B without R5 connected. As there are two parallel combinations of resistors, the voltage through each of them will be equal. This Voltage will be equal to the one that is driving the circuit, i. e. V, therefore the equivalent Thevenin Voltage can be obtained by using the formula VT = VDB = V [(R1 / {R1 + R3}) – (R2 / {R2 + R4})] So I5 can be worked out using the Thevenin equivalent Voltage and resistance along with R5. The bridge is balanced when VT is equal to zero.