A current wave is given by the expression i = 400sin(800t) milli-amperes. Calculate: (i) The frequency (ii)period (iii) the r.m.s value (iv) the time t when I = 250mA

Question

Asked on July 11, 2026Physics

This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.

Accepted Answer · 2026-07-11T15:06:17.140Z

b. A current wave is given by the expression i = 400(800t) milli-amperes. The general form of a sinusoidal current is i = I_m ( t), where I_m is the peak current and is the angular frequency. From the given expression, I_m = 400mA and = 800 rad/s. i) The frequency (f) The angular frequency is related to the frequency f by the formula = 2 f. f = ()/(2) f = 800 rad/s2 f ≈ 127.32 Hz ii) Period (T) The period T is the reciprocal of the frequency f. T = (1)/(f) T = (1)/(127.32 Hz) T ≈ 0.00785 s iii) The r.m.s. value (I_rms) For a sinusoidal waveform, the RMS value is the peak value divided by sqrt(2). I_rms = (I_m)/(sqrt(2)) I_rms = 400 mAsqrt(2) I_rms ≈ 282.84 mA iv) The time t when i = 250mA Substitute i = 250mA into the given expression: 250 = 400(800t) (800t) = (250)/(400) (800t) = 0.625 To find 800t, take the inverse sine (arcsin) of 0.625. Ensure your calculator is in radian mode for this step. 800t = (0.625) 800t ≈ 0.6751 rad t = 0.6751 rad800 rad/s t ≈ 0.000844 s c. A 240-V, 50-Hz, single phase supply is connected across a (15 + j50) load. Given: Voltage V = 240V (RMS) Frequency f = 50Hz Impedance Z = (15 + j50) i) The current and power factor Step 1: Calculate the magnitude and phase angle of the impedance. The magnitude of the impedance is |Z| = sqrt(R^2 + X_L^2). |Z| = sqrt(15^2 + 50^2) = sqrt(225 + 2500) = sqrt(2725) |Z| ≈ 52.20 The phase angle of the impedance is = ((X_L)/(R)). = ((50)/(15)) = (3.333) ≈ 73.30^ Step 2: Calculate the current. The magnitude of the current is |I| = (|V|)/(|Z|). |I| = 240V52.20 |I| ≈ 4.60 A The current lags the voltage because the load is inductive. Step 3: Calculate the power factor. The power factor (PF) is (). PF = (73.30^) PF ≈ 0.287 lagging ii) The value of the capacitor required across the load to improve the power factor to 0.95 lagging. Step 1: Calculate the initial active power (P) and reactive power (Q_1). Initial power factor PF_1 = 0.287. Initial phase angle _1 = 73.30^. Active power P = V I (_1) = 240V × 4.60A × 0.287 = 316.896 W. Alternatively, P = I^2 R = (4.60A)^2 × 15 = 21.16 × 15 W = 317.4 W. (Using I=4.60 for calculation) Let's use P = (V^2 R)/(|Z|^2) = (240^2 × 15)/(2725) = (57600 × 15)/(2725) = (864000)/(2725) ≈ 317.0 W. Initial reactive power Q_1 = V I (_1) = 240V × 4.60A × (73.30^) = 240V × 4.60A × 0.958 = 1057.6 VAR. Alternatively, Q_1 = I^2 X_L = (4.60A)^2 × 50 = 21.16 × 50 VAR = 1058 VAR. Step 2: Calculate the target phase angle (_2) and target reactive power (Q_2). Target power factor PF_2 = 0.95 lagging. _2 = (0.95) _2 ≈ 18.19^ The active power P remains the same. Target reactive power Q_2 = P (_2). Q_2 = 317.0 W × (18.19^) Q_2 = 317.0 W × 0.3287 Q_2 ≈ 104.2 VAR Step 3: Calculate the reactive power to be compensated by the capacitor (Q_c). The capacitor provides leading reactive power, so it reduces the lagging reactive power. Q_c = Q_1 - Q_2 Q_c = 1058 VAR - 104.2 VAR Q_c = 953.8 VAR Step 4: Calculate the capacitance (C). The reactive power of a capacitor is Q_c = (V^2)/(X_C) = V^2 C. Angular frequency = 2 f = 2 × 50 Hz = 100 rad/s. C = (Q_c)/(V^2 ) C = 953.8 VAR(240V)^2 × 100 rad/s C = (953.8)/(57600 × 100) F C = (953.8)/(18095573.68) F C ≈ 5.27 × 10^-5 F C ≈ 52.7 F 3a. Give four (4) comparisons between magnetic