Home
Class 12
PHYSICS
For the given circuit, If internal ...

For the given circuit,

If internal resistance of cell is 1.5`Omega`, then

A

`V_(P)-V_(Q)=0`

B

`V_(P)-V_(Q)=4V`

C

`V_(P)-V_(Q)=-4V`

D

`V_(P)-V_(Q)=-2.5V`

Text Solution

Verified by Experts

The correct Answer is:
D
Promotional Banner

Topper's Solved these Questions

  • CURRENT ELECTRICITY

    AAKASH SERIES|Exercise LECTURE SHEET (EXERCISE-III LEVEL- II (ADVANCED) (Linked Comprehension Type Questions)|2 Videos
  • CURRENT ELECTRICITY

    AAKASH SERIES|Exercise LECTURE SHEET (EXERCISE-IV LEVEL - I (MAIN) (Straight Objective Type Questions))|5 Videos
  • CURRENT ELECTRICITY

    AAKASH SERIES|Exercise LECTURE SHEET (EXERCISE-III LEVEL- I (MAIN) (Straight Objective Type Questions))|7 Videos
  • CURRENT ELECTRICITY

    AAKASH SERIES|Exercise PROBLEMS (LEVEL-II)|27 Videos
  • DUAL NATURE OF RADIATION AND MATTER

    AAKASH SERIES|Exercise PRACTICE EXERCISEX|42 Videos

Similar Questions

Explore conceptually related problems

For the given circuit, if internal resistance of cell is 1.5Omega then V_p-V_q is equal to:

Assetion : In the following circuit emf is 2 V and internal resistance of the cell is 1 Omega and R = 1 Omega , then reading of the voltmeter is 1 V . Reason : V = E - ir where E = 2 v, i= (2)/(2) = 1 A and r = 1 Omega

In the given circuit the internal resistance of the 18 V cell is negligible. If R_(1)=400 Omega and the reading of an ideal voltmeter across R_(4) is 5V, then the value of R_(2) will be :

In the circuit shown , the internal resistance of the cell is negligible . The steady stale current in the 2 Omega resistor is

A capacitor of 4 mu F is connected as shown in the circuit. The internal resistance of the battery is 0.5 Omega . The amount of charge on the capacitor plates will be .

A capacitor of 4 mu F is connected as shown in the circuit. The internal resistance of the battery is 0.5 Omega . The amount of charge on the capacitor plates will be .

In the circuit shows in Fig. 6.74, the internal resistance of the cell is negligible. For the value of R = 40//x Omega , no current flows through the galvanometer. What is x ?

In a mixed grouping of identical cells, five rows are connected in parallel and each row contains 10 cells. This combination sends a current I through an external resistance of 20Omega. if the emf and internal resistance of each cell is 1.5 V and 1 Omega, respectively then find the value of I.

The cell has an emf of 2 V and the internal resistance of this cell is 0.1 Omega , it is connected to resistance of 3.9 Omega , the voltage across the cell will be

In the given circuit the cells have zero internal resistance. The currents (in Amperes) passing through resistance R_(1) and R_(2) respectively, are :