In the circuit given in the figure, both batteries are ideal . Emf `E_1` of battery 1 has a fixed value, but emf `E_2` of battery 2 can be varied between `1.0V and 10.0 V` . The graph gives the currents through the two batteries as a function of `E_2` but are not marked as which plot corresponds to which battery. But for both plots, current is assumed to be negative when the direction of the current through the battery is opposite to the direction of that battery's emf (direction of emf is from negative to positive.)

The value of emf `E_1` is

Options

A. 8 V
B. 6 V
C. 4 V
D. 2 V

Answer

B

Explanation

b. `i_0 = 0.1 A, E_2 = 4 V, i_2 = 0`
As `0.1 R_1 + 0.1R_2 - E_1 =0`
`0.1 R_2 - 4V = 0`
`R_2 = 40 `Omega

(ii) Now `i_2 = 0.3A, i_1 = 0.1 A, E_2 = 8A`.
Now `0.1R_1 + E_1 - 8 =0 `
`0.1 + 4 -E_1 = 0`
`0.2 R_1 - 4 = 0`
or `R_1 = 4/0.2 = 20` Omega
`0.2 E_1 = 2+4 = 6V`.