Equivalent resistance is also known as the overall resistance.
For resistors in a series circuit, the total resistance is computed using the formula:
[tex] R_{T} = R_{1}+ R_{2}+ R_{3}... R_{n} [/tex]
In other words, you just add up the resistance of each resistor in the series circuit. In your case you only have two resistors. You have 2Ω and 4Ω. So all you need to do is add that up.
[tex] R_{T} = R_{1}+ R_{2} [/tex]
[tex] R_{T} = 2 + 4=6[/tex]
The total resistance of the series circuit is 6Ω
In a parallel circuit you get the total resistance using the formula:
[tex] \frac{1}{R_{T}} = \frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}...+\frac{1}{R_{n}}[/tex]
First you get the sum of all fractions and at the end take the reciprocal of the resulting fraction and divide. So let us take your problem into consideration where you have two resistors that have a resistance of 2Ω and 4Ω.
[tex] \frac{1}{R_{T}} = \frac{1}{R_{1}}+\frac{1}{R_{2}}[/tex]
[tex] \frac{1}{R_{T}} = \frac{1}{2}+\frac{1}{4}[/tex]
[tex] \frac{1}{R_{T}} = \frac{2}{4}+\frac{1}{4}[/tex]
[tex] \frac{1}{R_{T}} = \frac{3}{4}[/tex]
Get the reciprocal of the resulting fraction 3/4 and then divide. The reciprocal of 3/4 is 4/3.
4/3 = 1. 33Ω
So if you compare the equivalent resistance of the two circuits, the series circuit has a higher equivalent resistance.
