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JEE Main 2024 (5 Apr, Shift 1) · Combination of Resistors

Combination of Resistors question from JEE Main 2024 (5 Apr, Shift 1)

Q25JEE Main 2024 · 5 APR · SHIFT 1Medium

Combination of Resistors

In the given figure R1=10 Ω,R2=8 Ω,R3=4 ΩR_1 = 10\ \Omega, R_2 = 8\ \Omega, R_3 = 4\ \Omega and R4=8 ΩR_4 = 8\ \Omega. Battery is ideal with emf 12 V12\text{ V}. Equivalent resistance of the circuit and current supplied by battery are respectively:

Figure

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Answer: (d)

  1. R2R_2, R4R_4 and R3R_3 all connect the same top node to the same bottom wire, so they are in parallel.
  2. 1Rp=18+18+14=12⇒Rp=2 Ω\dfrac{1}{R_p} = \dfrac18 + \dfrac18 + \dfrac14 = \dfrac12 \Rightarrow R_p = 2\ \Omega
  3. Req=R1+Rp=10+2=12 ΩR_{eq} = R_1 + R_p = 10 + 2 = 12\ \Omega, I=1212=1I = \dfrac{12}{12} = 1 A
  4. So the answer is (d) 12 Ω12\ \Omega and 11 A.
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