Depicts the following scenario. In the circuit shown in the sketch below, the em

Question

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 Ohm. Now we will consider some slightly different related scenarios to Example 21-13. Refer back to Example 21-13. Suppose the three resistors in the circuit have the values R_1 = 130 Ohm, R_2 = 250 Ohm, and R_3 = 350 Ohm, and that the emf of the battery is 18.0 V. Find the potential difference across each resistor. Find the current that flows through each resistor. Refer back to Example 21-13. Suppose R_1 = R_2 = 225 Ohm and R_3 = 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. Find the value of R that gives a potential difference of 2.65 V across resistor 2.

Explanation / Answer

Part A:
R1 =130 ohms ,R2 = 250 ohms, R3 = 350 ohms

V = 18 V

R23 = 250+350 = 600 ohms

R123 = (250*600)/(250+600) = 176.5 ohms

in parallel combinaiton current is constant

V1 = 18 V

I1 = V1/R1 = 18/130

I1 = 0.1385 A

V23 = 18 V

I23 = 18/600 = 0.03 A

in series combination current is constant

I2 = 0.03 A, I3 = 0.03 A

V2 = 0.03*250 = 7.5 V

V3 = 0.03*350 = 10.5 V

DV1,DV2,DV3 = 18V, 7.5 V, 10.5 V

Part B: I1, I2 , I3 = 0.1385 A, 0.03 A, 0.03 A

Part C:

R1 = R2 = 225 ohms , R3 = R, V = 12 V

I = 0.075 A

R23 = 225 +R

R123 = (225+R)*225/(225+R+225)

V = IR

12 = 0.07*(225+R)*225/(225+R+225)

R = 495 ohms

Req = (225+495)*225/(225+225+495)

Req = 171.43 ohms

Part D: R23 = 225 +R

R123 = (225+R)*225/(225+R+225)

V23= V1 = 12 V

in parallel combination voltage is constant

V23 = I23*R23

I23 = V23/R23

in sereis combinaiton current is constant

I2 = I3

V2 = I23*R2 = (v23/R23)*R2

2.6 = (12/(225+R))*(225)

R = 813 ohms

Related Questions

Navigate

• Browse (All)
• Browse D
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.