The internal resistance of cell is 0.1Ω and its emf is 2V. When a current of 2A is being drawn from it, the potential difference across its terminals will be :

Question

The internal resistance of cell is 0.1Ω and its emf is 2V. When a current of 2A is being drawn from it, the potential difference across its terminals will be : (A) more than 2V (B) 2V (C) 1.8 V (D) none of the above.

Tardigrade Admin · Accepted Answer

V = Ir = xi . Here xi = 2V , r = 0.1 Ω and I = 2A. Hence V = ( 2V - 2 × 0.1) V = 1.8 V

Q. The internal resistance of cell is $0.1Ω$ and its emf is 2V. When a current of 2A is being drawn from it, the potential difference across its terminals will be :

more than 2V

2V

1.8 V

none of the above.

$V = I r = ξ$ . Here $ξ = 2 V, r$ = 0.1 $Ω$ and $I$ = 2A.
Hence V = ( 2V - 2 $\times$ 0.1) V = 1.8 V

Q. The internal resistance of cell is 0.1Ω and its emf is 2V. When a current of 2A is being drawn from it, the potential difference across its terminals will be :

more than 2V

2V

1.8 V

none of the above.

V=Ir=ξ . Here ξ=2V,r = 0.1 Ω and I = 2A. Hence V = ( 2V - 2 × 0.1) V = 1.8 V

Q. The internal resistance of cell is $0.1Ω$ and its emf is 2V. When a current of 2A is being drawn from it, the potential difference across its terminals will be :

$V = I r = ξ$ . Here $ξ = 2 V, r$ = 0.1 $Ω$ and $I$ = 2A.
Hence V = ( 2V - 2 $\times$ 0.1) V = 1.8 V