Question

please answer my question​

Answer 1

Answer:

R2 and R3 are in series So, resistance of that branch is 4+2 = 6 ohm

Then that branch is in parallel combination with the R1 branch, so then that parallel resistance will become (6)(6)/6+6 = 3 ohm

So, the net resistance of the circuit as observed by the 6 volt battery will be 3 ohm and hence the net current flowing through the battery will be = 6/3 = 2 ampere (By ohm's law)

So, since the ammeter has been connected directly in series with the battery and assuming that ammeter is ideal(has negligible voltage drop accross it) we can say that current measured by this ammeter will be the same current flowing through the battery)

Hence the current reading given by ammeter will be 2 ampere.

Answer 2

Answer:

Now in the fig  R₃ and R₂ are in series

R₂,₃ = R₃+R₂

=4Ω+2Ω

=6Ω

Now R₂,₃ and R₁ are in parallel

1/R= 1/ R₂,₃ +1/R₁

=1/6+1/6

=2/6

=1/3

∴R=3Ω

Now we need to find the value of current

V=6 v

V=IR

6=I x 6

∴I= 1 A

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