2.2k2 www. 1.5k2 3.10 v 3.9hr 1.2k2 Using the image above, calculate the following: 1) Theoretical values of Av's across the resistors 2) Theoretical values of vabsolute and I (current) at points A,B,C, D, and E 3) Calculate the power of the battery and the power of all the resistors

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2.2k2
www.
1.5k2
3.10 v
3.9hr
1.2k2
Using the image above, calculate the following:
1) Theoretical values of Av's across the resistors
2) Theoretical values of vabsolute and I (current) at points A,B,C, D, and E
3) Calculate the power of the battery and the power of all the resistors
Transcribed Image Text:2.2k2 www. 1.5k2 3.10 v 3.9hr 1.2k2 Using the image above, calculate the following: 1) Theoretical values of Av's across the resistors 2) Theoretical values of vabsolute and I (current) at points A,B,C, D, and E 3) Calculate the power of the battery and the power of all the resistors
Expert Solution
Step 1

Given in the circuit 

Resistance in AB R1 = 2.2 kΩ

Resistance in BC R2 = 1.5 kΩ

Resistance in CD R3 = 3.9 kΩ

Resistance in DE R4 = 1.2 kΩ

EMF of battery is E = 3.10 volt

 

Step 2

Part 1)

All resistances are connected in series , so net resistance of the circuit is

R =R1+R2+R3+R4

R = (2.2 + 1.5 + 3.9 + 1.2) kΩ

R = 8.8 kΩ

Now current in the circuit is 

I = ERI = 3.10 v8.8×103ΩI = 0.3523×10-3 AI = 0.3523 mA

 

All resistance are connected in series so the same current I is flowing in all resistances.

Now

Voltage drop across the R1 is

V1 = IR1

V1(0.3523×10-3 A)(2.2×103Ω)

V1 = 0.7751 volt

Voltage drop across the R2 is

V2 = IR2V2 = (0.3523×10-3A)(1.5×103Ω)V2 = 0.5285 volt

Voltage drop across R3 is

V3 = IR3V3 = (0.3523×10-3A)(3.9×103Ω)V3 = 1.374 volt

Voltage drop across R4 is

V4 = IR4V4 = (0.3523×10-3A)(1.2×103Ω)V4 = 0.423 volt

 

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