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State the laws of electrostatics. Two electric charges of +15µC and +1µC are spaced 50 cm apart in air. Calculate the force of attraction between the charges.

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Part 6· 6 steps

  • 6(a)(i): State the laws of electrostatics. The two fundamental laws of electrostatics are:

    • Coulomb's Law: States that the force between two point charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them. The force acts along the line joining the charges.
    • Law of Conservation of Charge: States that the net electric charge of an isolated system remains constant. Charge can neither be created nor destroyed, only transferred.

  • 6(a)(ii): Two electric charges of +15 μC+15\,\mu\text{C} and +10 μC+10\,\mu\text{C} are spaced 30 cm30\,\text{cm} apart in air. Calculate the force of attraction between the charges. Given: Charge q1=+15 μC=15×10−6 Cq_1 = +15\,\mu\text{C} = 15 \times 10^{-6}\,\text{C} Charge q2=+10 μC=10×10−6 Cq_2 = +10\,\mu\text{C} = 10 \times 10^{-6}\,\text{C} Distance r=30 cm=0.3 mr = 30\,\text{cm} = 0.3\,\text{m} Coulomb's constant k=9×109 N⋅m2/C2k = 9 \times 10^9\,\text{N}\cdot\text{m}^2/\text{C}^2

  1. Apply Coulomb's Law formula.

    $$
F = k \frac{q_1 q_2}{r^2}
$$

  2. Substitute the given values.

    $$
F = (9 \times 10^9\,\text{N}\cdot\text{m}^2/\text{C}^2) \frac{(15 \times 10^{-6}\,\text{C})(10 \times 10^{-6}\,\text{C})}{(0.3\,\text{m})^2}
$$
$$
F = (9 \times 10^9) \frac{150 \times 10^{-12}}{0.09}
$$
$$
F = (9 \times 10^9) \frac{1.5 \times 10^{-10}}{0.09}
$$
$$
F = \frac{13.5 \times 10^{-1}}{0.09}
$$
$$
F = \frac{1.35}{0.09}
$$
$$
F = 15\,\text{N}
$$
Since both charges are positive, the force is repulsive, not attractive.

The force between the charges is $\boxed{\text{15 N (repulsive)}}$.

    • 6(b): Calculate the phase currents and draw the vector diagram. Calculate the current in the neutral conductor. This question requires a circuit diagram or specific parameters (e.g., voltage, impedance, connection type) for a three-phase AC circuit, which are not provided in the image. Therefore, this question cannot be solved.

    • 6(c): For the diagram shown above, calculate the charge and voltage associated with each of the changes in the circuit. Show your answers in a tabular form. The circuit shows a 30V DC source connected to a network of capacitors. The capacitors are: C1=11 μFC_1 = 11\,\mu\text{F}, C2=12 μFC_2 = 12\,\mu\text{F}, C3=13 μFC_3 = 13\,\mu\text{F}, C4=14 μFC_4 = 14\,\mu\text{F}, C5=15 μFC_5 = 15\,\mu\text{F}, C6=16 μFC_6 = 16\,\mu\text{F}. The circuit consists of three parallel branches, with each branch containing two capacitors in series. The total voltage across each parallel branch is Vtotal=30 VV_{total} = 30\,\text{V}.

  3. Calculate the equivalent capacitance for each series branch.

    For series capacitors, $\frac{1}{C_{eq}} = \frac{1}{C_a} + \frac{1}{C_b}$.  
Branch 1 ($C_1, C_2$):
$$
\frac{1}{C_{12}} = \frac{1}{11\,\mu\text{F}} + \frac{1}{12\,\mu\text{F}} = \frac{12+11}{11 \times 12}\,\mu\text{F}^{-1} = \frac{23}{132}\,\mu\text{F}^{-1}
$$
$$
C_{12} = \frac{132}{23}\,\mu\text{F} \approx 5.739\,\mu\text{F}
$$
Branch 2 ($C_3, C_4$):
$$
\frac{1}{C_{34}} = \frac{1}{13\,\mu\text{F}} + \frac{1}{14\,\mu\text{F}} = \frac{14+13}{13 \times 14}\,\mu\text{F}^{-1} = \frac{27}{182}\,\mu\text{F}^{-1}
$$
$$
C_{34} = \frac{182}{27}\,\mu\text{F} \approx 6.741\,\mu\text{F}
$$
Branch 3 ($C_5, C_6$):
$$
\frac{1}{C_{56}} = \frac{1}{15\,\mu\text{F}} + \frac{1}{16\,\mu\text{F}} = \frac{16+15}{15 \times 16}\,\mu\text{F}^{-1} = \frac{31}{240}\,\mu\text{F}^{-1}
$$
$$
C_{56} = \frac{240}{31}\,\mu\text{F} \approx 7.742\,\mu\text{F}
$$

