the resistors as I1, I2, I3, respectively. (I is denoted as current)
Which of the following relationships hold for the circuit and explain your answer
i) I1 > I2 > I3
ii) I2=I3
iii) I3 > I2
iv) None of the above
b)To verify that your answer is correct, calculate all 3 currents
c)Calculate the potential difference across each resistor and the power dissipated by each resistor
HERE'S THE FIGURE
http://farm8.staticflickr.com/7204/6945498187_20b6ffdaa1_m.jpg
Which of the following relationships hold for the circuit and explain your answer
i) I1 > I2 > I3
ii) I2=I3
iii) I3 > I2
iv) None of the above
b)To verify that your answer is correct, calculate all 3 currents
c)Calculate the potential difference across each resistor and the power dissipated by each resistor
HERE'S THE FIGURE
http://farm8.staticflickr.com/7204/6945498187_20b6ffdaa1_m.jpg
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a) This is very simple, the current ( I1) will be greatest out of resistor one than the other two. Then I2 will be the next, since the current through parallel resistors are added, and that involves the problem, V/R = I, so since R2 is smaller than R3, I2 will be greater than I3.
b) Break it all down into 1 resistor. First you combine the R2 and R3, which is by putting it like this
R_eq = [(1/R1)+(1/R2)]^-1 which should come out as 4 ohms. Now you have R1 and R2,3 which are both 4 ohms. With series resistors, you just add them, and you get 8ohms. Now, with that, you find the amps that go through the WHOLE circuit, using I = V/R_all which should give you 1.5A.
Then, start to rebuild the circuit one at a time. So, bring it back into two series circuits, and since it's a series, the current (your amps) are the same. Now, you find that you have the same potential, since they are currently resistors of the same resistance (and resistors in series add up their voltage to have the same voltage from the battery).
Now you have one last rebuild left, and that is to put the second (combined) resistor into its original form of 6ohms and 12 ohms. Now, with parallel resistors, you have the same amount of voltage, but the current is summed between the two to get you 1.5A. The equation will then be I = V/R, which should get you I2 = 1A, and I3 = 0.5A.
Thus proving that I1 > I2 > I3 (I1 = 1.5A, I2 = 1A, I3 = 0.5A)
c) Now, since you went through part b, you now have ALL the voltage differences for each resistor, and you have all the variables needed to find the power dissipated. Now, the Power Dissipated formula is P_R = I*V_R = ((V_R)^2)/R which should give you P_R1 = 9, P_R2 = 6, and P_R3 = 3.
Hope this is helpful, and good luck with Dr. Menke's midterm today :P
b) Break it all down into 1 resistor. First you combine the R2 and R3, which is by putting it like this
R_eq = [(1/R1)+(1/R2)]^-1 which should come out as 4 ohms. Now you have R1 and R2,3 which are both 4 ohms. With series resistors, you just add them, and you get 8ohms. Now, with that, you find the amps that go through the WHOLE circuit, using I = V/R_all which should give you 1.5A.
Then, start to rebuild the circuit one at a time. So, bring it back into two series circuits, and since it's a series, the current (your amps) are the same. Now, you find that you have the same potential, since they are currently resistors of the same resistance (and resistors in series add up their voltage to have the same voltage from the battery).
Now you have one last rebuild left, and that is to put the second (combined) resistor into its original form of 6ohms and 12 ohms. Now, with parallel resistors, you have the same amount of voltage, but the current is summed between the two to get you 1.5A. The equation will then be I = V/R, which should get you I2 = 1A, and I3 = 0.5A.
Thus proving that I1 > I2 > I3 (I1 = 1.5A, I2 = 1A, I3 = 0.5A)
c) Now, since you went through part b, you now have ALL the voltage differences for each resistor, and you have all the variables needed to find the power dissipated. Now, the Power Dissipated formula is P_R = I*V_R = ((V_R)^2)/R which should give you P_R1 = 9, P_R2 = 6, and P_R3 = 3.
Hope this is helpful, and good luck with Dr. Menke's midterm today :P