Question

What is the total resistance of a parallel circuit that has two loads? Load one has a resistance of 10 ohms. Load two has a resistance of 24 ohms. (YOU MUST SHOW YOUR WORK)!!

Answer 1

The total resistance is
`7.0588\Omega`

Explanation:

Attached please find the circuit diagram. The circuit is composed by a voltage source and two resistors connected in parallel:
`R_1=10\Omega` and
`R_2=24\Omega`.

First step: find the total current

For finding the current that the voltage source can provide, you must find the current consumed by each load and then add both. To do that, take first into account that the voltage is the same for both resistors (
`R_1` and
`R_2`).

• 
  `I_(R_1)=(V_S)/(R_1)`
• 
  `I_(R_2)=(V_S)/(R_2)`

The total current is:

`I_(TOTAL)=I_(R_1)+I_(R_2)=(V_S)/(R_1)+(V_S)/(R_2)=(R_2\cdot V_S+R_1\cdot V_S)/(R_1\cdot R_2)`

`I_(TOTAL)=V_S\cdot (R_1+R_2)/(R_1\cdot R_2)`

Now, the total resistance (
`R_(TOTAL)`) would be the voltage divided by the total current:

`R_(TOTAL)=(V_S)/(I_(TOTAL))`

If you replace
`I_(TOTAL)` by the expression obtained previously, the total resistance would be:

`R_(TOTAL)=(V_S)/(V_S\cdot (R_1+R_2)/(R_1\cdot R_2))`

After simplifying the terms you should get:

`R_(TOTAL)=(R_1\cdot R_2)/(R_1 + R_2)}`

Now, you must replace the values of the resistors:

`R_(TOTAL)=((10\Omega )\cdot (24\Omega))/(10\Omega + 24\Omega)}=(120)/(17)\Omega=7.0588\Omega`

Thus, the total resistance is
`7.0588\Omega`