## Resistance

In the below resistor circuit diagram, what is the equivalent resistance at terminals A-B?

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__**Hint**

**Hint**

For resistors in Parallel:

$$$\frac{1}{R_{T}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}...+\frac{1}{R_{n}}$$$

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__**Hint 2**

**Hint 2**

For resistors in Series:

$$$R_{T}=R_{1}+R_{2}+R_{3}...+R_{n}$$$

The problem statement is asking to find the total resistance in the circuit, which has certain sections both in series and in parallel.

The equivalent resistance for
$$n$$
resistors in Parallel:

$$$\frac{1}{R_{T}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}...+\frac{1}{R_{n}}$$$

The equivalent resistance for
$$n$$
resistors in Series:

$$$R_{T}=R_{1}+R_{2}+R_{3}...+R_{n}$$$

Breaking down the diagram into smaller portions, let’s analyze the the highlighted section first.

These two resistors are in series, so
$$R_T=3+3=6\Omega$$
. The diagram can now be redrawn as:

These three resistors are in parallel, so
$$R_{T}=\frac{1}{\frac{1}{6}+\frac{1}{6}+\frac{1}{6}}=2\Omega$$
, so the redrawn diagram:

Finally, the simplified diagram of the original problem are two resistors in series, meaning the equivalent resistance at Terminals A-B is
$$R_T=2+2=4\:\Omega$$
.

4 Ω