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Feliz [49]
3 years ago
14

Someone please help me with finding the resistance of these circuits! I've been asking for an hour now. I will give brainliest i

f right!

Physics
1 answer:
vivado [14]3 years ago
8 0

Answer:

1. 59 Ω

2. 3 Ω

3. 0.625 kΩ

Explanation:

1. The total resistance in a series circuit is equal to the sum of the resistance.

R_T=R_1+R_2+R_3...\\R_T=20+19+20\\R_T=59

Therefore, the total resistance in the first circuit is 59 Ω.

2. The total resistance in a parallel circuit is equal to the sum of the reciprocals of the resistance.

\frac{1}{R_T} = \frac{1}{R_1} +\frac{1}{R_2} +\frac{1}{R_3} ...\\\frac{1}{R_T} = \frac{1}{6.0} +\frac{1}{12} +\frac{1}{36}+\frac{1}{18} \\\frac{1}{R_T} = \frac{1}{3} \\R_T=3

Therefore, the total resistance in the second circuit is 3 Ω.

3. This is another parallel circuit, so we use the same equation from above:

\frac{1}{R_T} = \frac{1}{R_1} +\frac{1}{R_2} +\frac{1}{R_3} ...\\\frac{1}{R_T} = \frac{1}{10} +\frac{1}{2} +\frac{1}{1} ...\\\frac{1}{R_T} =1.6\\R_T=\frac{1}{1.6}

Therefore, the total resistance in the third circuit is \frac{1}{1.6} kΩ, or 0.625 kΩ.

I hope this helps!

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