These problems pertain to sections 18.1, 18.2, and 18.3. You should be able to deal with all of them using combinations of series and parallel resistors.
Note that question 4 part C asks for the power delivered by the battery, but the dimension given with the answer is Amperes. That should be Watts. (Doesn't affect how you approach the problem -- I just mention it to avoid possible confusion.)
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Hello,
I am not grasping how to figure out part b for question 2. I understand how to find the current for the first point, but then it comes down to the second and third points, I get lost. I have tried doing this problem closely using example 18.3 from the text but the algebra the book uses is lost on me. I keep getting 6.14Amps for a 7ohm resistor. It doesn't click for me yet.
Help!
Hellooo..
For question 2 part b the second and third parts are closely related because they are resistors in parallel. There are 2 ways you can approach this:
1) You can either calculate the voltage drop (deltaV) for the parallel resistors and then solve using the resistances as shown in the diagram. You have an example of this in your lecture notes from Sept. 23rd.
OR
2) Since you know voltage drops for both resistors that run parallel are equal (V1=I1R1 and V2=I2R2, where V1 is equal to V2) then you can set up an equation with 2 unknown variables. Manipulate and solve. Keep in mind that the total current (Itotal) remains the same throughout the circuit and splits unevenly due to different resistor values where the circuit is parallel. Therefore the sum of these resistors will equal Itotal.
Prof. Stevens,
I am haing difficulty on question 3 figuring out the current in the 12ohm resistor. in my figure my R1 resistor is rated at 4.6 ohms and the R2 resistor is rated at 6.8 ohms. the voltage of the battery is 16volts. i assume in the figure everyone has the same 4ohm and 12ohm resistors in parallel and that is in series to the 2 ohm resistor. i began the problem by combining all of the resistors into a single resistor of 7.6769ohms. from there i used the current formula I=V/R where it was 16/7.6769 wich gave me 2.084amps for the total current. I then used the method that you discussed during thursday's lecture to solve for the current in a resistor that is in parallel. the 4ohm and 12ohm parallel resistors give a total resistance of 3ohms. i used the voltage equation V=IR to solve for the voltage across the resistors in parallel. V=(2.0841A)*(3ohms)=6.2525volts. i then proceded to use the current formula to solve for the current throught the 12ohm resistor I=(6.2525volts)/12ohms which gave me 0.52104amps. I would like to know whenre i am going wrong so that i can grasp the concept of this problem better... Thank you
Hey Chris...
Looks like you have the right idea but you just have to apply it twice. The circuit splits even before the 4ohm and 12ohm parallel resistors so the current used in your formula should be different. Hope this helps!
I just want to clarify that in the last sentence of my first post I meant to say the sum of the CURRENTS through the parallel resistors will equal Itotal.
Thank you ernest for the help. i had to re-think the problem. i realized that the total current is equal to the sums of the other currents.
Although I am having difficulty on question 4 part (d). It says to find the power delivered to the 50ohm resistor. In my example R1 is 29ohms and R2 is 73ohms, also i have a 12volt terminal voltage. i have solved the first 3 parts of the question. In par (a) i reduced the all of the resistors into a single resistor of 57.940ohms. Next in part (b) i determined that the current delived by the battery is 0.2071Amps by reducicing all of the resistors into one and using the formula I=V/R. For part (c) i determined the power delivered to the battery to be 2.48526Watts by using P=(I^2)*R = (.2071^2)*(57.94ohms)= 2.4852watts. However in part (d) i am having difficulty. In solving this part is it required to seperate the total current into its sums? or should the power formula be applied directly? i am at a loss right now on this problem and any assistance would point me in the right direction... thank you
Vito:
The 5:30 post is correct. Have a look at how we found the current through the 2 and 3 ohm resistors in lecture on Thursday.
Ernest, thanks for jumping in to help Chris. (Is that Ernest Batista, the TA? If so, fans can meet him MWF from 4:20 in the help room.)
Last Chris post: First, one small comment - for part c, the power is delivered by the battery - energy flows out of the battery to the resistors. To get part d, you have to work out how the total current splits between the R1+50 ohm branch and the R2 branch. Thats the same problem that started this stream, and you can get it by finding the voltage between the leftmost and rightmost parts of the drawing. Then Ohm's law on R1+50 will give you the current through that branch, and knowing the current through the 50 ohm resistor lets you work out the power dissapated there.
Thank You prof. Stephens, your direction helped me realize how the current splits and how the voltage changes. I solved by using I1R1=I2R2 and that total current I=I1+I2.
Howeve, Now i have some difficulty on question 5 part(a). In my question there are 3 60watt, 100volt lightbulbs conntected to a 100volt power source. The question asks to find the total power delivered to the 3 bulbs. I started this problem similarly to how we had lightbulb questions in lecture. First i solved for the resitance of the bulbs by R=(V^2)/P and (100^2)/60watts=166.667A. Then I solved for the current using I=V/R where 100/166.667=.5999Amps. Then used the power forluma P=(I^2)R where (.59999^2)*(166.667)=59.999watts. I know this is incorrect and i keep trying other methods of determining the power. Any advice would be much appreciated...thank you
Chris, you're on a good start with the resistances of the bulbs, but since none is connected across 100 V, none of them dissipates the 60 W power. Look at it as a network of 3 resistors of 166.7 Ohms (not A) each. Figure the equivalent resistance to find the current that comes from the 100 V circuit, and then determine the current through each bulb.
Hey professor,
I am stuck on number six.
My values are: R1 = 14.5 and R2 = 4.50
to find the potential difference (part a) I am simplifying the circuit to get "Req" by first finding the effective resistance in the parallel R1 and R2 resistors then the rest are in series so I simply add them up and result with a value of 42.4342 ohm.
To get the potential difference I am then finding the current in the resistors in between points a and b. Is that logic correct?
Any help would be great thanks!
-DM
and I am using that current to find the potential of course
Dan, I don't get the same total effective resistance that you do. Consolidate R2 and 20 Ohms in series, and then you have 3 resistors in parallel to get an effective resistance between a and b; call that Rab. That plus R1 is the total resistance, which tells you the current. Then that current through Rab will tell you the voltage between a and b.
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