circuit and electric circuit. Driving Force: In an electric circuit, the driving force is the electromotive force (EMF), measured in Volts. In a magnetic circuit, it is the magnetomotive force (MMF), measured in Ampere-turns. Flow: In an electric circuit, the flow is electric current (charge carriers), measured in Amperes. In a magnetic circuit, the flow is magnetic flux, measured in Webers. Opposition to Flow: In an electric circuit, the opposition to current flow is resistance, measured in Ohms. In a magnetic circuit, the opposition to flux is reluctance, measured in Ampere-turns/Weber. Governing Laws: Electric circuits are governed by Ohm's Law (V=IR) and Kirchhoff's Laws. Magnetic circuits are governed by Hopkinson's Law (analogous to Ohm's Law for magnetic circuits, MMF = S). 3b. Derive the equation for the reluctance of a uniform magnetic material in terms of dimensions of the materials and other relevant parameters. Step 1: Define MMF and Flux. The magnetomotive force (MMF) is given by MMF = NI, where N is the number of turns and I is the current. The magnetic flux () is given by = BA, where B is the magnetic flux density and A is the cross-sectional area. Step 2: Relate Flux Density to Magnetic Field Strength. The magnetic flux density B is related to the magnetic field strength H by B = H, where is the permeability of the material. The permeability can also be expressed as = _0 _r, where _0 is the permeability of free space and _r is the relative permeability. So, B = _0 _r H. Step 3: Relate Magnetic Field Strength to MMF. For a uniform magnetic circuit of length l, the magnetic field strength H is given by H = (MMF)/(l) = (NI)/(l). Step 4: Substitute and derive the expression for reluctance. Substitute the expression for H into the equation for B: B = _0 _r ((NI)/(l)) Now substitute this expression for B into the equation for : = (_0 _r (NI)/(l)) A = (_0 _r A NI)/(l) Reluctance (S) is defined as the ratio of MMF to flux: S = (MMF)/() Substitute the expressions for MMF and : S = (NI)/(_0 _r A NI)l Cancel NI from the numerator and denominator: S = (l)/(_0 _r A) Since = _0 _r, the equation can also be written as: S = (l)/( A) Where l is the length of the magnetic path, A is the cross-sectional area, and is the absolute permeability of the material. 3c. A closed magnetic circuit is made of material of relative permeability 600, of length 40cm and having a cross-sectional area of 100mm^2. A 400-turn coil is wound on the magnetic core. The coil carries a current of 500mA. Given: Relative permeability _r = 600 Length l = 40cm = 0.4m Cross-sectional area A = 100mm^2 = 100 × (10^-3m)^2 = 100 × 10^-6m^2 = 10^-4m^2 Number of turns N = 400 Current I = 500mA = 0.5A Permeability of free space _0 = 4 × 10^-7 H/m i) The reluctance of the magnetic circuit (S) Step 1: Calculate the absolute permeability (). = _0 _r = (4 × 10^-7 H/m) × 600 = 2400 × 10^-7 H/m ≈ 7.5398 × 10^-4 H/m Step 2: Calculate the reluctance (S). S = (l)/( A) S = 0.4m(7.5398 × 10^-4 H/m) × (10^-4m^2) S = (0.4)/(7.5398 × 10^-8) A-turns/Wb S ≈ 5.305 × 10^6 A-turns/Wb ii) The flux set up in the core () Step 1: Calculate the magnetomotive force (MMF). MMF = NI MMF = 400 × 0.5A = 200 A-turns Step 2: Calculate the flux (). = (MMF)/(S) = 200 A-turns5.305 × 10^6 A-turns/Wb ≈ 3.770 × 10^-5 Wb ≈ 37.70 Wb iii) The magneto motive force (MMF) MMF = NI MMF = 400 × 0.5A MMF = 200 A-turns IV) The inductance of the magnetic circuit (L) The inductance L can be calculated using the formula L = (N^2)/(S). L = ((400)^2)/(5.305 × 10^6 A-turns/Wb) L = (160000)/(5.305 × 10^6) H L ≈ 0.03016 H L ≈ 30.16 mH 4a. Define the term resonance as applied to ac circuits. Sketch a resonance curve as applied to RC circuits and hence deduce the equation for resonant frequency. Definition of Resonance: Resonance in an AC circuit is a condition that occurs when the inductive reactance (X_L) and capacitive reactance (X_C) are equal in magnitude, effectively canceling each other out. This leads to the circuit's impedance being purely resistive, resulting in maximum current (in series RLC circuits) or maximum voltage (in parallel RLC circuits) at a specific frequency called the resonant frequency. Sketch a resonance curve as applied to RC circuits: The question asks for a resonance curve for RC circuits. However, resonance, in the sense of X_L = X_C, does not occur in a simple RC circuit as it lacks an inductor. A resonance curve typically depicts the current or voltage response of an RLC circuit as frequency varies, showing a peak at the resonant frequency. If the question implies a series RLC circuit, the curve would look like this: (Imagine a graph with "Current (I)" on the y-axis and "Frequency (f)" on the x-axis. The curve starts low, rises to a distinct peak at a specific frequency (the resonant frequency f_0), and then falls back down.) Deduce the equation for resonant frequency: For a series RLC circuit, resonance occurs when the inductive reactance equals the capacitive reactance: X_L = X_C We know that X_L = L and X_C = (1)/( C), where = 2 f. L = (1)/( C) Multiply both sides by : ^2 L = (1)/(C) ^2 = (1)/(LC) Take the square root of both sides: = (1)/(sqrt(LC)) Substitute = 2 f_0 (where f_0 is the resonant frequency): 2 f_0 = (1)/(sqrt(LC)) Solve for f_0: f_0 = (1)/(2(LC)) 4b. In the figure drawn below, the circuit is at resonant at a frequency of 490Hz. The circuit consists of an 8 resistor, a 50mH inductor, and a capacitor C in series with a 10V source. Given: Resonant frequency f_0 = 490Hz Resistance R = 8 Inductance L = 50mH = 0.05H Voltage V = 10V (RMS) i. Current at resonance (I_0) At resonance, the impedance of a series RLC circuit is purely resistive, meaning Z = R. I_0 = (V)/(R) I_0 = 10V8 I_0 = 1.25 A ii. Value of C At resonance, f_0 = (1)/(2(LC)). We can rearrange this to solve for C. f_0^2 = (1)/(4^2 LC) C = (1)/(4^2 f_0^2 L) C = (1)/(4^2 (490 Hz))^2 (0.05 H) C = (1)/(4^2 (240100) (0.05)) F C = (1)/(47420.4 × 0.05) F C = (1)/(47420.4) F C ≈ 2.1088 × 10^-5 F C ≈ 21.09 F iii. Capacitor voltage at resonance (V_C) First, calculate the capacitive reactance X_C at resonance. X_C = (1)/(2 f_0 C) X_C = (1)/(2 (490 Hz)) (2.1088 × 10^-5 F) X_C = (1)/(0.06459) X_C ≈ 15.48 Now, calculate the capacitor voltage: V_C = I_0 X_C V_C = 1.25 A × 15.48 V_C ≈ 19.35 V iv. Inductor voltage at resonance (V_L) First, calculate the inductive reactance X_L at resonance. X_L = 2 f_0 L X_L = 2 (490 Hz) (0.05 H) X_L = 153.938 × 0.05 X_L ≈ 15.39 Now, calculate the inductor voltage: V_L = I_0 X_L V_L = 1.25 A × 15.39 V_L ≈ 19.24 V (Note: V_L and V_C should be equal at resonance, the slight difference is due to rounding in intermediate steps.) v. Quality factor of the circuit (Q) For a series RLC circuit, the quality factor is Q = (X_L)/(R) (or (X_C)/(R)). Q = (15.39)/(8) Q ≈ 1.92 vi. Bandwidth of the circuit (BW) The bandwidth of a series RLC circuit is given by BW = (f_0)/(Q). BW = 490 Hz1.92 BW ≈ 255.21 Hz 7a. Draw the circuit diagram of a 3-phase, 4-wire power system feeding a delta-connected load, show all relevant currents in the circuit. (Imagine a diagram