Three resistors R1 = 84.5 , R2 = 24.2 , R3 = 70.0 , and two batteries e m f1 = 4
ID: 1579578 • Letter: T
Question
Three resistors R1 = 84.5 , R2 = 24.2 , R3 = 70.0 , and two batteries e m f1 = 40.0 V, and e m f2 = 353 V are connected as shown in the diagram below.
(a) What current flows through R1, R2, and R3?
I1 =
I2 =
I3 =
(b) What is the absolute value of the potential difference across R1, R2, and R3?
|VR1| =
|VR2| =
|VR3| =
Explanation / Answer
let the current in R1 be I1 from bottom to top and let it be I2 in R2 from top to bottom
Applying KVL we got
E1+E2=I1*R1 +I2*R2=40+353=84.5I1+24.2I2..............(1)
current in R3 will be (I1-I2)
applying KVL we got
E2=I2*R2-(I1-I2)*R3=353=24.2I2-(I1-I2)*70
353=94.2I2-70I1...............(2)
solving (1) and (2) we got
I1=2.94 A
potential across R1=I1*R1=84.5x2.94=248.43 V
I2=5.93 A
potential across R2=I2*R2=24.2x5.93=143.50v
current in R3=I1-I2=5.93-2.94=2.99 V
potential across R3=I3xr3=2.99x70=209.3v
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