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Return to Blog# Series-Parallel Combinations of Resistors

## Analysis of Combination Circuits

The basic strategy for the analysis of combination circuits involves using the meaning of equivalent resistance for parallel branches to transform the combination circuit into a series circuit. Once transformed into a series circuit, the analysis can be conducted in the usual manner. In the previous tutorials, the method of determining the equivalent resistance of parallel is equal. The total or equivalent resistance of those branches is equal to the resistance of one branch divided by the number of branches. For combined resistors, the method of calculating their equivalent resistance is the same as that for any individual series or parallel circuit, because the resistors in series carry exactly the same current and resistors in parallel have exactly the same voltage across them.

## Procedure in Series-Parallel Computation

The total resistance in a combined circuit is obtained by reducing the different series and parallel combinations step-by-step to end up with a single equivalent resistance for the circuit which allows easily determining of the current. By undoing the reduction process, the current flowing through each resistor can be found. For doing the reduction process, below are the things to be considered:

1. Two or more resistors with their heads directly connected together and their tails directly connected together in parallel, and they can be reduced to one resistor using the equivalent resistance equation for resistors in parallel.

2. Two resistors connected together so that the tail of one is connected to the head of the next, with no other path for the current to take along the line connecting them, are in series and can be reduced to one equivalent resistor.

## Example

Calculate the total current (I) taken from the 12V supply

This may seem like a difficult task at first glance, but a closer look will reveal that the two resistors R2 and R3 are both connected together in a series combination, which can then be added up together. For this combination, the resultant resistance will be

R_2 + R_3 = 8 + 4 = 12 Ohms

So both the resistors R2 and R3 can now be replaced with a single resistance value: 12 Ohms.

A single resistor RA is now created that is parallel with the resistor R4. This can then be reduced to a single resistor value R(com) using the formula for two parallel connected resistors, as shown below.

\frac{1}{R_(*C_o_m_b_i_n_a_t_i_o_n*)} = \frac{1}{R_A}+ \frac{1}{R_4} = \frac{1}{12} + \frac{1}{12} = \frac{1}{R_(*c_o_m*)} = 6 \Omega

A new simplified circuit will now look like this:

The two remaining resistances, R1 and R(com) are now connected in series combination and can be added together in order to obtain the total circuit resistance between points A and B.

R_(*A*-_B_) = R_(*c_o_m*) + R_1 = 6 + 6 = 12 \Omega

The original four-resistor combinations circuit above can now be replaced by a single resistance of just 12 Ohms. The value of the circuit current (I) can be calculated as

Circuit \: Current (I) = V/R = 12/12 = 1 Ampere

**References**

http://www.electronics-tutorials.ws/resistor/res_5.html

http://www.physicsclassroom.com/class/circuits/u9l4e.cfm

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