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.
[tex]R_T=R_1+R_2+R_3...\\R_T=20+19+20\\R_T=59[/tex]
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.
[tex]\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[/tex]
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:
[tex]\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}[/tex]
Therefore, the total resistance in the third circuit is [tex]\frac{1}{1.6}[/tex] kΩ, or 0.625 kΩ.
I hope this helps!
