Please answer both questions. The current in series drcuit is 17.0 A, when an ad

Question

Please answer both questions. The current in series drcuit is 17.0 A, when an additional i7,0- resistor is inserted in series, the current drops to 11.5 A What is the resistance in the ariginal drcuit? Addtional Materials Section 20.6 Show My Workoon Submit Anower Save Progres points CJ10 20 P058 04 Submissions Used Two resistors have resistances R1 and R2, when the resisters are connected in series to 12.3-V battery, the arret from the battery is i. 95 A when the resistors are connected in parallel to the battery, the total current from the battery is 8.83 A. Determine Ry and R CEnter your answers from smalilest to largest) My

Explanation / Answer

1)

From the ohm's law,

V = IR

I = V / R

From this formula current (I) is inversely proportional toresistance (R).

I1/I2 = R2/R1

17/11.5 = (R + 17) / R

17R = 11.5(R + 17)

17R = 11.5R + 195.5

5.5R = 195.5

R = 35.55 ohms

2)

when resisters are connected in series then effectiveresistance is Rs = R1 +R2

current through this combination is I1 = 1.95A

EMF = E = 12.3 V

I1 = E/R1 + R2

I1 = 12.3 / (R1 +R2)

1.95 = 12 / (R1 +R2)

R1 + R2   = 6.15 ...........(1)

when resisters are connected in parallel then effectiveresistance is

Rp = R1R2 / (R1 +R2 )

current through this combination is I2 = 8.83 A

EMF = E = 12.3 V

I1 = E / R1R2/(R1 +R2 )

I1 = 12.3 / R1R2 / (R1 +R2)

8.83 = 12.3 / R1R2/(R1 +R2 )

8.83 =12.3 (R1 +R2 ) / R1R2

8.83 =12.3(6.15)/R1R2

R1R2 = 8.87 .................(2)

(R1-R2)2 = (R1+R2)2 - 4(R1R2)

= 37.8225 - 34.8

= 3.02

R1 - R2 = 1.74 .............(3)

solve (1) and (3)

R1 = 3.945 ohms

R2 = 2.205 ohms

         
  

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