  4. Calculate the total charge for each series branch (Q=CeqVtotalQ = C_{eq} V_{total}).

    Charge in Branch 1 ($Q_{12}$):
$$
Q_{12} = C_{12} V_{total} = \frac{132}{23}\,\mu\text{F} \times 30\,\text{V} = \frac{3960}{23}\,\mu\text{C} \approx 172.17\,\mu\text{C}
$$
Charge in Branch 2 ($Q_{34}$):
$$
Q_{34} = C_{34} V_{total} = \frac{182}{27}\,\mu\text{F} \times 30\,\text{V} = \frac{5460}{27}\,\mu\text{C} = \frac{1820}{9}\,\mu\text{C} \approx 202.22\,\mu\text{C}
$$
Charge in Branch 3 ($Q_{56}$):
$$
Q_{56} = C_{56} V_{total} = \frac{240}{31}\,\mu\text{F} \times 30\,\text{V} = \frac{7200}{31}\,\mu\text{C} \approx 232.26\,\mu\text{C}
$$
In a series connection, the charge on each capacitor is the same as the total charge of the branch.

  5. Calculate the voltage across each capacitor (V=Q/CV = Q/C).

    Branch 1:
$$
V_1 = \frac{Q_{12}}{C_1} = \frac{3960/23\,\mu\text{C}}{11\,\mu\text{F}} = \frac{360}{23}\,\text{V} \approx 15.65\,\text{V}
$$
$$
V_2 = \frac{Q_{12}}{C_2} = \frac{3960/23\,\mu\text{C}}{12\,\mu\text{F}} = \frac{330}{23}\,\text{V} \approx 14.35\,\text{V}
$$
(Check: $V_1 + V_2 = \frac{360+330}{23} = \frac{690}{23} = 30\,\text{V}$)

Branch 2:
$$
V_3 = \frac{Q_{34}}{C_3} = \frac{1820/9\,\mu\text{C}}{13\,\mu\text{F}} = \frac{140}{9}\,\text{V} \approx 15.56\,\text{V}
$$
$$
V_4 = \frac{Q_{34}}{C_4} = \frac{1820/9\,\mu\text{C}}{14\,\mu\text{F}} = \frac{130}{9}\,\text{V} \approx 14.44\,\text{V}
$$
(Check: $V_3 + V_4 = \frac{140+130}{9} = \frac{270}{9} = 30\,\text{V}$)

Branch 3:
$$
V_5 = \frac{Q_{56}}{C_5} = \frac{7200/31\,\mu\text{C}}{15\,\mu\text{F}} = \frac{480}{31}\,\text{V} \approx 15.48\,\text{V}
$$
$$
V_6 = \frac{Q_{56}}{C_6} = \frac{7200/31\,\mu\text{C}}{16\,\mu\text{F}} = \frac{450}{31}\,\text{V} \approx 14.52\,\text{V}
$$
(Check: $V_5 + V_6 = \frac{480+450}{31} = \frac{930}{31} = 30\,\text{V}$)

