Question

Help plzzzzzzzzzzzz ?

Answer 1

Explanation:

1. First, let's find the total resistance of the circuit. We begin by combining $R_{4}$, $R_{5}$ and $R_{6}$:

$R_{456}=R_{4} + \dfrac{R_{5}R_{6}}{R_{5} + R_{6}}$

$= 6\:Ω + \dfrac{(3\:Ω)(5\:Ω)}{3\:Ω+5\:Ω} = 7.9\:Ω$

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

$R_{23456} = \dfrac{R_{23}R_{456}}{R_{23}+R_{456}}=5.4\:Ω$

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

$I_{T} = \dfrac{V}{R_{T}} = \dfrac{20\:V}{15.4\:Ω} = 1.3\:A$

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

$I_{1} = \dfrac{V_{23}}{R_{23}} = \dfrac{7\:V}{17\:Ω} = 0.4\:A$

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

$I_{6} = \dfrac{1.6\:V}{5\:Ω} = 0.3\:A$