with the following elements and connections) Source Side: Three phase lines labeled R, Y, B. A neutral line labeled N. The source could be represented as three voltage sources (e.g., star-connected) providing phase voltages V_RN, V_YN, V_BN and line voltages V_RY, V_YB, V_BR. Load Side (Delta-connected): Three impedances Z_RY, Z_YB, Z_BR connected in a triangle (delta) configuration. Z_RY connected between R and Y lines. Z_YB connected between Y and B lines. Z_BR connected between B and R lines. Currents: Line Currents: I_R, I_Y, I_B flowing from the source lines to the load connection points. Phase Currents: I_RY (flowing through Z_RY), I_YB (flowing through Z_YB), I_BR (flowing through Z_BR). Neutral Current: I_N (flowing in the neutral line). For a delta load, the neutral line is typically not connected to the load directly, so I_N would be zero unless there's an imbalance in the source or other connections. However, if the question implies a 4-wire system feeding* the delta load, it means the 4 wires are available, but only 3 are used for the delta load. Diagram Description: Draw three vertical lines (R, Y, B) representing the phase conductors, and one more vertical line (N) for the neutral. Connect the R, Y, B lines to the corners of a triangle formed by three impedances. Connect Z_RY between R and Y. Connect Z_YB between Y and B. Connect Z_BR between B and R. Show arrows for line currents I_R, I_Y, I_B entering the delta load from the respective phase lines. Show arrows for phase currents I_RY, I_YB, I_BR flowing within the delta impedances. (For a balanced delta load, I_L = sqrt(3) I_P and V_L = V_P). 7b. A star-connected load having the following impedances, Z_r = (20 - j16), Z_y = (16 - j20) and Z_b = (20 - j0) are connected across a 3-phase, 4-wire system having a line voltage of 420V. Given: Z_r = (20 - j16) Z_y = (16 - j20) Z_b = (20 - j0) (purely resistive) Line voltage V_L = 420V 3-phase, 4-wire system. i) Draw the circuit diagram (Imagine a diagram with the following elements and connections) Source Side: Three phase lines labeled R, Y, B. A neutral line labeled N. The source could be represented as three voltage sources (e.g., star-connected) providing phase voltages V_RN, V_YN, V_BN and line voltages V_RY, V_YB, V_BR. Load Side (Star-connected): Three impedances Z_r, Z_y, Z_b connected from each phase line to a common neutral point (N'). Z_r connected between R line and N'. Z_y connected between Y line and N'. Z_b connected between B line and N'. The load neutral point N' is connected to the system neutral line N. Currents and Voltages: Line Currents: I_R, I_Y, I_B flowing from the source lines through the respective impedances. In a star connection, line currents are equal to phase currents. Phase Voltages: V_RN', V_YN', V_BN' across each impedance. Neutral Current: I_N flowing in the neutral line, which will be non-zero because the load is unbalanced (Z_r ≠ Z_y ≠ Z_b). Diagram Description: Draw three vertical lines (R, Y, B) representing the phase conductors, and one more vertical line (N) for the neutral. Draw a central point N' (load neutral). Connect Z_r between R and N'. Connect Z_y between Y and N'. Connect Z_b between B and N'. Connect N' to N. Show arrows for line currents I_R, I_Y, I_B flowing from the phase lines through the impedances towards N'. Show an arrow for neutral current I_N flowing in the neutral line. Show phase voltages V_RN', V_YN', V_BN' across each impedance. (For a star connection, V_P = V_L / sqrt(3) and I_L = I_P). Drop the next question.

