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## Homework Statement

## Homework Equations

Rparellel=1/R1+1/R2...

IR1=R2/R1+R2 x I

## The Attempt at a Solution

I am unsure of how to answer d) and e) using KVL because I count 4 junctions?

Where should I start?

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- Thread starter DevonZA
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- #1

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Rparellel=1/R1+1/R2...

IR1=R2/R1+R2 x I

I am unsure of how to answer d) and e) using KVL because I count 4 junctions?

Where should I start?

- #2

BvU

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And the 20 and 25 Ohm resistors are NOT parallel to e.g. the 12 Ohm !

- #3

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The total resistance for the parallel resistors still equals 15.18ohms but yes I should have done them separately:

(1/12+1/10+1/16)^-1 = 4.07ohms

(1/20+1/25)^-1= 11.11ohms

Do my values look correct?

- #4

BvU

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Your values in brackets are probably the book answers ?

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(1/12+1/10+1/16+1/20+1/25)^-1=15.18 ohms

Yes values are answers in the book

Yes values are answers in the book

- #6

BvU

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Then why ask if the values are correct ?

- #7

BvU

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No. $${1\over 12} + {1\over 10}+ {1\over 16}+ {1\over 20}+ {1\over 25} = {1\over 2.98}$$However,$$(1/12+1/10+1/16+1/20+1/25)^-1=15.18 ohms

{1\over {1\over 12} + {1\over 10}+ {1\over 16} } + {1 \over {1\over 20}+ {1\over 25} } = 4.07 + 11.11 = 15.18$$

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- #9

SammyS

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What sort of device is R7 that its resistance changes depending on whether the switch is opened or closed## Homework Statement

[ ATTACH=full]99511[/ATTACH]

- #10

BvU

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And it wouldn't be proof, just showing.

I should think they want you to show that 1.2 A * (4.07 + 60 + 11.11 + 24.82) Ohm = 120 V

- #11

CWatters

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+1

Although I prefer to make the voltages around the loop sum to zero.

Although I prefer to make the voltages around the loop sum to zero.

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Can you show this?+1

Although I prefer to make the voltages around the loop sum to zero.

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- #15

CWatters

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Although I prefer to make the voltages around the loop sum to zero.

Can you show this?

With reference to the equivalent circuit below... KVL says that going around a loop the voltages sum to zero.

If I arbitrarily choose to start at the -ve terminal of the battery and go around clockwise we have to prove that..

+120 + (-V1) + (-V2) + (-V3) + (-V4) = 0

V1 = 1.2 * 4.07 = 4.884V

V2 = 1.2 * 60 = 72V

V3 = 1.2 * 11.11 = 13.332V

V4 = 1.2 * 24.82 = 29.784V

Substitute..

+120 - 4.884 - 72 -13.332 - 29.784 = 0

- #16

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Thank you for the nice clear explanation CWattersWith reference to the equivalent circuit below... KVL says that going around a loop the voltages sum to zero.

View attachment 99596

If I arbitrarily choose to start at the -ve terminal of the battery and go around clockwise we have to prove that..

+120 + (-V1) + (-V2) + (-V3) + (-V4) = 0

V1 = 1.2 * 4.07 = 4.884V

V2 = 1.2 * 60 = 72V

V3 = 1.2 * 11.11 = 13.332V

V4 = 1.2 * 24.82 = 29.784V

Substitute..

+120 - 4.884 - 72 -13.332 - 29.784 = 0

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