Can any arbitrary combination of resistors be broken down into series and parallel combinations? See if you can draw a circuit diagram of resistors that cannot be broken down into combinations of series and parallel.
No, there are many ways to connect resistors that are not combinations of series and parallel, including loops and junctions. In such cases Kirchhoff’s rules, to be introduced in
Kirchhoff’s Rules , will allow you to analyze the circuit.
Problem-solving strategies for series and parallel resistors
Draw a clear circuit diagram, labeling all resistors and voltage sources. This step includes a list of the knowns for the problem, since they are labeled in your circuit diagram.
Identify exactly what needs to be determined in the problem (identify the unknowns). A written list is useful.
Determine whether resistors are in series, parallel, or a combination of both series and parallel. Examine the circuit diagram to make this assessment. Resistors are in series if the same current must pass sequentially through them.
Use the appropriate list of major features for series or parallel connections to solve for the unknowns. There is one list for series and another for parallel. If your problem has a combination of series and parallel, reduce it in steps by considering individual groups of series or parallel connections, as done in this module and the examples. Special note: When finding
, the reciprocal must be taken with care.
Check to see whether the answers are reasonable and consistent. Units and numerical results must be reasonable. Total series resistance should be greater, whereas total parallel resistance should be smaller, for example. Power should be greater for the same devices in parallel compared with series, and so on.
Test prep for ap courses
The figure above shows a circuit containing two batteries and three identical resistors with resistance
R . Which of the following changes to the circuit will result in an increase in the current at point
P ? Select
two answers.
Reversing the connections to the 14 V battery.
Removing the 2 V battery and connecting the wires to close the left loop.
Rearranging the resistors so all three are in series.
In a circuit, a parallel combination of six 1.6-kΩ resistors is connected in series with a parallel combination of four 2.4-kΩ resistors. If the source voltage is 24 V, what will be the percentage of total current in one of the 2.4-kΩ resistors?
If the circuit in the previous question is modified by removing some of the 1.6 kΩ resistors, the total current in the circuit is 24 mA. How many resistors were removed?
Two resistors, with resistances
R and 2
R are connected to a voltage source as shown in this figure. If the power dissipated in
R is 10 W, what is the power dissipated in 2
R ?