Two resistors have resistances R1 and R2, when the resistors are connected in se

Question

Two resistors have resistances R1 and R2, when the resistors are connected in series to a 11.6-V battery, the current from the battery is 1.82 A. when the resistors are connected in parallel to the battery, the total current from the battery is 8.91 A. Determine R1 and R2. (Enter your answers from smallest to largest.) Additional Materials Section 20.7 D Show My Work (Optional) Submit Answer Save Progress N. -10 points CJ10 20.P.062. 0/4 Submissions Used A 61.0-2 resistor is connected in parallel with a 119.0-2 resistor. This parallel group is connected in series with a 20.0-2 resistor. The total combination is connected across a 15.0-V battery (a) Find the current in the 119.0-2 resistor. (b) Find the power dissipated in the 119.0-2 resistor. Additional Materials

Explanation / Answer

In series

I1 = I2 = Is = 1.82 A

Vbattery = Is*(R1 + R2 )

11.6 = 1.82*(R1+R2)

R1+R2 = 6.37

R2 = 6.37-R1 ............(1)

In parallel

Vbattery = Ip*(R1*R2)/(R1+R2)

11.6 = 8.91*(R1*R2)/(R1+R2)......(2)

suing 1 in 2

11.6 = 8.91*(R1*(6.37-R1))/(6.37)

R1 = 1.82 ohms

R2 = 4.55 ohms

===================================

R1 = 61 ohms

R2 = 119 ohms

R3 = 20 ohms

R1 andR2 are in parallel

R12 = (R1*R2)/(R1+R2) = (61*119)/(61+119) = 40.33 ohms

R12 is in series to R3

R123 = R12 + R3 = 60.33 ohms

total current I = V/R123 = 15/60.33 = 0.25 A

potential across R12 = V12 = V1 = V2 = I*R12 = 0.25*40.33 = 10.1

current in R2 = V2/R2 = 10.1/119 = 0.085 A

power dissipate = V2^2/R2 = 10.1^2/119 = 0.857 W

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