Three identical cells each of emf and internal resistance of are connected in series to form a…

Resistors $R_4 = 30\,\Omega$, $R_5 = 40\,\Omega$, and $R_2 = 120\,\Omega$ are in parallel: $\frac{1}{R_6} = \frac{1}{30} + \frac{1}{40} + \frac{1}{120} = \frac{4+3+1}{120} = \frac{8}{120} = \frac{1}{15} \implies R_6 = 15\,\Omega$. Now $R_1 = 10\,\Omega$, $R_6 = 15\,\Omega$, and $R_3 = 14\,\Omega$ are in series: $R_{\text{ext}} = 10 + 15 + 14 = 39\,\Omega$. Total emf of three cells in series $= 3 \times 1\cdot5 = 4\cdot5\text{ V}$, total internal resistance $r = 3 \times 2 = 6\,\Omega$. Total resistance $R_{\text{total}} = 39 + 6 = 45\,\Omega$. Current delivered $I = \frac{E_{\text{total}}}{R + r} = \frac{4\cdot5}{45} = 0\cdot1\text{ A}$. Reading of ammeter $A = 0\cdot1\text{ A}$.