As all the resistors #R_1, R_2, R_3 and R_4# are in series, the same amount of current will flow through all. Let this current be #I#
# therefore I = I_1=I_2=I_3=I_4#
To calculate #I#, we must first calculate the equivalent resistance between the two terminals.
For a series circuit, #R_eq = R_1+R_2+ R_3 + R_4#
#R_(eq) = 4+6+ 8+ 10= 28 Omega#
As total voltage applied across the circuit is #20V#,
The current #I# can be calculated as:
#I = V/R = 20/28 = 5/7 = 0.7142857 A approx 71.4 mA #
SO, current through every resistor will be same = #71.4 mA or 0.714 A#
Now we can calculate the voltage drop and power across each resistor:
#V_1 = I xx R_1 = 0.714 xx 4= 2.856 V# and
#P_1 = V_1^2 / R_1 = 2.856^2 / 4 =2.039184 W #
#V_2 = I xx R_2 = 0.714 xx 6= 4.284 V# and
#P_2 = V_2^2 / R_2 = 4.284^2 / 6 =3.058776 W #
#V_3 = I xx R_3 = 0.714 xx 8= 5.712 V# and
#P_3= V_3^2 / R_3 = 5.712^2 / 8 = 4.078368 W #
#V_4 = I xx R_4 = 0.714 xx 10= 7.14 V # and
#P_4= V_4^2 / R = 7.14^2 / 10 =5.09796 W #
# I_1=I_2=I_3=I_4 = 0.714 A#
#V_1 = 2.856 V; P_1 = 2.039184 W#
#V_2 = 4.284 V ; P_2 = 3.058776 W#
#V_3 = 5.712 V; P_3= 4.078368 W #
#V_4 = 7.14 V And P_4 = 5.9296 W #
Note : #A# = amperes
#V#= Volts and #W# = Watts.
