Question

20 Find the total resistance in the diagram below.
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Answer 1

Given :

Three resistors of 24Ω, 8Ω and 4Ω are connected as shown in the figure.

To Find :

Equivalent resistance of the circuit.

Solution :

Do remeber following formulas :-

⧪ Equivalent resistance of parallel :

• 1/R = 1/R₁ + 1/R₂ + ... + 1/Rₙ

⧪ Equivalent resistance of series :

• R = R₁ + R₂ + ... + Rₙ

Now let's solve it step by step :

STEP - 1 :

It is clear from the attachment that, R₂ and R₃ are connected in parallel$.$

➠ 1/R₂₃ = 1/R₂ + 1/R₃

➠ 1/R₂₃ = 1/24 + 1/8

➠ 1/R₂₃ = 1/24 + 3/24

➠ R₂₃ = 24/4

➠ R₂₃ = 6Ω

STEP - 2 :

Finally R₂₃ and R₁ come in series.

➛ R = R₁ + R₂₃

➛ R = 4 + 6

➛ R = 10Ω

∴ Equivalent resistance of the circuit = 10Ω

Answer 2

$\bf{\underline{Given:-}}$

• $\sf{R_1 = 4\Omega}$
• $\sf{R_2 = 24\Omega}$
• $\sf{R_3 = 8\Omega}$

$\bf{\underline{To\:Find:-}}$

Total Resistance across the circuit.

$\bf{\underline{Solution:-}}$

$\sf{It\:can\:be\:clearly\:observed\:in\:the\:circuit\:R_2\:and\:R_3\:are\:connected\:in\:parallel\:combination}$

Hence Equivalent Resistance Across Resistors $\sf{R_2\:and\:R_3}$ is:-

$\sf{\dfrac{1}{R_p} = \dfrac{1}{R_2} + \dfrac{1}{R_3}}$

= $\sf{\dfrac{1}{R_p} = \dfrac{1}{24} + \dfrac{1}{8}}$

= $\sf{\dfrac{1}{R_p} = \dfrac{1+3}{24}}$

= $\sf{\dfrac{1}{R_p} = \dfrac{4}{24}}$

= $\sf{\dfrac{1}{R_p} = \dfrac{1}{6}}$

= $\sf{R_p = 6\Omega}$

Now,

$\sf{R_p\:and\:R_1\:are\:in\:series}$

Hence Equivalent Resistance across $\sf{R_1\:and\:R_p}$ is :-

$\sf{R= R_1 + R_p}$

= $\sf{R = 4\Omega + 6\Omega}$

= $\sf{R= 10\Omega}$

Therefore Total Current across The circuit is $\sf{10\Omega}$

$\bf{\underline{Important\:Notes}}$

Equivalent Resistance in Parallel combination:-

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

Equivalent Resistance in Series Combination:-

$\sf{R_s = R_1 + R_2.....}$