1. Solution

Given circuit is: a. The total resistance of the circuit (RTOTAL):

Since, R2 and R3 are in parallel, there equivalent resistance Req1 is given by,

\$\$frac{1}{Req1} = frac{1}{R2} + frac{1}{R3} = frac{1}{50} + frac{1}{30}\$\$

SoReq1 = 18.75 Ω

Similarly, resistors R6 and R7 being in series, their equivalent resistance is given by,

Req2 = R6 + R7 = 200 + 40 = 240 Ω

The equivalent circuit is shown below: Now,

Since R1, Req1 and R3 are in series, their equivalent resistance Req3 is obtained as:

Req3 = R1 + Req1 + R3 = 100 + 18.75 + 80 = 198.8 Ω

Also, equivalent resistance of parallel resistance R5 and Req2 is calculated as:

\$\$frac{1}{Req4} = frac{1}{R5} + frac{1}{Req2} = frac{1}{150} + frac{1}{240}\$\$

HenceReq4 = 92. 3 Ω

The equivalent circuit with Req3 and Req4 is shown below: Hence, the total resistance of the circuit (RTOTAL) can be calculated as:

RTOTAL = Req3 + Req4 = 198.8 + 92.3 = 291.1 Ω

So, it can be seen that the series combination of the resistors increases the resistance while the parallel combination decreases the resistance. Simplifying the parallel and series circuit into the equivalent resistance, we can obtain the total resistance of the circuit.

b. Current supplied by source (IS):

From Ohm’s law, V1 = IS×RTOTAL

Hence, the current supplied by source is:

\$\$I_{S} = frac{V1}{R_{text{TOTAL}}} = frac{18}{291.1} = 0.06183 A = 61.83 mA\$\$

So, the current supplied by the source depends on the total resistance of the circuit and for the given voltage, current is inversely proportional to resistance. For the given circuit, the current supplied by the source is 61.83 mA.

c. Current through R5:

Since, R5 and Req2 are in parallel, the current is divided between them. Hence, using the current divider rule, the current through R5 can be obtained as:

\$\$I_{R5} = I_{S} times frac{Req2}{R5 + Req2} = 0.06183 times frac{240}{150 + 240} = 0.038 A = 38 mA\$\$

Hence, current flowing through the resistor R5 is 38 mA. This indicates that, when the resistors are in parallel, the current between then is divided and the resistor with lower resistance takes on more current due to lower obstruction to the current flow.

Now, power dissipated in resistor R5 is given by;

PR5 = IR52 × R5 = 0.0382 × 150 = 0.2166 Watt

This power dissipated depends on the current flowing through the resistor and its resistance and is dissipated in the form of heat and sometimes light.

d. When R2 becomes short circuit: When R2 is short circuited, the current flows through the short circuited branch and hence no current flows through the resistor R3. So, R3 doesn’t contribute to the total resistance of the circuit. The equivalent resistance of the circuit is then given by:

Equivalent resistance, Re = (R1 + R4 + R5 // (R6 + R7)) = (R1 + R4 + Req4)

Re = 100 + 80 + 92.3 = 272.3 Ω

Current flowing in circuit with R2 short circuited:

\$\$I_{text{sc}} = frac{V1}{R_{e}} = frac{18}{272.3} = 0.0661 A = 66.1 mA\$\$

Current through the resistance R5:

\$\$I_{SCR5} = I_{text{SC}} times frac{Req2}{R5 + Req2} = 0.0661 times frac{240}{150 + 240} = 0.04067 A = 40.67 mA\$\$

Now, power dissipated in resistor R5 is given by;

PSCR5 = ISCR52 × R5 = 0.040672 × 150 = 0.248 Watt

Hence, power increases in the resistor R5 when resistor R2 is short circuited. The change in power is:

ΔPR5 = PSCR5 − PR5 = 0.248 − 0.2166 = 0.0314 Watt

Hence, when the resistor R2 is short circuited, the equivalent resistance of the whole circuit decreases due to which more current flows in the circuit and hence, more power generated in each resistance.

2. Solution

a. The current through resistor R2 using mesh analysis:

The given circuit is: Kirchhoff’s Voltage Law (KVL) states that the algebraic total of all voltages in the loop must be zero (Theraja & Theraja, 1999). Hence in loop 1, using KVL,

Vs1 − I1R1 − (I1I2)R2 − Vs2 = 0

15 − I1 × 104 − (I1I2) × 5 × 103 − 16 = 0

15 − I1 × 104 − I1 × 5 × 103 + I2 × 5 × 103 − 16 = 0

15 − 15 × 103 I1 + 5 × 103I2 − 16 = 0

 − 3I1 + I2 = 2 × 10 − 4 Eq. 1

Similarly, in loop 2, using KVL,

Vs2 − (I2I1)R2 − I2R3 − Vs3 = 0

16 − (I2I1) × 5 × 103 − I2 × 8 × 103 − 12 = 0

− 5 × 103 I2 + 5 × 103 I1 − 8 × 103I2 =  − 4

 5I1 − 13I2 =  − 4 × 10 − 3 Eq. 2

Solving Equations 1 and 2 for I1 and I2,

I1= 4.1176 × 10 − 5 A

I2= 3.235 × 10 − 4 A

Hence, current through resistance R2:

IR2 = I2 − I1= 3.235 × 10 − 4 − 4.1176 × 10 − 5 = 2.82 × 10 − 4 A

Hence, the current through resistance R2 is 2.82 × 10 − 4 A flowing in upward direction.

b. Current through resistor R2, using nodal analysis: \$\$I_{1} = frac{Vs1 – V_{1}}{R1} = frac{15 – V_{1}}{10^{4}}\$\$

\$\$I_{2} = frac{V_{1} – Vs2}{R2} = frac{V_{1} – 16}{5 times 10^{3}}\$\$

\$\$I_{3} = frac{V_{1} – Vs3}{R3} = frac{V_{1} – 12}{8 times 10^{3}}\$\$

In the given circuit, there is only one node denoted by “1” with three current paths. Let, V1 be the voltage at the node 1, I1, I2 and I3 be the current flowing through the resistor R1, R2 and R3 respectively.

