# Question 9f759

Aug 7, 2015

Circuit Simplification.

#### Explanation:

Start from the right side of the circuit. 20, 10, 20-ohm resistors are series connected, so their eq. resistance is

${R}_{\text{eq 1}} = 20 + 10 + 20 = 50 \Omega$.

Then this eq.resistance is connected in parallel with another 50-ohm resistor, so their eq.resistance is

${R}_{\text{eq 2}} = \frac{50}{2} = 25 \Omega$

This 25-ohm eq. resistor is again connected in series with a 25 ohm, so the eq. resistor once again becomes

${R}_{\text{eq 3}} = 25 + 25 = 50 \Omega$

But then another 50-ohm is connected with this eq. resistor in parallel, so eq. resistance becomes

${R}_{\text{eq 4}} = \frac{50}{2} = 25 \Omega$

At last, this 25-ohm eq. resistor is connected with the rest

${R}_{\text{eq 5}} = 10 + 5 + 25 = 40 \Omega$

The 2nd answer is a bit lengthy. Hopefully you don't get frustrated. I'm typing the solution. Until I post the 2nd answer, digest this 1st question.

Aug 30, 2015

For part (b) $V = \text{160 V}$

#### Explanation:

So, your circuit looks like this

You know that the power dissipated by the $10 \Omega$ resistor located in the lower right of the circuit is $\text{10 W}$.

What you have to do is use that power to calculate the intensity of the current that passes through that resistor

$\textcolor{b l u e}{P = {I}^{2} \cdot R \implies I = \sqrt{\frac{P}{R}}}$

In your case, you would get

I = sqrt( (10color(red)(cancel(color(black)("W"))))/(10color(red)(cancel(color(black)("W")))/"A"^2)) = "1 A"

Next, you need to work backward from this current to determine the current that works its way from the voltage source.

So, current ${I}_{1}$ is coming out of the source and going into point $\textcolor{b l u e}{A}$, where it splits into currents ${I}_{2}$ and ${I}_{3}$. Current ${I}_{3}$ goes into point $\textcolor{b l u e}{B}$ and plits into currents ${I}_{4}$ and ${I}_{5}$.

You know that ${I}_{4} = \text{1 A}$, since that's the current that goes through the $\text{10-W}$ resistor. Now, because the equivalent resistance of those three resistors in series is equal to $50 \Omega$, ${I}_{5}$ will be equal to ${I}_{4}$.

That happens because both currents pass through $50 \Omega$ resistors.

This means that

${I}_{3} = {I}_{4} + {I}_{5} = 2 \cdot {I}_{4} = 2 \cdot \text{1 A" = "2 A}$

The equivalent resistance between points $\textcolor{b l u e}{B}$ and $\textcolor{b l u e}{C}$ is equal to $25 \Omega$, since you have two $50 \Omega$ resistors in parralel.

This means that the equivalent resistance between points $\textcolor{b l u e}{A}$ and $\textcolor{b l u e}{C}$ will be $50 \Omega$, since you have two $25 \Omega$ resistors in series.

This means that ${I}_{3}$ is equal to ${I}_{2}$, since they both pass through $50 \Omega$ resistors. The current that's coming from the source, ${I}_{1}$, will thus be

${I}_{1} = {I}_{2} + {I}_{3} = 2 \cdot {I}_{3} = 2 \cdot \text{2 A" = "4 A}$

The equivalent resistance of the circuit is $40 \Omega$, which means that the voltage rise across the source will be

$V = I \cdot R$

V = 4color(red)(cancel(color(black)("A"))) * 40"V"/color(red)(cancel(color(black)("A"))) = color(green)("160 V")#