Question

Hello! My teacher gave us an optional problem for practice, and I'm really confused about part of it. I understand how to obtain the first two values, but the rest makes no sense to me. Thanks!

http://imgur.com/WSwYskQ

Answer 1

@Michele_Laino @robtobey @jhonyy9 @Kainui @jigglypuff314 @Preetha @imqwerty @IrishBoy123 @Somy @YanaSidlinskiy @rishavraj @sweetburger @ILovePuppiesLol @Jaynator495

Answer 2

its good that you are doing optional problems :-) I know i never did them

Answer 3

now the bad thing is i have absolutely no clue about this problem sorry :(

Answer 4

ok first of all whts the Req?? Did u find tht??

Answer 5

ya, the req is 9 ohms @rishavraj

Answer 6

thanks @ILovePuppiesLol :)

Answer 7

the itotal is 1 amp @rishavraj

Answer 8

yup correct...so whts ur doubt exactly ??

Answer 9

I dont understand how to calculate A1-3 and V1-3

Answer 10

do u know series circuit current remain same at all resistance..... u know tht??

Answer 11

yes

Answer 12

@rishavraj

Answer 13

so wht would be the reading of A2??

Answer 14

@NotSephiroth /?????

Answer 15

i dont know, im confused about that @rishavraj

Answer 16

u see A2 = 2 bcoz current in all resistances would be same... :))

Answer 17

first step:
we can redraw the circuit as below:

where:
\[\Large \begin{gathered}
{R_1} = \frac{1}{{\frac{1}{{12}} + \frac{1}{4}}} = ...? \hfill \\
\hfill \\
{R_2} = \frac{1}{{\frac{1}{2} + \frac{1}{3} + \frac{1}{6}}} = ...? \hfill \\
\end{gathered} \]

Answer 18

total current:
\[{I_{TOT}} = \frac{V}{{{R_{TOT}}}} = \frac{9}{9} = 1\;ampere\]

Answer 19

then:
\[{V_1} = {I_{TOT}}{R_1} = {I_{TOT}} \cdot \frac{1}{{\frac{1}{{12}} + \frac{1}{4}}} = ...?\]

Answer 20

Solve the A3 (ammeter 3) the same way you solved for A1. But use the equivalent resistance of 1 Ohm instead of 3, and of course the total current is 1 amp. You would then have a voltage drop of 1 volt across that network (V2), it is then simply a matter of Ohms Law.