Sinusoidal Voltage and Current
Determine the resistance if the circuit has a 0.2 H inductor with 200 volts,30 Hz?
SOLUTION
𝑮𝑰𝑽𝑬𝑵
𝐿 = 0.2 𝐻 𝑉 = 200 𝑉 f= 30 𝐻𝑧
𝑋𝐿 = 𝑤𝐿 = (2𝜋FL)
𝑋𝐿 = (2𝜋 𝑥 30 𝐻𝑧)(0.2𝐻)
𝑋𝐿 = 37.6991 𝑂ℎ𝑚𝑠
Three capacitor of 16, 15 and 12 µF, respectively are connected in series. What is the
total reactance of the capacitance assuming the source voltage is 200V, 60 Hz?
SOLUTION
𝑮𝑰𝑽𝑬𝑵
𝐶1 = 16 𝑢𝐹 𝐶2 = 15 𝑢𝐹 𝐶3 = 12 𝑢𝐹 𝑉 = 200 𝑉 𝑓 = 60 𝐻𝑧
1 1 1
𝐶𝑇 = 𝑋𝐶 = =
1 1 1 𝑤𝐶 2𝜋F𝐶
+ +
𝐶1 𝐶2 𝐶3
1
1 𝑋𝐶 =
𝐶𝑇 = 2𝜋(60hz)(4.7059 𝑢𝐹)
1 1 1
+ +
16𝑢𝐹 15𝑢𝐹 12𝑢𝐹 𝑋𝐶 = 563.6738 𝑂ℎ𝑚𝑠
−1 −1 −1 −1
𝐶𝑇 = ( 16𝑢𝑓 + 15𝑢𝐹 + 12𝑢𝐹 )
𝐶𝑇 = 4.7059 𝑢𝐹
A resistor of 25 Ω is connected in series with a capacitor of 45 μF. Calculate (a) the
impedance and (b) the current taken from a 240 V, 50 Hz supply. Find also the phase
angle between the supply voltage and the current
SOLUTION
𝑮𝑰𝑽𝑬𝑵
𝑅1 = 25 𝑂ℎ𝑚𝑠 𝐶1 = 45𝑢𝐹
𝑓 = 50𝐻𝑧 𝑉 = 240 𝑉
𝑨. 𝑰𝒎𝒑𝒆𝒅𝒂𝒏𝒄𝒆
1 1 𝑍= 𝑅 2 + 𝑋𝐿 − 𝑋𝐶 2
𝑋𝐶 = =
𝑤𝐶 2𝜋F𝐶
𝑍= 252 + 0 − 70.7355𝑜ℎ𝑚𝑠 2
1
𝑋𝐶 =
2𝜋(50hz)(45 𝑢𝐹) 𝑍 = 75.0234 𝑂ℎ𝑚𝑠
𝑋𝐶 = 70.7355 𝑂ℎ𝑚𝑠
A resistor of 25 Ω is connected in series with a capacitor of 45 μF. Calculate (a) the
impedance and (b) the current taken from a 240 V, 50 Hz supply. Find also the phase
angle between the supply voltage and the current?
SOLUTION
𝑮𝑰𝑽𝑬𝑵
𝑅1 = 25 𝑂ℎ𝑚𝑠 𝐶1 = 45𝑢𝐶
𝑓 = 50𝐻𝑧 𝑉 = 240 𝑉
𝑩. 𝑪𝒖𝒓𝒓𝒆𝒏𝒕 𝑪. 𝑷𝒉𝒂𝒔𝒆 𝑨𝒏𝒈𝒍𝒆
𝑉 𝑋𝐿 − 𝑋𝑐
𝐼= 𝑃ℎ𝑎𝑠𝑒 𝐴𝑛𝑔𝑙𝑒 = tan−1
𝑍 𝑅
240 𝑉 0 − 70.7355
𝐼= 𝑃ℎ𝑎𝑠𝑒 𝐴𝑛𝑔𝑙𝑒 = tan−1
75.0234 𝑂ℎ𝑚𝑠 25
𝑃ℎ𝑎𝑠𝑒 𝐴𝑛𝑔𝑙𝑒 = −70.5351 °
𝐼 = 3.199 𝐴
A resistor of 25 Ω is connected in series with a capacitor of 45 μF. Calculate (a) the
impedance and (b) the current taken from a 240 V, 50 Hz supply. Find also the phase
angle between the supply voltage and the current?
