1.

Find out the total resistance.

Answer»

Question:-

Find out the total resistance.

Note:-

  • Its important to name all the resistance in order to MAKE it easier to find the resistance. (refer to the attachment provided)

Required SOLUTION:-

From the attachment we can clearly see, we have:-

  • R₁ = 12.00 Ω
  • R₂ = 5.00 Ω
  • R₃ = 4.00 Ω
  • R₄ = 8.00 Ω
  • R₅ = 6.00 Ω

Also we can see that:-

  • R₁ and R₅ are in parallel combination
  • R₃ and R₄ are also in parallel combination.

We KNOW,

EFFECTIVE resistance in a parallel combination is calculated by the following formula:-

  • \sf{\dfrac{1}{R_p} = \dfrac{1}{R_1} + \dfrac{1}{R_2} . . . . . . . \dfrac{1}{R_n}}

Hence,

For resistors R₁ and R₅ effective resistance is as FOLLOWS:-

= \sf{\dfrac{1}{R_p} = \dfrac{1}{12} + \dfrac{1}{6}}

= \sf{\dfrac{1}{R_p} = \dfrac{1+2}{12}}

= \sf{\dfrac{1}{R_p} = \dfrac{3}{12}}

= \sf{\dfrac{1}{R_p} = \dfrac{1}{4}}

=> \sf{R_p = 4\Omega}

∴ Effective Resistance in the resistors R₁ and R₅ is 4Ω.

For resistors R₃ and R₄ effective resistance is as follows:-

= \sf{\dfrac{1}{R_t} = \dfrac{1}{4} + \dfrac{1}{8}}

= \sf{\dfrac{1}{R_t} = \dfrac{2+1}{8}}

= \sf{\dfrac{1}{R_t} = \dfrac{3}{8}}

=> \sf{R_t = \dfrac{8}{3}}

=> \sf{R_t = 2.7 \Omega}

∴ Effective Resistance in the resistors R₃ and R₄ is 2.7 Ω.

Now,

The resistors \sf{R_p,\:R_2\:and\:R_t} have come in series combination.

We know,

Effective resistance in a series combination is calculated by the following formula:-

  • \sf{R_s = R_1 + R_2 . . . . . . . . R_n}

Hence,

The effective resistance in resistors \sf{R_s,\:R_2\:and\:R_t} is as follows:-

\sf{R_s} = 4 + 5 + 2.7

=> \sf{R_s} = 11.7 Ω

Therefore, The total resistance across a and b is 11.7 Ω.

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