Accepted Answer · 2026-07-11T15:06:17.140Z

b. A current wave is given by the expression i = 400(800t) milli-amperes. The general form of a sinusoidal current is i = I_m ( t), where I_m is the peak current and is the angular frequency. From the given expression, I_m = 400mA and = 800 rad/s. i) The frequency (f) The angular frequency is related to the frequency f by the formula = 2 f. f = ()/(2) f = 800 rad/s2 f ≈ 127.32 Hz ii) Period (T) The period T is the reciprocal of the frequency f. T = (1)/(f) T = (1)/(127.32 Hz) T ≈ 0.00785 s iii) The r.m.s. value (I_rms) For a sinusoidal waveform, the RMS value is the peak value divided by sqrt(2). I_rms = (I_m)/(sqrt(2)) I_rms = 400 mAsqrt(2) I_rms ≈ 282.84 mA iv) The time t when i = 250mA Substitute i = 250mA into the given expression: 250 = 400(800t) (800t) = (250)/(400) (800t) = 0.625 To find 800t, take the inverse sine (arcsin) of 0.625. Ensure your calculator is in radian mode for this step. 800t = (0.625) 800t ≈ 0.6751 rad t = 0.6751 rad800 rad/s t ≈ 0.000844 s c. A 240-V, 50-Hz, single phase supply is connected across a (15 + j50) load. Given: Voltage V = 240V (RMS) Frequency f = 50Hz Impedance Z = (15 + j50) i) The current and power factor Step 1: Calculate the magnitude and phase angle of the impedance. The magnitude of the impedance is |Z| = sqrt(R^2 + X_L^2). |Z| = sqrt(15^2 + 50^2) = sqrt(225 + 2500) = sqrt(2725) |Z| ≈ 52.20 The phase angle of the impedance is = ((X_L)/(R)). = ((50)/(15)) = (3.333) ≈ 73.30^ Step 2: Calculate the current. The magnitude of the current is |I| = (|V|)/(|Z|). |I| = 240V52.20 |I| ≈ 4.60 A The current lags the voltage because the load is inductive. Step 3: Calculate the power factor. The power factor (PF) is (). PF = (73.30^) PF ≈ 0.287 lagging ii) The value of the capacitor required across the load to improve the power factor to 0.95 lagging. Step 1: Calculate the initial active power (P) and reactive power (Q_1). Initial power factor PF_1 = 0.287. Initial phase angle _1 = 73.30^. Active power P = V I (_1) = 240V × 4.60A × 0.287 = 316.896 W. Alternatively, P = I^2 R = (4.60A)^2 × 15 = 21.16 × 15 W = 317.4 W. (Using I=4.60 for calculation) Let's use P = (V^2 R)/(|Z|^2) = (240^2 × 15)/(2725) = (57600 × 15)/(2725) = (864000)/(2725) ≈ 317.0 W. Initial reactive power Q_1 = V I (_1) = 240V × 4.60A × (73.30^) = 240V × 4.60A × 0.958 = 1057.6 VAR. Alternatively, Q_1 = I^2 X_L = (4.60A)^2 × 50 = 21.16 × 50 VAR = 1058 VAR. Step 2: Calculate the target phase angle (_2) and target reactive power (Q_2). Target power factor PF_2 = 0.95 lagging. _2 = (0.95) _2 ≈ 18.19^ The active power P remains the same. Target reactive power Q_2 = P (_2). Q_2 = 317.0 W × (18.19^) Q_2 = 317.0 W × 0.3287 Q_2 ≈ 104.2 VAR Step 3: Calculate the reactive power to be compensated by the capacitor (Q_c). The capacitor provides leading reactive power, so it reduces the lagging reactive power. Q_c = Q_1 - Q_2 Q_c = 1058 VAR - 104.2 VAR Q_c = 953.8 VAR Step 4: Calculate the capacitance (C). The reactive power of a capacitor is Q_c = (V^2)/(X_C) = V^2 C. Angular frequency = 2 f = 2 × 50 Hz = 100 rad/s. C = (Q_c)/(V^2 ) C = 953.8 VAR(240V)^2 × 100 rad/s C = (953.8)/(57600 × 100) F C = (953.8)/(18095573.68) F C ≈ 5.27 × 10^-5 F C ≈ 52.7 F 3a. Give four (4) comparisons between magnetic circuit and electric circuit. Driving Force: In an electric circuit, the driving force is the electromotive force (EMF), measured in Volts. In a magnetic circuit, it is the magnetomotive force (MMF), measured in Ampere-turns. Flow: In an electric circuit, the flow is electric current (charge carriers), measured in Amperes. In a magnetic circuit, the flow is magnetic flux, measured in Webers. Opposition to Flow: In an electric circuit, the opposition to current flow is resistance, measured in Ohms. In a magnetic circuit, the opposition to flux is reluctance, measured in Ampere-turns/Weber. Governing Laws: Electric circuits are governed by Ohm's Law (V=IR) and Kirchhoff's Laws. Magnetic circuits are governed by Hopkinson's Law (analogous to Ohm's Law for magnetic circuits, MMF = S). 