calculate the equivalent resistance for both circuits.\nseries circuit:\n2 ω and 4 ω\nparallel circuit:\n2 ω…

calculate the equivalent resistance for both circuits.\nseries circuit:\n2 ω and 4 ω\nparallel circuit:\n2 ω and 4 ω\nwhich circuit has the higher equivalent resistance?\nthe series circuit has the higher equivalent resistance.\nthe parallel circuit has the higher equivalent resistance.\nthey have the same equivalent resistance.

calculate the equivalent resistance for both circuits.\nseries circuit:\n2 ω and 4 ω\nparallel circuit:\n2 ω and 4 ω\nwhich circuit has the higher equivalent resistance?\nthe series circuit has the higher equivalent resistance.\nthe parallel circuit has the higher equivalent resistance.\nthey have the same equivalent resistance.

Answer

Explanation:

Step1: Calculate series - circuit resistance

For a series circuit, the equivalent resistance $R_s$ is the sum of individual resistances. Given $R_1 = 2\Omega$ and $R_2=4\Omega$, so $R_s=R_1 + R_2=2 + 4=6\Omega$.

Step2: Calculate parallel - circuit resistance

For a parallel circuit with two resistances $R_1$ and $R_2$, the formula for equivalent resistance $R_p$ is $\frac{1}{R_p}=\frac{1}{R_1}+\frac{1}{R_2}$. Substituting $R_1 = 2\Omega$ and $R_2 = 4\Omega$, we have $\frac{1}{R_p}=\frac{1}{2}+\frac{1}{4}=\frac{2 + 1}{4}=\frac{3}{4}$. Then $R_p=\frac{4}{3}\Omega\approx1.33\Omega$.

Step3: Compare resistances

Since $6\Omega>\frac{4}{3}\Omega$, the series - circuit has a higher equivalent resistance.

Answer:

The series circuit has the higher equivalent resistance.