SOLUTION
𝑮𝑰𝑽𝑬𝑵
𝑅1 = 25 𝑂ℎ𝑚𝑠 𝐶1 = 45𝑢𝐶
𝑓 = 50𝐻𝑧 𝑉 = 240 𝑉
𝑨. 𝑰𝒎𝒑𝒆𝒅𝒂𝒏𝒄𝒆
𝑍 = 𝑅 ± 𝑗𝑋 𝑍 = 25 − 𝑗70.7355 𝑂ℎ𝑚𝑠
𝑍 = 𝑅 ± 𝑗𝑋𝑐 𝐶𝑜𝑛𝑣𝑒𝑟𝑡 𝑡𝑜 𝑝𝑜𝑙𝑎𝑟 𝑓𝑜𝑟𝑚
1 1 𝑍 = 75.0234∠ − 70.5351°𝑶𝒉𝒎𝒔
𝑋𝐶 = =
𝑤𝐶 2𝜋F𝐶
𝐼𝑚𝑝𝑒𝑑𝑎𝑛𝑐𝑒 𝐴𝑛𝑔𝑙𝑒 𝑃ℎ𝑎𝑠𝑒
1
𝑋𝐶 =
2𝜋(50hz)(45 𝑢𝐹)
𝑋𝐶 = 70.7355 𝑂ℎ𝑚𝑠
The following three impedances are connected in series across a 40 V, 20 kHz supply:
(i) a resistance of 8 Ω, (ii) a coil of inductance 130 μH and 5 Ω resistance, and (iii) a 10
Ω resistor in series with a 0.25 μF capacitor. Calculate (a) the circuit current, (b) the
circuit phase angle, and (c) the voltage drop across each impedance
SOLUTION
𝑮𝑰𝑽𝑬𝑵
𝑅𝐴 = 8 𝑂ℎ𝑚𝑠 𝐿𝐵 = 130𝑢𝐻 𝑅𝐵 = 5𝑜ℎ𝑚𝑠 𝑅𝐶 = 10 𝑜ℎ𝑚𝑠 𝐿𝐶 = 0.25𝑢𝐹
𝑓 = 20𝑘𝐻𝑧 𝑉 = 40 𝑉
1 1 𝑅𝑇 = 𝑅𝐴 + 𝑅𝐵 + 𝑅𝑐
𝑋𝐿 = 𝑤𝐿 = (2𝜋FL) 𝑋𝐶 = =
𝑤𝐶 2𝜋F𝐶
𝑅𝑇 = 8 + 10 + 5
𝑋𝐿 = (2𝜋 𝑥 20 𝑘𝐻𝑧)(130𝑢𝐻) 1
𝑋𝐶 =
2𝜋(20kHz)(0.25 𝑢𝐹) 𝑅𝑇 = 23 𝑜ℎ𝑚𝑠
𝑋𝐿 = 16.3362 𝑂ℎ𝑚𝑠
𝑋𝐶 = 31.8310 𝑂ℎ𝑚𝑠
The following three impedances are connected in series across a 40 V, 20 kHz supply:
(i) a resistance of 8 Ω, (ii) a coil of inductance 130 μH and 5 Ω resistance, and (iii) a 10
Ω resistor in series with a 0.25 μF capacitor. Calculate (a) the circuit current, (b) the
circuit phase angle, and (c) the voltage drop across each impedance
SOLUTION
𝑍= 𝑅2 + 𝑋𝐿 − 𝑋𝐶 2
𝑍= 232 + 31.8310 − 16.3362 2
𝑍 = 27.7325 𝑂ℎ𝑚𝑠
B.𝑷𝒉𝒂𝒔𝒆 𝑨𝒏𝒈𝒍𝒆
𝑨. 𝑪𝒖𝒓𝒓𝒆𝒏𝒕 𝑋𝐿 − 𝑋𝑐
𝑃ℎ𝑎𝑠𝑒 𝐴𝑛𝑔𝑙𝑒 = tan−1
𝑉 40 𝑉 𝑅𝑇
𝐼= 𝐼=
𝑍 27.7325 𝑂ℎ𝑚𝑠 31.8310 − 16.3362
−1
𝑃ℎ𝑎𝑠𝑒 𝐴𝑛𝑔𝑙𝑒 = tan
23
𝐼 = 1.4424 𝐴 𝑃ℎ𝑎𝑠𝑒 𝐴𝑛𝑔𝑙𝑒 = 33.9676 °
The following three impedances are connected in series across a 40 V, 20 kHz supply:
(i) a resistance of 8 Ω, (ii) a coil of inductance 130 μH and 5 Ω resistance, and (iii) a 10
Ω resistor in series with a 0.25 μF capacitor. Calculate (a) the circuit current, (b) the
