Answer to Question #237890 in Electric Circuits for Neilmar

Question

A resistance of 6Ω is connected in parallel with a 3Ω resistance. Both resistance are then
connected in series with 8Ω resistance. How much is the total resistance of the circuit?

Answer 1

Explanations & Calculations

• Start from one convenient end of the circuit, say from the parallel combination.
• The equivalent/total resistance of those $\small 6\Omega$ and $\small 4\Omega$ which are connected parallel is

$\qquad\qquad \begin{aligned} \small \frac{1}{R_e}&=\small \frac{1}{6\Omega}+\frac{1}{3\Omega}\\ \small R_e&=\small \frac{6\times3}{6+3}\\ &=\small 2\Omega \end{aligned}$

• Now, this equivalent resistance is connected in series with that $\small 8\Omega$. Then the equivalent of those two can be simply found by addition,

$\qquad\qquad \begin{aligned} \small R_{e-total}&=\small 2\Omega+8\Omega\\ &=\small \bold{10\,\Omega} \end{aligned}$

• More to know- The total resistance of a parallel combination is always lesser than the smallest resistance in that combination & greater than the largest one for a series combination.