Question

Find out the total resistance.

Answer 1

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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Answer 2

To calculate the total overall resistance of a number of resistors connected in this way you add up the individual resistances. This is done using the following formula: Rtotal = R1 + R2 +R3 and so on. Example: To calculate the total resistance for these three resistors in series..

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