Problem 21.107-21.108 Example 21-13 depicts the following scenario. In the circu

Question

Problem 21.107-21.108 Example 21-13 depicts the following scenario. In the circuit shown in the sketch below, the emf of the battery is 12.0 V, and each resistor has a resistance of 200.0 .

Part A

Refer back to Example 21-13. Suppose the three resistors in the circuit have the values R1 = 130 , R2 = 290 , and R3 = 360 , and that the emf of the battery is 18.0 V . Find the potential difference across each resistor.

Enter your answers numerically separated by commas. V1,V2,V3 = ________m V

Part B

Find the current that flows through each resistor.

Enter your answers numerically separated by commas.

I1,I2,I3=________mA

Part C

Refer back to Example 21-13. Suppose R1=R2=225 and R3=R . The emf of the battery is 12.0 V . Find the value of R such that the current supplied by the battery is 0.0750 A .

Req=__________    

Part D

Find the value of R that gives a potential difference of 2.65 V across resistor 2.

R=__________      

R R /g - e

Explanation / Answer

Given
Part A

   R1 = 130 ,R2 = 290 , and R3 = 360

battery V = 18 V

the potential difference across each resistor is

the resistors R1, (R2R3) are in parallel combination so the potential differnce issame across R1 and R2R3

Potential difference across R1 = 18 V

and across R2 is Dv2 = R3*V/(R2+R3) = 360*18/(290+360) = 9.97 V = 9970 mV

and across R3 is Dv3 = R2*V/(R2+R3) = 290*18/(290+360) = 8.03 V = 8030 mV

PArt B

equivalent resistance of the circuit is R = (R2+R3)(R1)/((R2+R3)+(R1))

               = (290+360)(130)/(360+290+130) ohm

                   = 108.33 ohm

the current in the circuit is I = V/R = 18/108.33 A = 0.1662 A

   current through R1 is I1 = 18/130 = 0.138462

and the current through R2,R3 is = 0.1662 -0.138462 A = 0.027738 V

Part C

   R1=R2=225 ohm

   R3 = R ohm
net resistace is R = (R2+R3)(R1)/((R2+R3)+(R1))

       = (225+R)(225)/(225+R+225)

v = 12 V

I = 0.0750 A    ==> I = V/R ==> R = V/I = 12/0.075 = 160 ohm

       (225+R)(225)/(225+R+225) = 160

       R = 328.85 ohm

PArt D

across R2 is Dv2 = R3*V/(R2+R3)

       2.65 = 328.85*12/(R2+328.85)

       R2 = 1160.282 ohm

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