I'm having trouble figuring out 21.36 and 21.92 of the homework.
21.36
When two resistors, R_1 and R_2, are connected in series across a 6.0V battery, the potential difference across R_1 is 4.1 V. When R_1 and R_2 are connected in parallel to the same battery, the current through R_2 is 0.46 A.
Fine values of R_1 and R_2.
21.92
The equivalent resistance between points A and B of the resistors shown in the figure is 25 ohms. Find the value of Resistance R.
I'm having a hard time because there's no voltage given. Am I supposed to incorporate Kirchoff's rules somehow?
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I think you switched the problem #, your first question is problem 21.92.
ReplyDeleteAnyway, I'll answer that first.
Here's what I did:
PARALLEL
-I know that the voltage across PARALLEL path are all the same (V1=V2).
-Since the current through R2 is given, I can calculate the value of R2 [R2= V/I2]
-Then I go to the series circuit where [V=V1+V2]
SERIES
-V batt=6, V1=4.1, find V2 (V2=V-V1)
-Now I have V2 from above, then take R2 from parallel (R2 is still the same) so I can calculate I2 [I2=V2/R2]
-I2 above is also true for I1 because the current through each resistor are the same (I1=I2)
-Last step is finding the R1 [R1=V1/I1]
21.36
-Think of the parallel path (55ohms & R) as one resistor.
-Now the 12ohm is in series with the combination (55ohms & R).
-R_eq of series = sum of resistors, so the R_comb is (25oms-12ohms).
-Put the 12ohms aside, just look the parallel
-Use the value above (R_comb) to find R. Use the parallel formula [1/R_comb= 1/55 + 1/R]
hope this helps
So the way that Michaela did that is correct. Do you completely understand how to get to the answer now?
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