Explanation:
1. First, let's find the total resistance of the circuit. We begin by combining [tex]R_{4}[/tex], [tex]R_{5}[/tex] and [tex]R_{6}[/tex]:
[tex]R_{456}=R_{4} + \dfrac{R_{5}R_{6}}{R_{5} + R_{6}}[/tex]
[tex]= 6\:Ω + \dfrac{(3\:Ω)(5\:Ω)}{3\:Ω+5\:Ω} = 7.9\:Ω[/tex]
Now time to combine [tex]R_{2}[/tex] and [tex]R_{3}[/tex] and they are connected in series so
[tex]R_{23} =R_{2} + R_{3} = 17\:Ω[/tex]
Note that [tex]R_{23}[/tex] and [tex]R_{456}[/tex] are connected in parallel so
[tex]R_{23456} = \dfrac{R_{23}R_{456}}{R_{23}+R_{456}}=5.4\:Ω[/tex]
Finally, [tex]R_{23456}[/tex] is connected in series with [tex]R_{1}[/tex] so the total resistance [tex]R_{T}[/tex] is
[tex]R_{T} = R_{1} + R_{23456} = 10\:Ω + 5.4\:Ω = 15.4\:Ω[/tex]
2. The total current in the circuit is
[tex]I_{T} = \dfrac{V}{R_{T}} = \dfrac{20\:V}{15.4\:Ω} = 1.3\:A[/tex]
3. The voltage drop across [tex]R_{1},\:V_{1}[/tex] is
[tex]V_{1} = I_{T}R_{1} = (1.3\:A)(10\:Ω) = 13\:V[/tex]
4. We can see that [tex]I_{T} = I_{1} + I_{2}[/tex]. To solve for [tex]I_{1}[/tex], we need [tex]V_{23}[/tex], which is just [tex]V_{T} - V_{1} = 20\:V - 13\:V = 7\:V[/tex] , which gives us
[tex]I_{1} = \dfrac{V_{23}}{R_{23}} = \dfrac{7\:V}{17\:Ω} = 0.4\:A[/tex]
5. From #2 & #4, [tex]I_{2} = 1.3\:A - 0.4\:A = 0.9\:A[/tex] and we also know that the voltage drop across [tex]R_{456}[/tex] is 7 V, the same as that of [tex]R_{23}[/tex]. The voltage drop across [tex]R_{4}[/tex] is
[tex]V_{4} = I_{2}R_{4} =(0.9\:A)(6\:Ω) = 5.4\:V[/tex]
This means that the voltage drop across [tex]R_{6}[/tex] is 7 V - 5.4 V = 1.6 V. Knowing this, the current through [tex]R_{6}[/tex] is
[tex]I_{6} = \dfrac{1.6\:V}{5\:Ω} = 0.3\:A[/tex]
