Two identical cells, each of emf E, having negligible internal resistance, are connected in parallel with each other across an external resistance R. What is the current through this resistance?

D.

0.04 A

The circuit is

As the lower diode attached to 100 Ω resistance is in reversed biased so, it is non- conducting.

Now, the circuit can be redrawn as,

∴ Current through the circuit, $\mathrm{I} = \frac{\mathrm{V}}{\mathrm{R}}$  

             $= \frac{10}{250} = 0.04 \mathrm{A}$ 

C.

4 : 1

As we know that, ${\mathrm{R}}_1 = \mathrm{\rho }\frac{\mathrm{l}}{\mathrm{a}} = \mathrm{\rho } \frac{{\mathrm{l}}^2}{\mathrm{V}}$

where, l = length of wire

          a = area of cross-section of the wire

and    V = volume of the wire R₁ ∝ l²

    $$\begin{gathered} \Rightarrow \frac{{\mathrm{R}}_1}{{\mathrm{R}}_2} = {\left(\frac{{\mathrm{l}}_1}{{\mathrm{l}}_2}\right)}^2 = {\left(\frac{1}{2}\right)}^2 \\ \Rightarrow {\mathrm{R}}_2 : {\mathrm{R}}_1 = 4 : 1 \end{gathered}$$

C.

E_A = 1.875 V, E_B = 1.625 V

The figure can be redrawn as

The current through the circuit

       $$\begin{gathered} \mathrm{i} = \frac{\mathrm{net} \mathrm{emf}}{\mathrm{effective} \mathrm{resistance}} \\ = \frac{2 - 1.5}{5 + 5 + 10} = \frac{0.5}{20} = \frac{1}{40} \\ = 0.025 \mathrm{A} \end{gathered}$$

The terminal potential difference of the batteries

         V_A = ε_A − ir_A = 2 − 0.025 × 5

              = 2 − 0.0125 = 1.875 V

and    V_B = ε_B + ir_B = 1.5 + 0.025 × 5

              = 1.5 + 0.0125

              = 1.625 V

A.

1.33 A

The circuit diagram can be redrawn as

For the loop ABCDA,

   +2 − 2I' + 2 − 2I = 0           .......(i)

For the loop ABFEA,

   2 − 2I + 2 − 2(I − I') = 0

          4 − 2I − 2I + 2I' = 0

                 4 − 4I + 2I' = 0

                                 2 = 2I − I'       ...... (ii)

From Eqs. (i) and (ii), we get

         2 = 2I − I' ,  2 = I + I'

         4 = 3I ,         I = 4/3 = 1.33 A