circuit phase angle, and (c) the voltage drop across each impedance
SOLUTION
𝑪. 𝑽𝒐𝒍𝒕𝒂𝒈𝒆 𝒅𝒓𝒐𝒑 𝒂𝒄𝒓𝒐𝒔𝒔 𝒊𝒎𝒑𝒆𝒅𝒂𝒏𝒄𝒆
𝑉𝐴 = (𝐼𝑇 )(𝑅𝐴 ) 𝑉𝐵 = (𝐼𝑇 )(𝑍𝐵 ) 𝑉𝐶 = (𝐼𝑇 )(𝑍𝐶 )
𝑉𝐴 = (1.4424𝐴)(8 𝑂ℎ𝑚𝑠)
𝑍𝐵 = 𝑅𝐵2 + 𝑋𝐿𝐵 − 𝑋𝐶𝐵 2 𝑍𝐶 = 𝑅𝐶2 + 𝑋𝐿𝐶 − 𝑋𝐶𝐶 2
𝑉𝐴 = 11.5329 𝐴
𝑍𝐵 = 52 + 16.3362 − 0 2 𝑍𝐶 = 102 + 0 − 31.8310 2
𝑍𝐵 = 17.0842 𝑜ℎ𝑚𝑠 𝑍𝐶 = 33.3648 𝑜ℎ𝑚𝑠
𝑉𝐵 = (1.4424𝐴)(17.0842 𝑂ℎ𝑚𝑠) 𝑉𝐶 = (1.4424𝐴)(33.3648𝑂ℎ𝑚𝑠)
𝑉𝐵 = 24.6423 𝑉 𝑉𝐶 = 48.1254 𝑉
Three impedance Z1 = 1 –j4, Z2 = -j6 and Z3 = 4 + j3 ohms respectively are connected
in series-parallel. Z1 is connected in series with the parallel combinations of Z2 and
Z3. If this circuit is connected across a 230 V, 60 Hz source, determine the voltage
across the parallel combination of Z2 and Z3.
SOLUTION 230 𝑉
𝑉
𝑍2 𝑍3 𝐼𝑇 = 𝐼𝑇 =
𝑍23 = 𝑍 8.8294
𝑍2 + 𝑍3
𝐼𝑇 = 26.0490
(−𝑗6)(4 + 𝑗3)
𝑍23 =
−𝑗 + 4 + 𝑗3
𝑉23 = 𝐼𝑇 𝑍23
𝑍23 = 5.76 − 𝑗1.68 𝑜ℎ𝑚
𝑍23 = 6∠ − 16.2602° 𝑂ℎ𝑚𝑠 𝑉23 = (26.0490)(6)
𝑍𝑇 = 𝑍1 + 𝑍23 𝑉23 = 156.294 𝑉
𝑍𝑇 = 1 − 𝑗4 + (5.76 − 𝑗1.68)
𝑍𝑇 = 6.76 − 𝑗5.68
𝑍𝑇 = 8.8294∠ − 40.0382° 𝑂ℎ𝑚𝑠
Three impedance Z1 = 1 –j4, Z2 = -j6 and Z3 = 4 + j3 ohms respectively are connected
in series-parallel. Z1 is connected in series with the parallel combinations of Z2 and
Z3. If this circuit is connected across a 230 V, 60 Hz source, determine the voltage
across the parallel combination of Z2 and Z3.
SOLUTION 230 𝑉
𝑉
𝑍2 𝑍3 𝐼𝑇 = 𝐼𝑇 =
𝑍23 = 𝑍 6.76 − 𝑗5.68 𝑂ℎ𝑚𝑠
𝑍2 + 𝑍3
(−𝑗6)(4 + 𝑗3) 𝐼𝑇 = 19.9436 + 𝑗16.7573𝐴
𝑍23 =
−𝑗 + 4 + 𝑗3
𝐼𝑇 = 26.0490∠40.0382°𝐴
𝑍23 = 5.76 − 𝑗1.68 𝑜ℎ𝑚
𝑉23 = 𝐼𝑇 𝑍23
𝑍𝑇 = 𝑍1 + 𝑍23 𝑉23 = (5.76 − 𝑗1.68 𝑜ℎ𝑚𝑠)(26.0490∠40.0382°𝐴)
𝑍𝑇 = 1 − 𝑗4 + (5.76 − 𝑗1.68) 𝑉23 = 156.2943∠23.7780°𝑉
𝑍𝑇 = 6.76 − 𝑗5.68
