Question

Three resistors R1 = 5 Ω, R2 = 8 Ω and R3 = 12 Ω are connected in series with a

battery of 4 V. Find the value of current in the circuit and the voltage drop

across each resistor. (V = 4Volts)​

Answer 1

total resistance= r1+r2+r3

=5+8+12=25 Ohms

voltage= 4 V

V=IR

Current= V/R

= 4/25

=0.16 Amperes

I am sorry but I don't know how to calculate voltage drop.

Current is calculated :))

Answer 2

R1=5

R2=8

R3=12

V=4V

In series

Rs=R1+R2+R3

Rs=5+8+12

Rs=25ohm

using ohms formula

V=IR

I=4/25

I=0.16A

In series combination Current(I) is same

and Potential difference is different

V1=IR1

V1=0.16×5=0.8A

V2=IR2

V2=0.16×8=1.28A

V3=IR3

V3=0.16×12=1.92A