Solved problems on Thevenin’s Theorem

1) Find current flowing through 1 $\Omega$ resistor.

Calculate $V_{th}$

$130-0.2I-0.2I-110 = 0$

$-0.4I = -20$

$I = 50 A$

$-V_{th} + 110 +0.2I = 0$

$V_{th} = 120 V$

$Calculate R_{th}$

$0.2 || 0.2 = 0.1 \Omega$

0.1 + 0.2 = 0.3 $\Omega$

$R_{th} = 0.3 \Omega$

Thevenin’s equivalent circuit.

$I_{1\Omega} = \dfrac{V_{th}}{R_{th} + R_L} = \dfrac{120}{0.3+1} = 92.3076 A$

Solved Problems on Superposition Theorem

1) Find Current flowing through $1 \Omega$ resistor by using superposition theorem.

Consider 130 V Voltage source only.

$-0.2I_1-0.2(I_1-I_2)+130 = 0$

$-0.4I_1 +0.2I_2 = -130 \ldots\ldots (1)$

$-I_1-0.2(I_2-I_1)-0.2I_2=0$

$0.2I_1-1.4I_2=0 \ldots\ldots (2)$

$I_1 = 350 A$

$I_2 = 50 A$

$I’_{1\Omega} = 50 A (\downarrow)$

Consider 110 V only.

$-0.2I_1 -0.2(I_1-I_2)-110 = 0$

$-0.4I_1+0.2I_2 = 110 \ldots\ldots(3)$

$-I_2+110-0.2(I_2-I_1)-0.2I_2=0$

$0.2I_1 – 1.4I_2 = -110 \ldots\ldots (4)$

$I_1 = -253.8467 A$

$I_2 = 42.3076 A$

$I’’_{1\Omega} = 42.3076 A (\downarrow)$

$\therefore I_{1\Omega} = I’_{1\Omega} + I’’_{1\Omega} = 50 + 42.3076 = 92.3076 A (\downarrow)$

2) Find current flowing through 5 $\Omega$.

$2-0.1I_1-10I_1-5(I_1-I_2)=0$

$15.1I_1 + 5I_2=-2 \ldots\ldots (1)$

$-20I_1-0.2I_2-5(I_2-I_1)=0$

$5I_1-25.2I_2=0 \ldots\ldots (2)$

$I_1 = 0.1417 A$

$I_2 = 0.02812 A$

$I’_{5\Omega} = I_1-I_2 = 0.1158 A (\downarrow)$

Consider 4 V only.

$-0.1I_1-10I_1-5(I_1-I_2) = 0$

$-15.1I_1+5I_2 = 0 \ldots\ldots (3)$

$20I_2-0.2I_2-4-5(I_2-I_1)=0$

$5I_1-25.2I_2 = 4 \ldots\ldots (4)$

$I_1 = -0.05685 A$
$I_2 = -0.1698 A$

$I’’_{5 \Omega} = I_2-I_1 = 0.1130 A (\downarrow)$

$I_{5 \Omega} = I’_{5\Omega} + I’’_{5\Omega} = 0.2288 A (\downarrow)$

3) Determine current through 1 $\Omega$ resistor.

Consider a 10 V voltage source only.

$-2(I_1-I_2)-2I_1 = 0$

$-4I_1+2I_2=0 \ldots\ldots (1)$

$-10 -2(I_2-I_1)-I_2=0$

$2I_1-3I_2=10 \ldots\ldots (2)$

$I_1 = -2.5 A$

$I_2 =-5 A$

$I’_{1\Omega} = -5 A (\rightarrow)$

Consider a 6 V voltage source only.

$-2I_1-2(I_1-I_2)-6 = 0$

$-4I_1+2I_2=6 \ldots\ldots (3)$

$-I_2+6-2(I_2-I_1)=0$

$2I_1 – 3I_2 =-6 \ldots\ldots (4)$

$I_1 = -0.75 A$

$I_2 =1.5 A$

$I’’_{1\Omega} = 1.5 A (\rightarrow)$

Consider 8 V only.

$8-2I_1-2(I_1-I_2)=0$

$-4I_1+2I_2=-8 \ldots\ldots (5)$

$-I_2-2(I_2-I_1)=0$

$2I_1-3I_2=0 \ldots\ldots (6)$

$I_1 = 3 A$

$I_2 = 2 A$

$I’”_{1\Omega} = 2 A (\rightarrow)$

$Total \; \; current \; \; I_{1 \Omega}=I’_{1 \Omega}+I’’_{1 \Omega}+I’’’_{1 \Omega}$

$I_{1 \Omega}= 2+1.5+(-5) = -1.5 A (\rightarrow)$

$I_{1 \Omega} = 1.5 A (\leftarrow)$

4) Find current through 20 $\Omega$ resistor.

