# Solve for the currents through each resistor? And the voltages across each resistor? Including the each power?

Feb 5, 2018

${I}_{1} = {I}_{2} = {I}_{3} = {I}_{4} = 0.714 A$
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
${V}_{4} = 7.14 V A n d {P}_{4} = 5.9296 W$

#### Explanation:

As all the resistors ${R}_{1} , {R}_{2} , {R}_{3} \mathmr{and} {R}_{4}$ 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, ${R}_{e} q = {R}_{1} + {R}_{2} + {R}_{3} + {R}_{4}$

${R}_{e q} = 4 + 6 + 8 + 10 = 28 \Omega$

As total voltage applied across the circuit is $20 V$,
The current $I$ can be calculated as:

$I = \frac{V}{R} = \frac{20}{28} = \frac{5}{7} = 0.7142857 A \approx 71.4 m A$

SO, current through every resistor will be same = $71.4 m A \mathmr{and} 0.714 A$
Now we can calculate the voltage drop and power across each resistor:

${V}_{1} = I \times {R}_{1} = 0.714 \times 4 = 2.856 V$ and

${P}_{1} = {V}_{1}^{2} / {R}_{1} = {2.856}^{2} / 4 = 2.039184 W$

${V}_{2} = I \times {R}_{2} = 0.714 \times 6 = 4.284 V$ and

${P}_{2} = {V}_{2}^{2} / {R}_{2} = {4.284}^{2} / 6 = 3.058776 W$

${V}_{3} = I \times {R}_{3} = 0.714 \times 8 = 5.712 V$ and

${P}_{3} = {V}_{3}^{2} / {R}_{3} = {5.712}^{2} / 8 = 4.078368 W$

${V}_{4} = I \times {R}_{4} = 0.714 \times 10 = 7.14 V$ 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
${V}_{4} = 7.14 V A n d {P}_{4} = 5.9296 W$

Note : $A$ = amperes
$V$= Volts and $W$ = Watts.