Respuesta :

Given :

  • [tex]R_1 = 40 ohms[/tex]

  • [tex]R_2 = 25.4 ohms[/tex]

  • [tex]R_3 = 70.8 ohms[/tex]

Solution :

The resistors [tex]R_1 and R_2 [/tex]are in series connection, so the equivalent resistance of [tex]R_1 and R_2[/tex] will be their sum.

[tex] \boxed{ \mathrm{R_t = R_1 + R_2}}[/tex]

  • [tex]R_t= 40 + 25.4[/tex]

  • [tex]R_t= 65.4 \: \: ohms[/tex]

Now, the equivalent resistance of [tex](R_1 and R_2)[/tex] and [tex]R_3 [/tex] is the total resistance of the circuit, and since they are in parallel connection, Total resistance :

  • [tex] \mathrm{ \dfrac{1}{ R_{eq}}} = \dfrac{1}{65.4} + \dfrac{1}{70.8} [/tex]

  • [tex] \dfrac{1}{R_{eq}} = \dfrac{65.4 + 70.8}{4630.32} [/tex]

  • [tex] \frac{1}{R_{eq}} = \dfrac{136.2}{4630.32} [/tex]

  • [tex]R_{eq} = \dfrac{4630.32}{136.2} [/tex]

  • [tex]R_{eq} = 33.996 \: \: ohms[/tex]

  • [tex] \boxed{ \mathrm{R_{eq} = 34 \: ohms}} \: (approx)[/tex]

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