For Mesh 1:

$10-10I_1+10I_2=0$

$-10I_1+10I_2=-10 \ldots \ldots(1)$

For Mesh 2:

$-10I_2+10I_1-I_2-20I_2=0$

$10I_1-31I_2=0 \ldots\ldots(2)$

$I_1 = 1.4761 A$

$I_2 = 0.4761 A$

$I’_{20\Omega} = 0.4761 A (\downarrow)$

Consider 8 V voltage source only.

$-I-8-20I=0$

$-21I = 8$

$I’’_{20\Omega} = -0.3809 A (\downarrow)$

Consider a 12 V voltage source only.

$-12-8I_1+8I_2=0$

$-8I_1+8I_2=12 \ldots\ldots(3)$

$-I_2-8I_2+8I_1-20I_2=0$

$8I_1-29I_2=0 \ldots\ldots(4)$

$I_1=-2.0714 A$

$I_2=-0.5714 A$

$I’’’_{20\Omega} = -0.5714 (\downarrow)$

$I_{20\Omega} = I’_{20\Omega}+I’’_{20\Omega}+I’’’_{20\Omega}$

$I_{20\Omega} = 0.4761+(-0.3809)+(-0.5714)$

$I_{20\Omega} = -0.4762 A (\downarrow)$

$I_{20\Omega} = 0.4762 A (\uparrow)$

Single Phase Transformer Solved Problems

1)A single phase 90 kVA, 3.2 kV/220 V, 50Hz transformer has 89$\%$ efficiency at unity power factor both at full load and half load. Determine the efficiency at 70$\%$ of full load and 0.8 power factor.

$\eta_{100\%}=\eta_{50\%}$

$\dfrac{VIcos\phi}{VIcos\phi+P_i+P_{cu}}=\dfrac{xVIcos\phi}{xVIcos\phi+P_i+x^2P_{cu}}$

$\dfrac{VIcos\phi}{VIcos\phi+P_i+P_{cu}}=\dfrac{0.5VIcos\phi}{0.5VIcos\phi+P_i+0.5^2P_{cu}}$

$P_i+(0.5)^2P_{cu}=0.5P_i+0.5P_{cu}$

$P_i-0.5P_i=0.5P_{cu}-(0.5)^2P_{cu}$

$P_{cu}=2P_i$

Efficiency at $100\%$

$\eta_{100\%}= \dfrac{VIcos\phi}{VIcos\phi+P_i+P_{cu}}$

$\eta_{100\%}= \dfrac{VIcos\phi}{VIcos\phi+3P_i}$

$\eta_{100\%}= \dfrac{90\times 10^3}{90\times 10^3+3P_i}$

$P_i = 3707.8651 W$

$P_{cu} =2 P_i = 7415.7302 W$

Now efficiency at $70\%$ of full load

$\eta_{70\%} = \dfrac{xVIcos\phi}{xVIcos\phi+P_i+x^2P_{cu}}$

$\eta_{70\%} = \dfrac{0.7\times 90\times 10^3 \times 0.8}{0.7\times 90\times 10^3 \times 0.8+3707.8651+0.7^2 \times 7415.7302}$

$\eta_{70\%}$= 87.28$\%$

SOLVED PROBLEMS ON CAPACITOR

1) The plate area of parallel plate capacitor is 0.01 $m^2$. The distance between the plate is 2.5 cm. The insulating medium is air. Find capacitance, if the space between the plates is filled with an insulating material of relative permittivity 5?

$C = \dfrac{\epsilon_0 \epsilon_r A }{d}$

$C = \dfrac{8.854 \times 10^{-12} \times 1 \times 0.01}{2.5 \times 10^{-1}}$

C = $3.5416 \times 10^{-12}$ F

for $\epsilon_r = 5$
$C = \dfrac{\epsilon_0 \epsilon_r A }{d}$

$C = \dfrac{8.854 \times 10^{-12} \times 5 \times 0.01}{2.5 \times 10^{-1}}$

$C = 1.7708 \times 10^{-11}$

2) A parallel plate capacitor has a plate area of 500 $cm^2$, a plate separation of 0.15 mm and air as a dielectric. Find the capacitance. Also find the voltage between the plates the electric flux density and electric field strength, if the capacitor has a charge of 2 $\mu$ C.

