![]() ![]() In the next tutorial about Resistors, we will look at the electrical potential difference (voltage) across two points including a resistor. With Chris ODonnell, Daniela Ruah, Eric Christian Olsen, Barrett Foa. These more complex circuits need to be solved using Kirchhoff’s Current Law, and Kirchhoff’s Voltage Law which will be dealt with in another tutorial. Parallel Resistors: Directed by Eric Laneuville. However, calculations of more complex T-pad Attenuator and resistive bridge networks which cannot be reduced to a simple parallel or series circuit using equivalent resistances require a different approach. This step will allow us to reduce the complexity of the circuit and help us transform a complex combinational resistive circuit into a single equivalent resistance remembering that series circuits are voltage dividers and parallel circuits are current dividers. When solving any combinational resistor circuit that is made up of resistors in series and parallel branches, the first step we need to take is to identify the simple series and parallel resistor branches and replace them with equivalent resistors. Then the complex combinational resistive network above comprising of ten individual resistors connected together in series and parallel combinations can be replaced with just one single equivalent resistance ( R EQ ) of value 10Ω. ![]() Resistors in Series and Parallel Example No2įind the equivalent resistance, R EQ for the following resistor combination circuit. Lets try another more complex resistor combination circuit. Then continue to replace any series or parallel combinations until one equivalent resistance, R EQ is found. It is sometimes easier with complex resistor combinations and resistive networks to sketch or redraw the new circuit after these changes have been made, as this helps as a visual aid to the maths. This therefore gives a total supply current, I T of: 0.5 + 0.5 = 1.0 amperes as calculated above. Resistors are connected in parallel by connecting them side by side across one another, as illustrated in Figure 6. Since the resistive values of the two branches are the same at 12Ω, the two branch currents of I 1 and I 2 are also equal at 0.5A (or 500mA) each. This can then be reduced to a single resistor value R(com) using the formula for two parallel connected resistors, as shown below. A single resistor RA is now created that is parallel with the resistor R4. We can take this one step further by using Ohms Law to find the two branch currents, I 1 and I 2 as shown. So both the resistors R2 and R3 can now be replaced with a single resistance value: 12 Ohms. Then we can see that any complicated resistive circuit consisting of several resistors can be reduced to a simple single circuit with only one equivalent resistor by replacing all the resistors connected together in series or in parallel using the steps above.
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