Question

Help plzzzzzzzzzzzz ?

Answer 1

Step-by-step explanation:

1. First, let's find the total resistance of the circuit. We begin by combining
`R_(4)`,
`R_(5)` and
`R_(6)`:

`R_(456)=R_(4) + (R_(5)R_(6))/(R_(5) + R_(6))`

`= 6\:Ω + ((3\:Ω)(5\:Ω))/(3\:Ω+5\:Ω) = 7.9\:Ω`

Now time to combine
`R_(2)` and
`R_(3)` and they are connected in series so

`R_(23) =R_(2) + R_(3) = 17\:Ω`

Note that
`R_(23)` and
`R_(456)` are connected in parallel so

`R_(23456) = (R_(23)R_(456))/(R_(23)+R_(456))=5.4\:Ω`

Finally,
`R_(23456)` is connected in series with
`R_(1)` so the total resistance
`R_(T)` is

`R_(T) = R_(1) + R_(23456) = 10\:Ω + 5.4\:Ω = 15.4\:Ω`

2. The total current in the circuit is

`I_(T) = (V)/(R_(T)) = (20\:V)/(15.4\:Ω) = 1.3\:A`

3. The voltage drop across
`R_(1),\:V_(1)` is

`V_(1) = I_(T)R_(1) = (1.3\:A)(10\:Ω) = 13\:V`

4. We can see that
`I_(T) = I_(1) + I_(2)`. To solve for
`I_(1)`, we need
`V_(23)`, which is just
`V_(T) - V_(1) = 20\:V - 13\:V = 7\:V` , which gives us

`I_(1) = (V_(23))/(R_(23)) = (7\:V)/(17\:Ω) = 0.4\:A`

5. From #2 & #4,
`I_(2) = 1.3\:A - 0.4\:A = 0.9\:A` and we also know that the voltage drop across
`R_(456)` is 7 V, the same as that of
`R_(23)`. The voltage drop across
`R_(4)` is

`V_(4) = I_(2)R_(4) =(0.9\:A)(6\:Ω) = 5.4\:V`

This means that the voltage drop across
`R_(6)` is 7 V - 5.4 V = 1.6 V. Knowing this, the current through
`R_(6)` is

`I_(6) = (1.6\:V)/(5\:Ω) = 0.3\:A`