  6. Present the answers in a tabular form.

    | Capacitor | Capacitance ($C$) | Charge ($Q$) | Voltage ($V$) |
| :-------- | :------------------ | :------------- | :------------ |
| $C_1$     | $11\,\mu\text{F}$  | $172.17\,\mu\text{C}$ | $15.65\,\text{V}$ |
| $C_2$     | $12\,\mu\text{F}$  | $172.17\,\mu\text{C}$ | $14.35\,\text{V}$ |
| $C_3$     | $13\,\mu\text{F}$  | $202.22\,\mu\text{C}$ | $15.56\,\text{V}$ |
| $C_4$     | $14\,\mu\text{F}$  | $202.22\,\mu\text{C}$ | $14.44\,\text{V}$ |
| $C_5$     | $15\,\mu\text{F}$  | $232.26\,\mu\text{C}$ | $15.48\,\text{V}$ |
| $C_6$     | $16\,\mu\text{F}$  | $232.26\,\mu\text{C}$ | $14.52\,\text{V}$ |

Part 7· 8 steps

  • 7(a): Give a sketch of a two-winding transformer and label the main parts. A two-winding transformer consists of a primary winding and a secondary winding wound around a common laminated soft iron core.

          +---------------------------------+
      |                                 |
Vp ---+   +-------------------------+   +--- Vs
      |   |                         |   |
      |   |  Laminated Iron Core    |   |
      |   |                         |   |
Np ---+   +-------------------------+   +--- Ns
      |                                 |
      +---------------------------------+
    • Primary Winding (Np): The coil connected to the input AC voltage source (VpV_p).
    • Secondary Winding (Ns): The coil connected to the load, where the output voltage (VsV_s) is induced.
    • Laminated Soft Iron Core: Provides a low reluctance path for the magnetic flux, linking the primary and secondary windings. It is laminated to reduce eddy current losses.

  • 7(b): State the types of losses that occur in a transformer and explain how these losses can be minimized. The main types of losses in a transformer are:

    1. Copper Losses (I²R Losses): These losses occur in the primary and secondary windings due to the resistance of the copper wire. They are proportional to the square of the current.
      • Minimization: Use thicker copper wires (lower resistance) for the windings and design the transformer for optimal current density.
    2. Core Losses (Iron Losses): These losses occur in the transformer core and consist of two components:
      • Hysteresis Loss: Caused by the repeated magnetization and demagnetization of the core material as the AC current flows.
        • Minimization: Use core materials with low hysteresis loop area, such as silicon steel.
      • Eddy Current Loss: Caused by circulating currents induced in the core material due to the changing magnetic flux.
        • Minimization: Use a laminated core (thin sheets of core material insulated from each other) to increase the resistance to eddy current flow.

  • 7(c): A two winding transformer has primary to secondary turn ratio 5:35:3. The input voltage is 400 V400\,\text{V}, primary winding leakage resistance is 3 Ω3\,\Omega, secondary winding resistance is 1 Ω1\,\Omega and the load impedance is 6 Ω6\,\Omega.

    • (i) Draw circuit diagram of the transformer. The equivalent circuit diagram of a transformer referred to the primary side, assuming a purely resistive load:

      V1 ---[R1]-----[Ideal Transformer]-----[R2']-----[ZL']---
              |                      |
              |                      |
              |                      |
              |                      |
              ------------------------

      Where: V1=400 VV_1 = 400\,\text{V} (Input Voltage) R1=3 ΩR_1 = 3\,\Omega (Primary Winding Resistance) R2′R_2' = Secondary Winding Resistance referred to primary ZL′Z_L' = Load Impedance referred to primary Ideal Transformer has turn ratio N1:N2=5:3N_1:N_2 = 5:3.

    • (ii) Calculate the Primary current, secondary current and secondary voltage. Given: Turn ratio N1N2=53\frac{N_1}{N_2} = \frac{5}{3} Input voltage V1=400 VV_1 = 400\,\text{V} Primary resistance R1=3 ΩR_1 = 3\,\Omega Secondary resistance R2=1 ΩR_2 = 1\,\Omega Load impedance ZL=6 ΩZ_L = 6\,\Omega (assumed resistive)

