What is the total resistance of the circuit?

Circuit diagram: Feb 26, 2017

The total resistance of the circuit is $\frac{175}{2} \Omega$ or $87.5 \approx 88 \Omega$.

Explanation:

This is a combination circuit. The best method of analysis in determining the total resistance of the circuit is to reduce it to its simplest representation, i.e. add up each of the individual resistances of each resistor. How we do this will depend on whether the resistors are connected in series or in parallel. Where every resistor has resistance $R = 100$ $\Omega$.

Let's start from the top of the circuit, with resistors $1$ and $2$. We can see that ${R}_{1}$ and ${R}_{2}$ are connected in parallel. The total resistance for resistors connected in parallel is found by:

${R}_{12} = {\left(\frac{1}{R} _ 1 + \frac{1}{R} _ 2 + \ldots + \frac{1}{R} _ n\right)}^{-} 1$

$= {\left(\frac{1}{100} \Omega + \frac{1}{100} \Omega\right)}^{-} 1$

$= {\left(\frac{1}{50} \Omega\right)}^{-} 1$

$= 50 \Omega$

We can now reduce ${R}_{1}$ and ${R}_{2}$ to one resistor with $R = 50 \Omega$. Next, let's reduce ${R}_{4}$ and ${R}_{5}$. Again, they are connected in parallel.

${R}_{45} = {\left(\frac{1}{R} _ 4 + \frac{1}{R} _ 5\right)}^{-} 1$

$= {\left(\frac{1}{100} \Omega + \frac{1}{100} \Omega\right)}^{-} 1$

$= {\left(\frac{1}{50} \Omega\right)}^{-} 1$

$= 50 \Omega$ Now let's reduce ${R}_{3}$ and ${R}_{45}$. These resistors are connected in series. For resistors connected in series, the total resistance is found by simply adding up the resistances of each resistor.

For ${R}_{3}$ and ${R}_{45}$:

${R}_{345} = {R}_{3} + {R}_{45}$

$= 100 \Omega + 50 \Omega$

$= 150 \Omega$ Next, let's reduce ${R}_{345} , {R}_{6} ,$ and ${R}_{7}$. These three resistors are all connected in parallel.

${R}_{34567} = {\left(\frac{1}{R} _ \left(345\right) + \frac{1}{R} _ 6 + \frac{1}{R} _ 7\right)}^{-} 1$

$= {\left(\frac{1}{150 \Omega} + \frac{1}{100 \Omega} + \frac{1}{100 \Omega}\right)}^{-} 1$

$= {\left(\frac{2}{75} \Omega\right)}^{-} 1$

$= \frac{75}{2} \Omega$ Lastly, we can add ${R}_{12}$ and ${R}_{34567}$ in series.

${R}_{t o t} = {R}_{12} + {R}_{34567}$

$= 50 \Omega + \frac{75}{2} \Omega$

$= \frac{175}{2} \Omega$

$\therefore$ The total resistance of the circuit is $\frac{175}{2} \Omega$ or $87.5 \approx 88 \Omega$.