Question

a battery of 9 V is connected in series with the resistance of 0.2 ohm, 0,.3 ohm ,0.4 ohm 0.5 and 12 ohm respectively. how much current would flow through that 12 ohm resistor? ​

Answer 1

Answer:

Explanation:

v=ir

.2+.3+.4+.5+12=13.4

v=9v

i=v/r

i=9/13.4

i=.671

in series connection current flowing is same

i.e .671

Answer 2

Explanation:

R = 0.2 + 0.3 + 0.4 + 0.5 + 12

R = 12.14 ohm

Now, V = IR

9 = I x 12.14

9 / 12.14 = I

or, I = 0.74 ampere.