$C = \dfrac{\epsilon_0 \epsilon_r A }{d}$

$C = \dfrac{8.854 \times 10^{-12} \times 1 \times 500 \times 10^{-4}}{0.15 \times 10^{-3}}$

$C = 2.9513 \times 10^{-9} F$

$Q = CV$

$V = \dfrac{Q}{C}$

$V = \dfrac{2 \times 10^{-6}}{2.9513 \times 10^{-9} }$

$V = 677.6674$ V

$D = \dfrac{2 \times 10^{-6}}{500 \times 10^{-4}}$

$D = 4 \times 10^{-5} \; C/m^2$

$E = \dfrac{V}{d}$

$E = \dfrac{677.6674}{0.15 \times 10^{-3}}$

$E = 4517782.667 V/m$

3) Two parallel metal discs, each 100 mm in diameter, are spaced 1 mm apart, the dielectric being air. Calculate the electric field intensity and capacitance.

$A = \dfrac{\pi d^2}{4}$

A = 0.007854 $m^2$

$E = \dfrac{V}{d} = \dfrac{10^3}{10^{-3}}$

$E = 10^6$

$C = \dfrac{\epsilon_0 \epsilon_r A }{d}$

$C = \dfrac{8.854 \times 10^{-12} \times 1 \times 0.007854}{10^{-3}}$

$C = 6.9539 \times 10^{-11} F$

4) The plate area of a parallel plate capacitor is 0.01 $m^2$. The distance between the plates is 2.5 cm. The insulating medium is air. Find its capacitance. What would be its capacitance, if the space between the plates is filled with an insulating material of relative permittivity 5?

$C = \dfrac{\epsilon_0 \epsilon_r A }{d}$

$C = \dfrac{8.854 \times 10^{-12} \times 1 \times 0.01}{2.5 \times 10^{-2}}$

$C = 3.5416 \times 10^{-12} F$

For $\epsilon_r = 5$

$C = \dfrac{\epsilon_0 \epsilon_r A }{d}$

$C = \dfrac{8.854 \times 10^{-12} \times 5 \times 0.01}{2.5 \times 10^{-1}}$

$C = 1.7708 \times 10^{-11} F$

5) A 5 $\mu$F capacitor is charged to a potential difference of 110 V and then connected in parallel with an uncharged 3 $\mu$F capacitor. Calculate the potential difference across the parallel capacitor.

$Q_1 = C_1 \times V_1 = 5 \times 10^{-6} \times 110 = 5.5 \times 10^{-4} C$

Both the capacitors are connected in parallel. Therefore voltage across both the capacitor will remain same.

$V_1 = V_2$

$\dfrac{q_1}{C_1} = \dfrac{q_2}{C_2}$

Now capacitor $C_2$ is uncharged, therefore the charge will flow from $C_1$ to $C_2$ and this flow of charge will stop when the potential difference across $C_1$ and $C_2$ will become equal i.e. $V_1 = V_2$. Capacitor $C_1$ will loose charge say $q_2$ and capacitor $C_2$ will gain that charge $q_2$. Hence new charge on $C_1$ will become $q_1$.

$\therefore$ Charge on $C_1$ will be $q_1 = Q_1 – q_2$

$\dfrac{Q_1 – q_2}{C_1} = \dfrac{q_2}{C_2}$

$\dfrac{(5.5 \times 10^{-4}) – q_2}{5 \times 10^-6} = \dfrac{q_2}{3 \times 10^-6}$

$q_2 = 2.06525 \times 10^{-4}$

$V_1 = V_2 = \dfrac{q_2}{C_2} = 68.75 V$

Method 2

$Q_1 = C_1 \times V_1 = 5.5 \times 10^{-4} C$

When the capacitors are connected in parallel, the total capacitance is 8 $\mu F$ and the charge of 0.01 C is divided between the two capacitor.

$Q = CV$

$V = \dfrac{Q}{C}$

$V = \dfrac{5.5 \times 10 ^{-4}}{8 \times 10^{-6}}$

$V = 68.75 V$

6) A 2 $\mu F$ capacitor is charged to 100 V and 3 $\mu F$ capacitor to 200 V.  After disconnecting the supply capacitor terminal of similar polarity are connected together calculate the potential difference between the terminal.

Initial charge on each capacitor.

$Q_{1i} = C_1V_1 = 2 \times 10^{-6} \times 100 = 2 \times 10^{-4} C$

$Q_{2i} = C_2V_2 = 3 \times 10^{-6} \times 100 = 6 \times 10^{-4} C$

Now, $V_1 = V_2$

$\dfrac{Q_{1i} + Q}{C_1} = \dfrac{Q_{2i} – Q}{C_2}$

$\dfrac{2 \times 10^{-4} + Q}{ 2 \times 10^{-6}} = \dfrac{6 \times 10^{-4} – Q}{ 3 \times 10^{-6}}$

$Q = 1.2 \times 10^{-4} C$

$V_1 = V_2 = \dfrac{Q_{1i} + Q}{C_1} = \dfrac{( 2 \times 10^{-4} )+(1.2 \times 10^{-4})}{ 2 \times 10^{-6} } = 160 \;V$

Note:  In the above numerical $Q_{1i} < Q_{2i}$, therefore after connecting similar terminals of the capacitors capacitor $C_2$will lose some charge and $C_1$ will gain some charge

CHARGING AND DISCHARGING OF CAPACITOR

1) A capacitor having capacitance of 4 $\mu F$ is connected in series with a resistance of 1 M$\Omega$ across 200 V DC supply. Find – 1) time constant 2) The initial charging current 3) The time taken by capacitor to raise up to 160 V.