Kirchhoff’s Current Law (KCL) states that, all currents entering and leaving a node must add up to zero algebraically (Theraja & Theraja, 1999). So, using KCL at node 1,

I1 = I2 + I3

\$\$frac{15 – V_{1}}{10^{4}} = frac{V_{1} – 16}{5 times 10^{3}} + frac{V_{1} – 12}{8 times 10^{3}}\$\$

On solving for V1,

V1 = 14.588 V

Hence,

\$\$I_{2} = frac{V_{1} – Vs2}{R2} = frac{14.588 – 16}{5 times 10^{3}} = – 2.82 times 10^{- 4}text{ A}\$\$

Hence, the current in resistor R2 is 2.82 × 10 − 4 A in upward direction which is same as obtained from the mesh analysis.

This shows that any of the two methods i.e. mesh or nodal analysis can be used to solve the given circuit.

3. Solution

The given circuit is: a. Total resistive component (RTOTAL):

Since, R2, R3 and R4 are in parallel their equivalent resistance Req is obtained as:

\$\$frac{1}{text{Req}} = frac{1}{R2} + frac{1}{R3} + frac{1}{R4} = frac{1}{400} + frac{1}{700} + frac{1}{580}\$\$

HenceReq = 176.9 Ω

The equivalent circuit is shown below: Further, R1 and Req is in series and hence the total resistive component of the circuit is given by,

RTOTAL = R1 + Req = 250 + 176.9 = 426.9 Ω

b. Total capacitive component (CTOTAL): Capacitor C2, C3 and C4 are in series connection with their equivalent capacitance (Ceq) as:

\$\$frac{1}{text{Ceq}} = frac{1}{C2} + frac{1}{C3} + frac{1}{C4} = frac{1}{50} + frac{1}{150} + frac{1}{320}\$\$

HenceCeq = 33.56 μF

Further, Ceq and C1 are in parallel. So, the total capacitive component of the circuit is given by;

CTOTAL = C1 + Ceq = 10 + 33.56 = 43.56 μF

c. Total inductive component (LTOTAL): Equivalent inductance of L2 and L3 in parallel connection is:

\$\$frac{1}{text{Leq}} = frac{1}{L2} + frac{1}{L3} = frac{1}{80} + frac{1}{160}\$\$

HenceLeq = 53.33 mH

Leq and L1 are now in series connection. Hence, the total inductive component of the circuit is given by:

LTOTAL = L1 + Leq = 10 + 53.33 = 63.33 mH

d. The total impedance in complex form (ZTOTAL):

The equivalent circuit can be represented as: Inductive reactance:

XL = 2πfLTOTAL = 2π × 50 × 63.33 × 10 − 3 = 19.89 Ω

Capacitive reactance:

\$\$X_{C} = frac{1}{2pi fC_{text{TOTAL}}} = frac{1}{2pi times 50 times 43.56 times 10^{- 6}} = 73.073 Omega\$\$

Total impedance:

ZTOTAL = RTOTAL + j(XLXC) = 426.9 + j(19.89−73.073) = 426.9 − j53.183

e. Total impedance in polar form (ZTOTAL):

ZTOTAL = 426.9 − j53.183 = x + jy

In polar form,

ZTOTAL = |ZTOTAL| < θ

\$\$left| Z_{text{TOTAL}} right| = sqrt{x^{2} + y^{2}} = sqrt{{426.9}^{2} + {( – 53.183)}^{2}} = 430.2 Omega\$\$

\$\$theta = tan^{- 1}left( frac{y}{x} right) = tan^{- 1}left( frac{- 53.183}{426.9} right) = – 7.101{^circ}\$\$

Hence, total impedance in polar form:

ZTOTAL = 430.2 <  − 7.101 Ω

The negative phase angle is due to higher value of capacitive reactance as compared to the inductive reactance.

f. Current supplied in polar form (IS):

Given peak voltage (Vpk) = 230 V

Hence, RMS voltage (V) = \$frac{V_{text{pk}}}{sqrt{2}} = frac{230}{sqrt{2}}\$ = 162.63 V

Then, current supplied to the circuit:

\$\$I_{S} = frac{V}{Z_{text{TOTAL}}} = frac{162.63 < 0{^circ}}{430.2 < – 7.101{^circ}} = 0.378 < 7.101{^circ} A\$\$

Hence, for the given circuit, the current supplied by the given source in polar form is 0.378 < 7.101 A. The positive phase angle of the current IS indicates that the current supplied in the circuit leads the supplied voltage by an angle of 7.101. This is due to higher capacitive resistance in the circuit as in the capacitor, the current leads the voltage.

Theraja, B. L., & Theraja, A. K. (1999). In A Textbook of Electrical Technology (Vol. I). S Chand & Co Ltd.

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