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

Options

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 `

Answer

C

Explanation

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.`