Components that are in parallel, on the other hand, share the same two nodes.
Nodes are connection points between components.
The nodes are labelled "a", "b", and "c". This basic knowledge forms the basis for circuit analysis.
Series components have the same current flowing through them, and parallel components have the same voltage across them. The following example shows the steps involved in using this knowledge to solve electrical circuit network problems.
Step (1):
Simplify the circuit in a step-by-step fashion by combining groups of resistors in series or parallel to an equivalent single resistor, thereby producing an equivalent circuit which can be more easily solved.
Simplify the circuit in a step-by-step fashion by combining groups of resistors in series or parallel to an equivalent single resistor, thereby producing an equivalent circuit which can be more easily solved.
For the example shown,two combinations will be required.
Step (2):
Solve the simplified circuit by application of the basic rules for either a series or a parallel circuit.
Solve the simplified circuit by application of the basic rules for either a series or a parallel circuit.
For the example, applying the basic rules of series circuits and using Ohm's Law, we can solve for the current flow through and voltage drop across each element.
Applying Ohm's Law for the whole circuit,
Recalling the Rule for a simple series circuit from the Series-Resistance section
Then applying Ohm's Law to each element
Step (3):
Using the information from the equivalent circuit, work backwards in a step by step process towards the original circuit. For the example problem this will require two steps.
Using the information from the equivalent circuit, work backwards in a step by step process towards the original circuit. For the example problem this will require two steps.
(3a) Knowing the voltage drop across R234 = 30 V, we see the voltage across the two parallel resistors R2 and R34 is 30 V. Therefore, we can solve for the current flow through each of the resistors.
(3b) Knowing the current flow through R34 = 1.5 amp, we now know the current flow through each of the two resistors (R3 and R4) in the series must be 1.5 amp. Therefore, we can solve for the voltage drop across each of the two resistors.
We have now successfully solved for the current flow through and the voltage drop across each element of the series-parallel combination circuit.
it nw makes much more sense than the period yenzara iya.. Thanks man..
ReplyDeleteWhen things are meant to be passed they appear to be difficult. The good approach to education is to aim to understand then you definitely will pass
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