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Current Electricity

JEE Main 2024 (Online) 27th January Evening ShiftMCQEasy📈 Topic in 10% papers+4−1
Ques. 1JEE Main 2024, 27 January (Evening Shift)

Three voltmeters, all having different internal resistances are joined as shown in figure. When some potential difference is applied across A and B, their readings are V1, V2 and V3. Choose the correct option. [figure]

  1. V1=V2
  2. V1 V3-V2
  3. V1+V2=V3
  4. V1+V2>V3

Answer

(3) V1+V2=V3

Solution

Upper branch: V1 and V2 in series; lower branch: V3 alone; both branches are across A and B.

V1+V2=VAB and V3=VAB (KVL round the loop: V1+V2-V3=0)

V1+V2=V3

(A): V1=V2 needs equal resistances (series, same current) (A) is wrong

(B): V1=V3-V2 actually holds (B) is wrong

(C): V1+V2=V3 (C) is correct

(D): V1+V2 equals V3, not greater (D) is wrong

Answer: (3)

Keep practising — Current Electricity

Ques. 2JEE Main 2024, 27 January (Evening Shift)

Wheatstone bridge principle is used to measure the specific resistance (S1) of given wire, having length L, radius r. If X is the resistance of wire, then specific resistance is ; S1=X ( r2L ). If the length of the wire gets doubled then the value of specific resistance will be :

  1. S14
  2. 2 S1
  3. S12
  4. S1

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