Resistance
In the below resistor circuit diagram, what is the equivalent resistance at terminals A-B?
Expand Hint
For resistors in Parallel:
$$$\frac{1}{R_{T}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}...+\frac{1}{R_{n}}$$$
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
Time Analysis
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