Consider the circuit shown below. The internal resistance of the battery is 9.00

Question

Consider the circuit shown below. The internal resistance of the battery is 9.00 , and other resistances are: R1= 40.0 , R2= 15.0 , R3= 45.0 , R4= 30.0 , R5= 35.0 , R6= 40.0 , and R7= 35.0 . The emf of the battery is 200 V .

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Question

( There are a ton of parts to this, so thank you in advance :D )Consider the circuit shown below. The internal resistance of the battery is 9.00 , and other resistances are: R1= 40.0 , R2= 15.0 , R3= 45.0 , R4= 30.0 , R5= 35.0 , R6= 40.0 , and R7= 35.0 . The emf of the battery is 200 V .

Part A

What is the equivalent resistance of the circuit?

Part B

What is the total current put out by the battery?

Part C

What is the power consumed by the battery?

Part D

What is the current through R1 ?

Part E

What is the voltage drop across resistor 1?

Part F

What is the current through the internal resistance of the battery?

Part G

What is the voltage drop across the internal resistance of the battery?

Part H

What is the voltage the battery is actually giving to our circuit, when we account for the internal resistance of the battery?

Part I

What is the current through R2 ?

Part J

What is the voltage drop across R2 ?

Part K

What is the current through R3 ?

Part L

What is the voltage drop across R3 ?

Part M

What is the current through R5 ?

Part N

What is the voltage drop across R5 ?

Explanation / Answer

R2 and R3 are in series. their equivalent is R' then

R' = R2 + R3 = 15 + 45 = 60 ohm

now R' and R5 are in parallel connection,

1/R" = 1/R' + 1/R5 = 1/60 + 1/35

R" = 22.1 ohm

R6 and R7 are in series, r' = R6 + R7 = 40 + 35 = 75 ohm

now r' and R4 are in parallel connection,

1/r" = 1/r' + 1/R4 = 1/75 + 1/30

r" = 21.4 ohm

now R1, R" and r" are in series.

Req = R1 + R" + r" + r = 40 + 22.1 + 21.4 + 9 = 92.5 ohm .........Ans (A)

B)

I = V/Req = 200/92.5 = 2.16 A

C)

P = VI = 200 x 2.16 = 432 W

D)

current through R1, I1 = I = 2.16 A   

e) V1 = I1 R1 = 2.16 x 40 = 86.4 volt

f) I = 2.16 A

g) Vr = 9 x 2.16 = 19.44 volt

h) V = e - Vr =200 - 19.44 = 180.56 volt

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