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1000 11. Determine the equivalent resistance of the circuit. Then determine the current in each path, and the voltage across
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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 Req1, Req2, 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 Req1 is Ieq1 and voltage across it is Veq1.

Now, we can imagine as our circuit consists of one resistor Req 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 I_{eq}=I_1 =I_{eq1}= I_{eq2} =I_7= I_8 = 0.33A

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,

V_{eq1}=V_2=V_3=9.9V, \ and \ also \ V_{eq2}=V_4=V_5=V_6=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_4 =P_5=P_6 = 0.73Watt

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)

-I_2(R_2)+E_1-I_2r_1-I_1R_1 = 0

\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)

I_1R_1+I_3R_3+I_3r_3-E_2 = 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 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

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