3b. Derive the equation for the reluctance of a uniform magnetic material in terms of dimensions of the materials and other relevant parameters. Step 1: Define MMF and Flux. The magnetomotive force (MMF) is given by MMF = NI, where N is the number of turns and I is the current. The magnetic flux () is given by = BA, where B is the magnetic flux density and A is the cross-sectional area. Step 2: Relate Flux Density to Magnetic Field Strength. The magnetic flux density B is related to the magnetic field strength H by B = H, where is the permeability of the material. The permeability can also be expressed as = _0 _r, where _0 is the permeability of free space and _r is the relative permeability. So, B = _0 _r H. Step 3: Relate Magnetic Field Strength to MMF. For a uniform magnetic circuit of length l, the magnetic field strength H is given by H = (MMF)/(l) = (NI)/(l). Step 4: Substitute and derive the expression for reluctance. Substitute the expression for H into the equation for B: B = _0 _r ((NI)/(l)) Now substitute this expression for B into the equation for : = (_0 _r (NI)/(l)) A = (_0 _r A NI)/(l) Reluctance (S) is defined as the ratio of MMF to flux: S = (MMF)/() Substitute the expressions for MMF and : S = (NI)/(_0 _r A NI)l Cancel NI from the numerator and denominator: S = (l)/(_0 _r A) Since = _0 _r, the equation can also be written as: S = (l)/( A) Where l is the length of the magnetic path, A is the cross-sectional area, and is the absolute permeability of the material. 3c. A closed magnetic circuit is made of material of relative permeability 600, of length 40cm and having a cross-sectional area of 100mm^2. A 400-turn coil is wound on the magnetic core. The coil carries a current of 500mA. Given: Relative permeability _r = 600 Length l = 40cm = 0.4m Cross-sectional area A = 100mm^2 = 100 × (10^-3m)^2 = 100 × 10^-6m^2 = 10^-4m^2 Number of turns N = 400 Current I = 500mA = 0.5A Permeability of free space _0 = 4 × 10^-7 H/m i) The reluctance of the magnetic circuit (S) Step 1: Calculate the absolute permeability (). = _0 _r = (4 × 10^-7 H/m) × 600 = 2400 × 10^-7 H/m ≈ 7.5398 × 10^-4 H/m Step 2: Calculate the reluctance (S). S = (l)/( A) S = 0.4m(7.5398 × 10^-4 H/m) × (10^-4m^2) S = (0.4)/(7.5398 × 10^-8) A-turns/Wb S ≈ 5.305 × 10^6 A-turns/Wb ii) The flux set up in the core () Step 1: Calculate the magnetomotive force (MMF). MMF = NI MMF = 400 × 0.5A = 200 A-turns Step 2: Calculate the flux (). = (MMF)/(S) = 200 A-turns5.305 × 10^6 A-turns/Wb ≈ 3.770 × 10^-5 Wb ≈ 37.70 Wb iii) The magneto motive force (MMF) MMF = NI MMF = 400 × 0.5A MMF = 200 A-turns IV) The inductance of the magnetic circuit (L) The inductance L can be calculated using the formula L = (N^2)/(S). L = ((400)^2)/(5.305 × 10^6 A-turns/Wb) L = (160000)/(5.305 × 10^6) H L ≈ 0.03016 H L ≈ 30.16 mH 4a. Define the term resonance as applied to ac circuits. Sketch a resonance curve as applied to RC circuits and hence deduce the equation for resonant frequency. Definition of Resonance: Resonance in an AC circuit is a condition that occurs when the inductive reactance (X_L) and capacitive reactance (X_C) are equal in magnitude, effectively canceling each other out. This leads to the circuit's impedance being purely resistive, resulting in maximum current (in series RLC circuits) or maximum voltage (in parallel RLC circuits) at a specific frequency called the resonant frequency. Sketch a resonance curve as applied to RC circuits: The question asks for a resonance curve for RC circuits. However, resonance, in the sense of X_L = X_C, does not occur in a simple RC circuit as it lacks an inductor. A resonance curve typically depicts the current or voltage response of an RLC circuit as frequency varies, showing a peak at the resonant frequency. If the question implies a series RLC circuit, the curve would look like this: (Imagine a graph with "Current (I)" on the y-axis and "Frequency (f)" on the x-axis. The curve starts low, rises to a distinct peak at a specific frequency (the resonant frequency f_0), and then falls back down.) Deduce the equation for resonant frequency: For a series RLC circuit, resonance occurs when the inductive reactance equals the capacitive reactance: X_L = X_C We know that X_L = L and X_C = (1)/( C), where = 2 f. L = (1)/( C) Multiply both sides by : ^2 L = (1)/(C) ^2 = (1)/(LC) Take the square root of both sides: = (1)/(sqrt(LC)) Substitute = 2 f_0 (where f_0 is the resonant frequency): 2 f_0 = (1)/(sqrt(LC)) Solve for f_0: f_0 = (1)/(2(LC)) 4b. In the figure drawn below, the circuit is at resonant at a frequency of 490Hz. The circuit consists of an 8 resistor, a 50mH inductor, and a capacitor C in series with a 10V source. Given: Resonant frequency f_0 = 490Hz Resistance R = 8 Inductance L = 50mH = 0.05H Voltage V = 10V (RMS) i. Current at resonance (I_0) At resonance, the impedance of a series RLC circuit is purely resistive, meaning Z = R. I_0 = (V)/(R) I_0 = 10V8 I_0 = 1.25 A ii. Value of C At resonance, f_0 = (1)/(2(LC)). We can rearrange this to solve for C. f_0^2 = (1)/(4^2 LC) C = (1)/(4^2 f_0^2 L) C = (1)/(4^2 (490 Hz))^2 (0.05 H) C = (1)/(4^2 (240100) (0.05)) F C = (1)/(47420.4 × 0.05) F C = (1)/(47420.4) F C ≈ 2.1088 × 10^-5 F C ≈ 21.09 F iii. Capacitor voltage at resonance (V_C) First, calculate the capacitive reactance X_C at resonance. X_C = (1)/(2 f_0 C) X_C = (1)/(2 (490 Hz)) (2.1088 × 10^-5 F) X_C = (1)/(0.06459) X_C ≈ 15.48 Now, calculate the capacitor voltage: V_C = I_0 X_C V_C = 1.25 A × 15.48 V_C ≈ 19.35 V iv. Inductor voltage at resonance (V_L) First, calculate the inductive reactance X_L at resonance. X_L = 2 f_0 L X_L = 2 (490 Hz) (0.05 H) X_L = 153.938 × 0.05 X_L ≈ 15.39 Now, calculate the inductor voltage: V_L = I_0 X_L V_L = 1.25 A × 15.39 V_L ≈ 19.24 V (Note: V_L and V_C should be equal at resonance, the slight difference is due to rounding in intermediate steps.) v. Quality factor of the circuit (Q) For a series RLC circuit, the quality factor is Q = (X_L)/(R) (or (X_C)/(R)). Q = (15.39)/(8) Q ≈ 1.92 vi. Bandwidth of the circuit (BW) The bandwidth of a series RLC circuit is given by BW = (f_0)/(Q). BW = 490 Hz1.92 BW ≈ 255.21 Hz 7a. Draw the circuit diagram of a 3-phase, 4-wire power system feeding a delta-connected load, show all relevant currents in the circuit. (Imagine a diagram with the following elements and connections) Source Side: Three phase lines labeled R, Y, B. A neutral line labeled N. The source could be represented as three voltage sources (e.g., star-connected) providing phase voltages V_RN, V_YN, V_BN and line voltages V_RY, V_YB, V_BR. Load Side (Delta-connected): Three impedances Z_RY, Z_YB, Z_BR connected in a triangle (delta) configuration. Z_RY connected