$\tau = RC = 1 \times 10^{6} \times 4 \times 10^{6} = 4 sec$

$I_0 = \dfrac{V}{R} = \dfrac{200}{10^6} = 200 \times 10^{-6} A$

$V_C = V (1 – e^{\frac{-t}{\tau}})$

$160 = 200(1-e^{\frac{-t}{4}})$

$t = 6.4377 sec$

Solved problems on Kirchhoff’s Voltage Law (KVL) / Mesh Analysis

Get solved examples on Kirchhoff’s voltage law. We will find current by using KVL in the given circuit. This problem solving technique is also known as Mesh analysis.

Magnetic circuit with an air gap

1) An iron ring has its mean length of flux path as 60 cm and its cross-sectional area as 15 cm². If relative permeability is 500. Find the current required to be passed through a coil of 300 turns wound uniformly around it to produce a flux density of 1.2 T. What would be the flux density with the same current if the iron ring is replaced by an air core.

Given: l = 60 cm; a = 15 cm²; $\mu_r = 500$ ; N = 300; B=1.2 T.

$MMF(F) = \dfrac{Bl}{\mu_0 \mu_r}$

$F = \dfrac{1.2 \times 60 \times 10^{-2}}{4\pi \times 10^{-7}\times 500}$

F = 1145.9155 AT

If iron ring is replaced by air core.

$MMF(F) = \dfrac{B_{air}l}{\mu_0 \mu_r}$

$B_{air} = 2.4 \times 10^{-3} T$

2) A coil of 600 turns and resistance of 10 $\Omega$ is wound uniformly over a steel ring of mean circumference of 30 cm and cross sectional area of 9 $cm^2$. It is connected to a supply of 20 V if the relative permeability of the ring is 1600. Find 1) Reluctance 2) Magnetic field intensity 3) MMF 4) Flux.

i) Reluctance:

$S = \dfrac{l}{\mu_0 \mu_r a}$

$S = \dfrac{30 \times 10^{-2}}{4 \times 10^{-7} \times 1600 \times 9 \times 10^{4}}$

$S = 165786.3991 \; AT/Wb$

ii) Magnetic field intensity:

$I = \dfrac{V}{R}$

$I = \dfrac{20}{10}$

$I = 2 \; A$

$H = \dfrac{NI}{l}$

$H = \dfrac{600 \times 2}{30 \times 10^{-2}}$

$H = 4000 AT/m$

iii) MMF = NI = 600 $\times$ 2 = 1200 AT.

iv) Flux:

$\phi = \dfrac{600 \times 2}{165786.3991}$

$\phi = 7.2382 \times 10^{-3} \; Wb$

3) A ring shaped core is made up of two parts of the same material. Part one is a magnetic path of length 25 cm and with cross sectional area $4 cm^2$, whereas part two is of length 10 cm and cross section area of 6 $cm^2$. The flux density in part two is 1.5 T. If the current through the coil, wound over core is 0.5 A. Calculate the number of turns of coil. Assume $\mu_r$for material is 1000.

Given: $l_1$ = 25 cm; $l_2$ = 10 cm;$a_1$= 4 $cm^2$; $a_2$= 6 $cm^2$; B = 1.5 T; $\mu_r$= 1000

$S = S_1 + S_2$

$S = \dfrac{l_1}{\mu_0 \mu_r A_1} + \dfrac{l_2}{\mu_0 \mu_r A_2}$

$S = \dfrac{1}{4\pi \times 10^{-7} \times{10^3}} \left[ \dfrac{0.25}{4\times 10^{-4}} + \dfrac{0.1}{6\times 10^{-4}} \right]$

S = 629988.3164 AT/Wb.

$B_2 = \dfrac{\phi}{A_2}$

$\phi = 1.5 \times 6 \times 10^{-4} = 9 \times 10^{-4} Wb.$

$F = \phi \times S$

$NI = \phi \times S$

$N = \dfrac{S \phi}{I} = \dfrac{629988.3164 \times 9 \times 10^{-4}}{0.5}$

N = 1133.98 = 1134 T.