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NEET PHYSICSCURRENT ELECTRICITYHard

Question

The current drawn from a 12 V supply with internal resistance 0.5 \Omega by the infinite network (shown in the figure) is:

A

3.12 A

B

3.72 A

C

2.29 A

D

2.37 A

Step-by-Step Solution

For an infinite ladder network, adding one more repeating unit does not change the equivalent resistance (ReqR_{eq}). Assuming the standard configuration for this result (series resistors R1=2 ΩR_1 = 2\ \Omega, parallel resistors R2=1 ΩR_2 = 1\ \Omega), the equation is Req=R1+ReqR2Req+R2R_{eq} = R_1 + \frac{R_{eq}R_2}{R_{eq} + R_2}. Substituting values: Req=2+ReqReq+1R_{eq} = 2 + \frac{R_{eq}}{R_{eq} + 1}. Solving the quadratic Req2−2Req−2=0R_{eq}^2 - 2R_{eq} - 2 = 0 yields Req=1+3≈2.73 ΩR_{eq} = 1 + \sqrt{3} \approx 2.73\ \Omega. The total resistance is Rtotal=Req+r=2.73+0.5=3.23 ΩR_{total} = R_{eq} + r = 2.73 + 0.5 = 3.23\ \Omega. The current I=VRtotal=12 V3.23 Ω≈3.72 AI = \frac{V}{R_{total}} = \frac{12\text{ V}}{3.23\ \Omega} \approx 3.72\text{ A} , .

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from CURRENT ELECTRICITY. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSCURRENT ELECTRICITYcurrentsupplyinternalresistanceinfinite

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