Respuesta :
[tex]\huge\boxed{\fcolorbox{black}{red}{ QUESTION}}[/tex]
There are two sets of resistors joined in series in a circuit. Set 1 has two resistors of 10Ω and 40Ω in parallel combination. Set 2 has three resistors of 20 Ω, 30 Ω and 60 Ω in parallel combination. A potential difference of 12 V is applied across the combination.
[tex] \orange{\underline{\huge{\bold{\textit{\green{\bf{GIVEN}}}}}}}[/tex]
Two set of Parallel resistors
Contains two resistors of --> 10Ω and 40Ω
Contains three resistors of --> 20 Ω, 30 Ω and 60Ω
Potential Difference Of 12 V
[tex]{\bold{\orange {To \: Find }}}[/tex]
Draw a circuit diagram to represent this arrangement Calculate
(a) the total resistance and
(b) the total current flowing in the circuit.
[tex] \huge\mathbb{\red A \pink{N}\purple{S} \blue{W} \orange{ER}}[/tex]
FOR DIAGRAM REFER ATTACHMENT!
[tex]{\bold{\green{\star{\red {Part \: 1:- The \: Total \: Resistance }}}}}[/tex]
First we calculate the total resistance of each parallel combination:-
FIRST SET Contains two resistors of 10Ω and 40Ω
so total resistance-->
[tex] \frac{1}{Rs1} = \frac{1}{R1} + \frac{1}{R2} \\ \\ \frac{1}{Rs1} = \frac{R1 + R2}{R1R2} \\ \\ \frac{1}{Rs1} = \frac{10 + 40}{400} \\ \\ \frac{1}{Rs1} = \frac{50}{400} \\ \\ \frac{1}{Rs1} = \frac{1}{8} \\ \\ Rs1 = 8 Ω[/tex]
SECOND SET Contains three resistors of 20 Ω, 30 Ω and 60Ω
so total resistance-->
[tex] \frac{1}{Rs2} = \frac{1}{R1} + \frac{1}{R2} + \frac{1}{R3} \\ \\ \frac{1}{Rs2} = \frac{R1 + R2 +R3}{R1R2R3} \\ \\ \frac{1}{Rs2} = \frac{3+ 2 + 1}{60} \\ \\ \frac{1}{Rs2} = \frac{6}{60} \\ \\ \frac{1}{Rs2} = \frac{1}{10} \\ \\ Rs2 = 10Ω[/tex]
Now by having Total Resistance of Set one and set two we can add them by series formula.
[tex]R = Rs1 + Rs2 \\ R = 8 + 10 = 18 \: Ω[/tex]
[tex]{\bold{\green{\star{\orange{Part \: 2:- The \: Total \: Current \: Flowing \: In \: The \: Circuit. }}}}}[/tex]
Total Resistance of the circuit = 18 Ω
Potential Difference Of the circuit = 12 V
So Total current-->
[tex] V = IR \\ 12 = 18 \times I \\ \\ I = 0.66 \: A[/tex]