# 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

As the image is not clear I'm taking resistance as R

_{1}= 1Ω parallel resistances as R

_{2}= 6Ω R

_{3}=6Ω and last resitance R

_{4}= 4Ω

Now

$\frac{1}{{R}_{parallel}}=\frac{1}{6}+\frac{1}{6}\phantom{\rule{0ex}{0ex}}{R}_{paralle}=3\Omega \phantom{\rule{0ex}{0ex}}{R}_{eq}=3+4+1=8\Omega $

V= IR

So I = V/R

_{eq }

I = 24/8 = 3A

so Ammeter reading = 3A

now voltage across

${V}_{R1}=I{R}_{1}=3\times 1=3V\phantom{\rule{0ex}{0ex}}{V}_{R4}=I{R}_{4}=3\times 4=12V\phantom{\rule{0ex}{0ex}}{V}_{R2}={V}_{R3}={V}_{{R}_{paralle}}=I{R}_{parallel}=3\times 3=9V$

Regards

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