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Current Electricity

JEE Main 2024 (Online) 29th January Evening ShiftNumericalEasy📈 Topic in 40% papers+4
Ques. 1JEE Main 2024, 29 January (Evening Shift)

In the given circuit, the current flowing through the resistance 20 is 0.3 A, while the ammeter reads 0.9 A. The value of R1 is . [figure]

Answer

30

Solution

R1, 20 Ω and 15 Ω are in parallel (same p.d. V); the ammeter reads the total current.

V=0.3 20=6 V

I15= 615=0.4 A

IR1=0.9-(0.3+0.4)=0.2 A

R1= 60.2=30

Answer: 30

Keep practising — Current Electricity

Ques. 2JEE Main 2024, 29 January (Evening Shift)

In the given figure, the charge stored in 6 capacitor, when points A and B are joined by a connecting wire is . [figure]

Ques. 3JEE Main 2024, 27 January (Morning Shift)

A wire of length 10 cm and radius 7 10-4 m is connected across the right gap of a meter bridge. When a resistance of 4.5 is connected on the left gap by using a resistance box, the balance length is found to be at 60 cm from the left end. If the resistivity of the wire is R 10-7 , then value of R is :

  1. 63
  2. 70
  3. 66
  4. 35
Ques. 4JEE Main 2024, 27 January (Morning Shift)

A wire of resistance R and length L is cut into 5 equal parts. If these parts are joined parallely, then resultant resistance will be :

  1. 125R
  2. 15R
  3. 25 R
  4. 5 R
Ques. 5JEE 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
Ques. 6JEE 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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