The figure alongside shows a circuit diagram having a battery of 24 V and negligible resistance. Calculate 1.reading of ammeter 2) p.d across each resistor

Dear Student 

As the image is not clear I'm taking resistance as R₁= 1Ω parallel resistances as R₂ = 6​Ω R₃=6​Ω  and last resitance R₄ = 4​Ω 

Now
 $$\begin{gathered} \frac{1}{R_{parallel}}= \frac{1}{6}+\frac{1}{6} \\ {R}_{paralle }= 3\Omega \\ {R}_{eq}= 3 + 4 +1 = 8\Omega \end{gathered}$$

V= IR
So I = V/R_eq 
I = 24/8 = 3A

so Ammeter reading = 3A 
now voltage across 
$$\begin{gathered} V_{R1}= I{R}_1 = 3\times 1 = 3V \\ {V}_{R4}=I{R}_4 = 3\times 4 = 12 V \\ {V}_{R2} =V_{R3} = V_{R_{paralle}}= I{R}_{parallel} = 3\times 3 = 9 V \end{gathered}$$

Regards