Question

In the circuit given below, 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 $1.0\text{V}\text{and}10.0\text{V}.$ 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 the direction of that battery's emf. (Direction of emf is from negative to positive). FInd the value of $R_2.$

Answer 1

From the graph, When $E_2$ is increased.

$I_2$ is increasing.

$\therefore I_2$ belongs to current in cell of emf $E_2$

When $E_2=4\text{V}:I_2=0\text{A}\&I_1=0.1\text{A}$


$\text{Voltage across}{R}_2=4\text{V}$

$\Rightarrow 0.1R_2=4\Rightarrow R_2=40\Omega$

Ans: 40 , 40.0 , 40.00