Question

1000 11. Determine the equivalent resistance of the circuit. Then determine the current in each path, and the voltage across each resistor. Then determine the power dissipated by each resistor and the total power dissipated by the circuit. Assume the voltage of the source is 110V. 120.02 W R, R. 60.22 w R R, 60312 R, 602 4012 w R, R, 8012 두 R. w 1001 12. Determine the cach of the currents in each path, and the voltages across each resistor. E, 18 V b 0.52 R 2.50 6.022 R, $1.52 0.52 E = 45V

Answer 1

11.

Resistances in series : The equivalent resistance of the circuit is found by simply adding up the resistance values of the individual resistors. Current flowing through each resistance is same.

Resistances in parallel : The equivalent resistance is given by

$\frac{1}{R_{eq}}= \frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+........+\frac{1}{R_n}$. Voltage across each resistor is same.

We will also be using ohm's Law : $V=I\times R$

R₂ and R₃ are in parallel :

$R_{eq1} = \frac{R_2R_3}{R_2+R_3} = \frac{120\times 40}{120\Omega + 40\Omega} =30 \Omega$

Also, R₄, R₅ and R₆ are in parallel

$R_{eq2} = \frac{R_4R_5R_6}{R_4R_5+R_5R_6+R_6R_4} = \frac{60\times 60\times 60}{(60\times 60)+(60\times 60)+(60\times 60)} \Omega$

$\Rightarrow R_{eq\ 2} = \frac{60}{3} \Omega= 20\Omega$

Now, R₁ R_eq1, R_eq2, R₇ and R₈ are in series.

$\Rightarrow R_{eq}=R_1+ R_{eq1}+ R_{eq2}+R_7 + R_8=330\Omega$

Let Power dissipated, Current and Voltage across each resistor be denoted by the same subscript to avoid confusion eg. current through R_eq1 is I_eq1 and voltage across it is V_eq1.

Now, we can imagine as our circuit consists of one resistor R_eq and a voltage source of 110V. Therefore,

$V_{eq} = 110V, \ I_{eq} = \frac{V_{eq}}{R_{eq}}=\frac{110}{330}=0.33A$

Since, current flowing through resistances in series is the same.

Hence

Voltage across each resistor :

$V_{1} = I_1R_1 = 0.33A\times 100\Omega=33V$

$V_{eq1} = I_{eq1}R_{eq1} = 0.33A\times 30\Omega=9.9V$

$V_{eq2} = I_{eq2}R_{eq2} = 0.33A\times 20\Omega=6.6V$

$V_{7} = I_7R_7 = 0.33A\times 80\Omega=26.4V$

$V_{8} = I_8R_8 = 0.33A\times 100\Omega=33V$

Since, voltage across resistances in parallel is same.

Hence,

Veqi = V2 = V3 = 9.9V, and also Veq2 = V4 = V5 = V6 = 6.6V

Current flowing through each resistor is given by,

$I_2 = \frac{V_2}{R_2} = \frac{9.9V}{120\Omega}= 0.0825A$

$I_3 = \frac{V_3}{R_3} = \frac{9.9V}{40\Omega} =0.2475\ A$

$I_4 = \frac{V_4}{R_4} = \frac{6.6V}{60\Omega}=0.11A$

Since, resistances R₄, R₅ and R₆ are equal and voltage across them is also equal.

Hence, Current flowing through them will also be equal i.e. I₄ = I₅ = I₆ = 0.11A

$I_7 = \frac{V_7}{R_7} = \frac{26.4V}{80\Omega}=0.33A$

$I_8 = \frac{V_8}{R_8} = \frac{33V}{100\Omega}=0.33A$

Power dissipated by any resistor can be found using formula :

Power dissipated by resistor = current flowing through it $\times$ Voltage across it

$P_1 = I_1V_1 = 33V\times 0.33A = 10.89Watt$

$P_2 = I_2V_2 = 9.9V\times 0.082A = 0.81675Watt$

$P_3 = I_3V_3 = 9.9V\times 0.2475A = 2.45Watt$

$P_4 = I_4V_4 = 6.6V\times 0.11A = 0.73Watt$

Since, current and voltage across R₄, R₅ and R₆ are equal hence, power dissipated is also equal i.e. 0.73 W.

$P_7 = I_7V_7 = 26.4V\times 0.33A = 8.712Watt$

$P_8 = I_8V_8 = 33V\times 0.33A = 10.89Watt$

Total power dissipated by the circuit is the sum of all the dissipated power

$P = 10.89W+0.81675W+2.47W+0.73\times 3W+8.712W+10.89W$

$P \approx 35.95W$

12.  

We will be using,

Kirchhoff's Current Law(KCL): Algebraic sum of all the currents at a junction is zero.

Sign Convention : Take currents coming into the junction as positive and currents leaving the junction as negative.

Kirchhoff's Voltage Law (KVL):  The algebraic sum of all voltages within the loop must be equal to zero.

Sign Convention : If potential increases during traversing any element, put a positive sign in the voltage drop. If potential decreases during traversing any element, put a negative sign in the voltage drop.

Always remember that potential decreases in the direction of flow of current.

Applying KVL in loop abcdea, (in clockwise direction)

$\Rightarrow -I_2(2.5\Omega)+18V-I_2(0.5\Omega)-I_1(6\Omega) = 0$

$\Rightarrow 3I_2+6I_1 = 18 .....................(1)$

Applying KVL in loop aefgha,  (in clockwise direction)

I1R1 + 13R3 + 1373 E2 = 0

$\Rightarrow I_1(6\Omega)+I_3(1.5\Omega)+I_3(0.5\Omega)-45 = 0$

$\Rightarrow 6I_1+2I_3=45............(2)$

Also, applying KCL at junction a,

I ₁ = I ₂ + I ₃ .................(3)

We have 3 equations and 3 variables, we can easily solve them using elimination method.

Substitute Il from (3) into (1) and (2) 312 + 6( 12 + I3) = 18 9 Iz + 6 I3 =18 3 I2 + 2 I3 = 6 - (4) 6 (I2 + I3) + 2 I3 -45 6 I2 + 8 I3 = 45 2 4 13 13 multiplying equation (4) by 2 subt & then Subtracting from (5) 6 Iz + 8 I3 - (3 I2 4 2 I3 ) x 2 = 45 - 6x2 652 +8 13 - 642 - 4 13 = 45-12 33 = 8.25 A Putting Is in equation (u), 3 12 + 2 ( 8.25) = 6 3 I2 = - 10.5 Iz = - 3.5A. negative sign represents that direction of flow of current Iz is opposite to what have assumed. ube

I₂ flows from b to a not a to b.

Putting I 2 and I 3 in equation (3)

I ₁ = -3.5A + 8.25A = 4.75A

Voltage across each resistor can be found using ohm's law:

$V_{R2} = I_2 R_2 = 3.5A \times 2.5 \Omega = 8.75V$

$V_{r1} = I_2 r_1 = 3.5A \times 0.5 \Omega = 1.75V$

$V_{R1} = I_1 R_1 = 4.75A \times 6.0 \Omega = 28.5V$

$V_{R3} = I_3 R_3 = 8.25A \times 1.5 \Omega = 12.37V$

$V_{r2} = I_3 r_2 = 8.25A \times 0.5 \Omega = 4.13V$