Question #9f759

3 Answers

Answer:

Circuit Simplification.

Explanation:

enter image source here

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

#R_"eq 1" =20+10+20=50Omega#.

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

#R_"eq 2" = 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_"eq 3" = 25+25= 50Omega#

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

#R_"eq 4" = 50/2=25Omega#

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

#R_"eq 5" = 10+5 + 25 = 40Omega#

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

Answer:

For part (b) #V= "160 V"#

Explanation:

So, your circuit looks like this

enter image source here

You know that the power dissipated by the #10Omega# resistor located in the lower right of the circuit is #"10 W"#.

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

#color(blue)(P = I^2 * R implies I = sqrt(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.

enter image source here

So, current #I_1# is coming out of the source and going into point #color(blue)(A)#, where it splits into currents #I_2# and #I_3#. Current #I_3# goes into point #color(blue)(B)# and plits into currents #I_4# and #I_5#.

You know that #I_4 = "1 A"#, since that's the current that goes through the #"10-W"# resistor. Now, because the equivalent resistance of those three resistors in series is equal to #50Omega#, #I_5# will be equal to #I_4#.

That happens because both currents pass through #50Omega# resistors.

This means that

#I_3 = I_4 + I_5 = 2 * I_4 = 2 * "1 A" = "2 A"#

The equivalent resistance between points #color(blue)(B)# and #color(blue)(C)# is equal to #25Omega#, since you have two #50Omega# resistors in parralel.

This means that the equivalent resistance between points #color(blue)(A)# and #color(blue)(C)# will be #50Omega#, since you have two #25Omega# resistors in series.

This means that #I_3# is equal to #I_2#, since they both pass through #50Omega# resistors. The current that's coming from the source, #I_1#, will thus be

#I_1 = I_2 + I_3 = 2 * I_3 = 2 * "2 A" = "4 A"#

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

#V = I * R#

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