Circuit Challenge #25

Circuit Challenge #25

Circuits are great fun. I even made my own.

The Setup

Total voltage: 24 V

The five resistors have resistances that are the prime numbers greater than or equal to 5.

Total Resistance

The first logical step would be to calculate the total resistance. The equations for calculating equivalent resistances are:

• in series: Req = R1 + R2 + R3 + …
• in parallel: 1/Req = 1/R1 + 1/R2 + 1/R3 + …

Because this circuit is a compound circuit, we will be using both of these equations. First, we would want to calculate the Req of the parallel circuit, which, altogether, can then be treated as “in series” with R1 and R2:

Total Current

Once we know the total resistance, we can calculate the total current because we already know the total voltage. In order to calculate the total current, we will use Ohm’s Law:

• V = I * R; (voltage = current * resistance) 
  • solved for I, it becomes: I = V/R

Individual Resistors - In Series

We can begin to calculate the voltage and current across each individual resistor now that we know the total voltage, total resistance, and total current.

We’ll start with the resistors in series: R1 and R2. In series, the current is the same everywhere (Ibattery = I1 = I2 = I3 = …). We can use the given R and I (which is equal to the total current) to calculate the voltages across each resistor with Ohm’s Law.

We can fill this information into our circuit diagram.

Individual Resistors - In Parallel

The whole parallel circuit can be treated as a single equivalent resistor that is in series with R1 and R2. In series, the total voltage is equivalent to the sum of all the individual voltages across each resistor (Vbattery = V1 + V2 + V3 + …).

Vbattery = V1 + V2 + Vparallel

24 = 7.31 + 10.234 + Vparallel

Vparallel = 6.456 V

In a parallel circuit, the voltage is the same everywhere (Vbattery = V1 = V2 = V3 = …), so we know the voltages across R3, R4, and R5 are all 6.456 V. We also know the resistances, so we can use Ohm’s Law (V = I * R or I = V/R) to calculate the current that passes through each of these resistors.

R3 R = 11 ohms V = 6.456 V I = V/R = 0.587 A

R4 R = 13 ohms V = 6.456 V I = V/R = 0.497 A

R5 R = 17 ohms V = 6.456 V I = V/R = 0.380 A

We can check that these currents are correct by adding them up to make sure they are approximately equal to the total current:

0.587 + 0.497 + 0.380 = 1.464 A ~ 1.462 A :)

All Done

We can now complete our circuit diagram:

Done. Done. Done, Heidi, SHHHHHHHHHHHHHHHHHHH. #fanggang 

Extra

Here’s a screenshot of the circuits made on an iPad:

 

We could also modify the circuit slightly. For example, we can make R3 have a resistance of 110 ohms. If we were to do this, and recalculated everything, we would observe the following:

• the Req of the parallel circuit would be larger
• the total resistance of the whole circuit would also be larger
• the total current would be smaller because resistance and current are inversely related according to Ohm’s Law (V = I * R)
• the currents across all resistors would be smaller
• if these resistors were light bulbs, they would all be dimmer
• the voltage across R1 and R2 would decrease because of the smaller current

• the voltage across the parallel circuit (R3, R4, R5) would increase in order to make the total voltage = 24 V