Question

find the current drawn from the battery by the network of four resistors shown in fig​

Answer 1

Answer:

we know that two resistors is in parallel

Explanation:

for parallel 1/R=1/R1+1/R2

1/R=1/10+1/101

1/R=2/10

2R=10

R=5ohm

Now one resistor is in series

5+10=15ohm

net resistance is 15ohm

given voltage=3v

ACC. to ohm law

v=IR

3=I(15)

I=3/15

I=0.2ampere

verification=v=IR

v=0.2(15)

v=3.0

hence verified

Answer 2

Answer:

we know that two resister is in parrallel

Explanation:

for parrallel 1/R= 1/R1 + 1/R2

1/R= 1/10 + 1/101

1/R = 2/10

2R=10

R= 5 ohm

now, one resister is in series

5+10=15 ohm

Net resister is 15 ohm

given, voltage = 3v

ACC, to ohm law

v =1R

3= 1(15)

1= 3/15

1= 0.2 ampere

verification= v = 1R

v = 0.2(15)

v=3.0

hence, verified

I hope is it helpful for you