In the given figure R₁ = 10 Ω, R₂ = 8 Ω, R₃ = 4 Ω and R₄ = 8 Ω… | JEE Main 2024 (5 Apr, Shift 1)

Q25JEE Main 2024 · 5 APR · SHIFT 1Medium

Combination of Resistors

In the given figure $R_1 = 10\ \Omega, R_2 = 8\ \Omega, R_3 = 4\ \Omega$ and $R_4 = 8\ \Omega$ . Battery is ideal with emf $12\text{ V}$ . Equivalent resistance of the circuit and current supplied by battery are respectively:

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Answer: (d)

$R_2$ , $R_4$ and $R_3$ all connect the same top node to the same bottom wire, so they are in parallel.
$\dfrac{1}{R_p} = \dfrac18 + \dfrac18 + \dfrac14 = \dfrac12 \Rightarrow R_p = 2\ \Omega$
$R_{eq} = R_1 + R_p = 10 + 2 = 12\ \Omega$ , $I = \dfrac{12}{12} = 1$ A
So the answer is (d) $12\ \Omega$ and $1$ A.

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