  1. Calculate the secondary resistance and load impedance referred to the primary side.

        $$
    R_2' = R_2 \left(\frac{N_1}{N_2}\right)^2 = 1\,\Omega \times \left(\frac{5}{3}\right)^2 = 1 \times \frac{25}{9} = \frac{25}{9}\,\Omega
    $$
    $$
    Z_L' = Z_L \left(\frac{N_1}{N_2}\right)^2 = 6\,\Omega \times \left(\frac{5}{3}\right)^2 = 6 \times \frac{25}{9} = \frac{150}{9} = \frac{50}{3}\,\Omega
    $$

  2. Calculate the total equivalent impedance referred to the primary side.

        $$
    Z_{eq,primary} = R_1 + R_2' + Z_L'
    $$
    $$
    Z_{eq,primary} = 3 + \frac{25}{9} + \frac{50}{3} = \frac{27}{9} + \frac{25}{9} + \frac{150}{9} = \frac{27+25+150}{9} = \frac{202}{9}\,\Omega
    $$

  3. Calculate the primary current (I1I_1).

        $$
    I_1 = \frac{V_1}{Z_{eq,primary}} = \frac{400\,\text{V}}{202/9\,\Omega} = \frac{400 \times 9}{202} = \frac{3600}{202} = \frac{1800}{101}\,\text{A}
    $$
    $$
    I_1 \approx 17.82\,\text{A}
    $$

  4. Calculate the secondary current (I2I_2) using the current ratio.

        $$
    \frac{I_1}{I_2} = \frac{N_2}{N_1} \implies I_2 = I_1 \frac{N_1}{N_2}
    $$
    $$
    I_2 = \frac{1800}{101}\,\text{A} \times \frac{5}{3} = \frac{600 \times 5}{101} = \frac{3000}{101}\,\text{A}
    $$
    $$
    I_2 \approx 29.70\,\text{A}
    $$

  5. Calculate the secondary voltage (V2V_2).

        $$
    V_2 = I_2 Z_L = \frac{3000}{101}\,\text{A} \times 6\,\Omega = \frac{18000}{101}\,\text{V}
    $$
    $$
    V_2 \approx 178.22\,\text{V}
    $$

    Primary current $I_1 = \boxed{\frac{1800}{101}\,\text{A} \approx 17.82\,\text{A}}$  
    Secondary current $I_2 = \boxed{\frac{3000}{101}\,\text{A} \approx 29.70\,\text{A}}$  
    Secondary voltage $V_2 = \boxed{\frac{18000}{101}\,\text{V} \approx 178.22\,\text{V}}$

*   **(iii) Calculate the input power, output power and the efficiency of the transformer.**

  6. Calculate the input power (PinP_{in}).

    Since the load is resistive, the power factor is 1. Pin=V1I1=400 V×1800101 A=720000101 W P_{in} = V_1 I_1 = 400\,\text{V} \times \frac{1800}{101}\,\text{A} = \frac{720000}{101}\,\text{W} Pin≈7128.71 W P_{in} \approx 7128.71\,\text{W}

  7. Calculate the output power (PoutP_{out}).

        $$
    P_{out} = V_2 I_2 = \frac{18000}{101}\,\text{V} \times \frac{3000}{101}\,\text{A} = \frac{54000000}{10201}\,\text{W}
    $$
    $$
    P_{out} \approx 5293.60\,\text{W}
    $$
    Alternatively, $P_{out} = I_2^2 Z_L = \left(\frac{3000}{101}\right)^2 \times 6 = \frac{9000000}{10201} \times 6 = \frac{54000000}{10201}\,\text{W}$.

  8. Calculate the efficiency (η\eta).

        $$
    \eta = \frac{P_{out}}{P_{in}} \times 100\%
    $$
    $$
    \eta = \frac{54000000/10201\,\text{W}}{720000/101\,\text{W}} \times 100\% = \frac{54000000}{10201} \times \frac{101}{720000} \times 100\%
    $$
    $$
    \eta = \frac{54000000 \times 101}{101 \times 101 \times 720000} \times 100\% = \frac{54000000}{101 \times 720000} \times 100\%
    $$
    $$
    \eta = \frac{540}{101 \times 7.2} \times 100\% = \frac{540}{727.2} \times 100\% \approx 74.25\%
    $$

    Input power
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