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A route for the flow of electricity that has elements of both parallel and series circuits.

Combination Circuits:. A route for the flow of electricity that has elements of both parallel and series circuits. Example: find the current in each resistor . The first step is to simplify the circuit by making the parallel resistors into a single resistance.

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A route for the flow of electricity that has elements of both parallel and series circuits.

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  1. Combination Circuits: A route for the flow of electricity that has elements of both parallel and series circuits.

  2. Example: find the current in each resistor

  3. The first step is to simplify the circuit by making the parallel resistors into a single resistance. • Resistance in parallel can be calculated by 1 / Rpar = 1 / R1 + 1 / R2 + 1 / R3 + ... The new resistance is found to be (1/(8+8) + ¼)^-1 = 3.2 Ohms The new resistance is found to be (1/6 + ¼ + ¼ + 1/3)^-1 = 1 Ohm

  4. Now label the currents throughout your circuit • Our simplified circuit now looks like this;

  5. Loop 1 Loop 2 • In order to solve for the currents in each branch it is easiest to separate the two loops into individual loops and set up systems

  6. Loop 2 Loop 1 • Now, using Ohm’s Law (V=IR), we can set up systems to solve for I1 and I2 10 = I2(13 + 7 +12 + 3.2 + 6) - 6I1 10 = I1(1 + 6) - 6I2 10 = 41.2I2 - 6I1 10 = 7I1 - 6I2

  7. Now we use algebra to solve the systems for I1 and I2 10 = 7I1 - 6I2 10 = -6I1 + 41.2I2 x6(10 = 7I1 + 6I2) =60 = 42I1 +36I2 x7(10 = 6I1 + 41.2I2) = 70 = -42I1 + 288.4I2 60= 42I1 +36I2 70 = -42I1 + 288.4I2 130 = 324.4I2I2 = .401 A 10 = 7I1+ 6(.401) I1 = 1.085 A

  8. Now we can find the current in each resistor. • Since we know that in series the current stays the same we have the current for almost all the resistors, the tricky part comes with the resistors in parallel. • Let’s de-simplify our parallel resistors and solve for the current in each branch!

  9. 3.2 Ohms 1 Ohm The original parallel systems are shown above, as well as their simplified resistance values. Using Ohm’s law, and the calculated I1 and I2, we can find the total voltage in each parallel system. V=IR V1 = 1.085*1 = 1.085 V V2 = .401*3.2 = 1.283 V

  10. 3.2 Ohms 1.283 V 1 Ohm 1.085 V Now, knowing that voltage is equal in all branches in parallel, as well as knowing the value for each resistor, we can calculate the current travelling through each resistor through Ohm’s law. R1 =1.085 = I*6Ω I = .181 R2 = 1.085=I*4Ω I = .271 R3 = 1.085= I*3Ω I = .362 R4 = 1.085= I*4Ω I = .271 R5 = 1.283= I*4Ω I = .321 R6 = 1.283= I*16Ω I = .080

  11. Since we know current stays constant through series circuits we know the value of the current through R8, R9, and R10. Being the value of I2, .401 A. For R11, however, it is shared by both loops, thus both currents must be accounted for. I1 is stronger than I2 and they are travelling in opposite directions, because of this you must subtract I2 from I1, getting a value of .882 A.

  12. Summary: R7 = .040 A R8 = .401 A R9 = .401 A R10 = .401 A R11 = .882 A R1 = .181 A R2 = .271 A R3 = .362 A R4 = .271 A R5 = .321 A R6 = .040 A

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