Home
Class 12
PHYSICS
Two cells A and B of e.m.f.2 V and 1.5 V...

Two cells A and B of e.m.f.2 V and 1.5 V respectively, are connected as shown in figure through an external resistance 10 `Omega` The internal resistance of each cell is 5 `Omega`. The potential difference `E_A and E_B` across the terminals of the cells A and B respectively are

A

`E_A = 2.0 v,E_B = 1.5 V`

B

`E_A = 2.125 V,E_B = 1.375 V`

C

`E_A =1.875 V,E_B=1.625 V`

D

`E_A = 1.875 V,E_B =1.375 V `

Text Solution

Verified by Experts

The correct Answer is:
C
Suppose current I is flowing through the circuit using Kirchhoff.s voltage law in the circuit
`2-1.5 -5I-10 I-5I =0`
`implies 0.5 = 20 I implies I=(0.5 )/(20) =(1)/(40 )A `
terminal potential differnce across the cell A
` E_A = 2 -Ir =2 -(1)/(40 ) xx 5 =1.875 V.`
terminal potential difference across the cell B
` E_B = 1.5 + Ir = 1.5 +(1)/(40)xx 5 = 1.875 V.`
terminal potential differnce across the cell B.
` E_B = 1.5 +Ir = 1.5 +(1)/(40) xx 5 =1.625 V.`
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • BULK PROPERTIES OF MATTER

    MTG-WBJEE|Exercise WB JEE PREVIOUS YEARS QUESTIONS (ONE OR MORE THAN ONE OPTION CORRECT TYPE)|1 Videos

  • ELECTROMAGNETIC INDUCTION AND ALTERNATING CURRENT

    MTG-WBJEE|Exercise WB JEE Previous Years Questions |14 Videos

Similar Questions

Explore conceptually related problems

Seven identical cells each of e.m.f. E and internal resistance r are connected as shown . The potential difference between A and B is

Two cells having emf's of 10 V and 8 V, respectively, are connected in series with a resistance of 24Omega In the external circuit. Their internal resistances are 2Omega// V (proportional to their emf's). Find the current in the circuit

Knowledge Check

  • A cell of e.m. E and internal resistance r is connected across a resistancer. The potential difference between the terminals of the cell must be

    A
    E
    B
    `E//2`
    C
    `E//4`
    D
    `3//2`

  • A cell of e.m.f. E is connected with an external resistance R , then p.d. across cell is V . The internal resistance of cell will be

    A
    `((E-V)R)/(E)`
    B
    `((E-V)R)/(V)`
    C
    `((V-E)R)/(V)`
    D
    `((V-E)R)/(E)`

  • A cell of emf 1.5 V is connected with an external resistance 20 Omega . The potential difference falls to 1.0 V. The internal resistance of cell is

    A
    `1.5 Omega`
    B
    `1.0 Omega`
    C
    `10.0Omega`
    D
    `2.0Omega`

    • Similar Questions

    Explore conceptually related problems

    Two cells e_(1) and e_(2) connected in opposition to each other as shown in figure. The cell of emf 9 V and internal resistance 3Omega the cell is of emf 7V and internal resistance 7Omega . The potential difference between the points A and B is

    A cell of emf 1.5 Vis connected with an external resistance 2Omega . The potential difference falls to 1.0 v. The internal resistance of cell is

    A cell of emf 1.5 Vis connected with an external resistance 2Omega . The potential difference falls to 1.0 v. The internal resistance of cell is

    A cell has an emf 1.5V. When connected across an external resistance of 2Omega , the terminal potential difference falls to 1.0V. The internal resistance of the cell is:

    A cell E_(1) of emf6 Vand internal resistance 2 Omega is connected in opposition, as shown in figure, with a cell E_(2) of emf 4 V and internal resistance 8Omega . The potential difference across points A and B is

    Home
    Profile