between R and Y lines. Z_YB connected between Y and B lines. Z_BR connected between B and R lines. Currents: Line Currents: I_R, I_Y, I_B flowing from the source lines to the load connection points. Phase Currents: I_RY (flowing through Z_RY), I_YB (flowing through Z_YB), I_BR (flowing through Z_BR). Neutral Current: I_N (flowing in the neutral line). For a delta load, the neutral line is typically not connected to the load directly, so I_N would be zero unless there's an imbalance in the source or other connections. However, if the question implies a 4-wire system feeding* the delta load, it means the 4 wires are available, but only 3 are used for the delta load. Diagram Description: Draw three vertical lines (R, Y, B) representing the phase conductors, and one more vertical line (N) for the neutral. Connect the R, Y, B lines to the corners of a triangle formed by three impedances. Connect Z_RY between R and Y. Connect Z_YB between Y and B. Connect Z_BR between B and R. Show arrows for line currents I_R, I_Y, I_B entering the delta load from the respective phase lines. Show arrows for phase currents I_RY, I_YB, I_BR flowing within the delta impedances. (For a balanced delta load, I_L = sqrt(3) I_P and V_L = V_P). 7b. A star-connected load having the following impedances, Z_r = (20 - j16), Z_y = (16 - j20) and Z_b = (20 - j0) are connected across a 3-phase, 4-wire system having a line voltage of 420V. Given: Z_r = (20 - j16) Z_y = (16 - j20) Z_b = (20 - j0) (purely resistive) Line voltage V_L = 420V 3-phase, 4-wire system. i) Draw the circuit diagram (Imagine a diagram with the following elements and connections) Source Side: Three phase lines labeled R, Y, B. A neutral line labeled N. The source could be represented as three voltage sources (e.g., star-connected) providing phase voltages V_RN, V_YN, V_BN and line voltages V_RY, V_YB, V_BR. Load Side (Star-connected): Three impedances Z_r, Z_y, Z_b connected from each phase line to a common neutral point (N'). Z_r connected between R line and N'. Z_y connected between Y line and N'. Z_b connected between B line and N'. The load neutral point N' is connected to the system neutral line N. Currents and Voltages: Line Currents: I_R, I_Y, I_B flowing from the source lines through the respective impedances. In a star connection, line currents are equal to phase currents. Phase Voltages: V_RN', V_YN', V_BN' across each impedance. Neutral Current: I_N flowing in the neutral line, which will be non-zero because the load is unbalanced (Z_r ≠ Z_y ≠ Z_b). Diagram Description: Draw three vertical lines (R, Y, B) representing the phase conductors, and one more vertical line (N) for the neutral. Draw a central point N' (load neutral). Connect Z_r between R and N'. Connect Z_y between Y and N'. Connect Z_b between B and N'. Connect N' to N. Show arrows for line currents I_R, I_Y, I_B flowing from the phase lines through the impedances towards N'. Show an arrow for neutral current I_N flowing in the neutral line. Show phase voltages V_RN', V_YN', V_BN' across each impedance. (For a star connection, V_P = V_L / sqrt(3) and I_L = I_P). Drop the next question.

A current wave is given by the expression i = 400sin(800t) milli-amperes. Calculate: (i) The frequency (ii)period (iii) the r.m.s value (iv) the time t when I = 250mA

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A current wave is given by the expression i = 400sin(800t) milli-amperes. Calculate: (i) The frequency (ii)period (iii) the r.m.s value (iv) the time t when I = 250mA

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