Question

16 - Two resistance of 40 ohm
each are connected parallel and
this arrangement is connected
with series with a resistance R. If
the applied voltage is 200 volt
and current drawn from the
battery is 5 ampere then the
value of Rand potential across R
is respectively in their SI UNIT​

Answer 1

Answer:we know that,

Resistance each 40ohms are connected in parallel,

Now 1/Req=1/R1+1/R2

1/Req=1/40+1/40

1/Req=2/40

1/Req=1/20

Req=20ohms

Now this 20ohms resistance is again connected in series with R ohms resistance..

Req=R1+R2

Req=(20+R)ohms

Voltage=200volt

Current=5amperes..

Now, from ohms law,

V=iR

R=v/I

20+R=200/5

R+20=40

R=20ohms.......

So the value of R is 20ohms

Explanation:

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Answer 2

Answer:

R=39.95Ω

Explanation:

Let the final resistance be R

According to ohm's law,

V=IR

Therefore,

200=5*R

R=200/5

R=40Ω

Now,

R=1/40+1/40+r

40=2/40+r

40=(1+20r)/20

800=1+20r

799=20r

r=39.95Ω

Ans.