Question

which setup below has the least overall resistance?

three 150-ohm resistors in series
a parallel circuit of two 400-ohm resistors in series in one path and a 400-ohm resistor in another
three 600-ohm resistors in parallel
two 150-ohm resistors in parallel followed by one 150-ohm resistor in series

Explanation:

Calculate option 1 resistance

Three \(150\,\Omega\) resistors in series:

$$R_{\text{eq}} = 150 + 150 + 150 = 450\,\Omega$$

Calculate option 2 resistance

One path has two \(400\,\Omega\) resistors in series:

$$R_1 = 400 + 400 = 800\,\Omega$$

The other path has one \(400\,\Omega\) resistor:

$$R_2 = 400\,\Omega$$

These two paths are in parallel:

$$R_{\text{eq}} = \frac{R_1 \cdot R_2}{R_1 + R_2} = \frac{800 \cdot 400}{800 + 400} = \frac{320000}{1200} \approx 266.7\,\Omega$$

Calculate option 3 resistance

Three \(600\,\Omega\) resistors in parallel:

$$\frac{1}{R_{\text{eq}}} = \frac{1}{600} + \frac{1}{600} + \frac{1}{600} = \frac{3}{600}$$

$$R_{\text{eq}} = \frac{600}{3} = 200\,\Omega$$

Calculate option 4 resistance

Two \(150\,\Omega\) resistors in parallel:

$$R_p = \frac{150}{2} = 75\,\Omega$$

Followed by one \(150\,\Omega\) resistor in series:

$$R_{\text{eq}} = R_p + 150 = 75 + 150 = 225\,\Omega$$

Compare overall resistances

Comparing the calculated equivalent resistances:

• Option 1: \(450\,\Omega\)
• Option 2: \(266.7\,\Omega\)
• Option 3: \(200\,\Omega\)
• Option 4: \(225\,\Omega\)

The setup with the least overall resistance is Option 3.

Answer:

• Three 150-ohm resistors in series
• A parallel circuit of two 400-ohm resistors in series in one path and a 400-ohm resistor in another
• Three 600-ohm resistors in parallel (Correct answer)
• Two 150-ohm resistors in parallel followed by one 150-ohm resistor in series