Kirchhoff's Law Problem -- Resistors in Series and Parallel and a Switch

Check your first equation ! It's not for R_palrallel but for 1/=R_palrallel !

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

 

I only know $${1\over 12} + {1\over 10}+ {1\over 16}+ {1\over 20}+ {1\over 25} \ne 15.82$$
Your values in brackets are probably the book answers ?

 

Then why ask if the values are correct ?

 

No. $${1\over 12} + {1\over 10}+ {1\over 16}+ {1\over 20}+ {1\over 25} = {1\over 2.98}$$However,$$
{1\over {1\over 12} + {1\over 10}+ {1\over 16} } + {1 \over {1\over 20}+ {1\over 25} } = 4.07 + 11.11 = 15.18$$

 

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

 

The parts d and e are a bit weird, because basically you use KVL to calculate R₇ in a and b.
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

 

+1

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

 

@DevonZA ,

Are you saying that this problem is from a textbook?

What book? Who is the author ?

 

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