Determine: i. the equivalent resistance. ii. the supply current. iii. the current flowing in each resistor. (d) Differentiate between voltage divider and current divider with the aid of diagram.

$$\frac{1}{R_{34}} = \frac{1}{R_3} + \frac{1}{R_4}$$ $$\frac{1}{R_{34}} = \frac{1}{20\, \Omega} + \frac{1}{80\, \Omega}$$ $$\frac{1}{R_{34}} = \frac{4}{80\, \Omega} + \frac{1}{80\, \Omega} = \frac{5}{80\, \Omega}$$ $$R_{34} = \frac{80}{5}\, \Omega = 16\, \Omega$$

$$R_{eq} = R_1 + R_2 + R_{34} + R_5$$ $$R_{eq} = 2.4\, \Omega + 50\, \Omega + 16\, \Omega + 110\, \Omega$$ $$R_{eq} = 178.4\, \Omega$$

The equivalent resistance is $\boxed{\text{178.4 } \Omega}$.

ii. The supply current.

$$I_S = \frac{V_S}{R_{eq}}$$ $$I_S = \frac{100\, \text{V}}{178.4\, \Omega}$$ $$I_S \approx 0.5605\, \text{A}$$

The supply current is $\boxed{\text{0.5605 A}}$.

iii. The current flowing in each resistor.

Since $R_1$, $R_2$, and $R_5$ are in series with the voltage source, the current flowing through them is the total supply current.

$$I_1 = I_2 = I_5 = I_S \approx 0.5605\, \text{A}$$

The voltage across the parallel combination ($R_{34}$) is $V_{34} = I_S \times R_{34}$.

$$V_{34} = 0.5605\, \text{A} \times 16\, \Omega$$ $$V_{34} \approx 8.968\, \text{V}$$

The current through $R_3$ is $I_3 = \frac{V_{34}}{R_3}$.

$$I_3 = \frac{8.968\, \text{V}}{20\, \Omega} \approx 0.4484\, \text{A}$$

The current through $R_4$ is $I_4 = \frac{V_{34}}{R_4}$.

$$I_4 = \frac{8.968\, \text{V}}{80\, \Omega} \approx 0.1121\, \text{A}$$

The currents flowing in each resistor are:
$I_1 = \boxed{\text{0.5605 A}}$
$I_2 = \boxed{\text{0.5605 A}}$
$I_3 = \boxed{\text{0.4484 A}}$
$I_4 = \boxed{\text{0.1121 A}}$
$I_5 = \boxed{\text{0.5605 A}}$

iv. Differentiate between voltage divider and current divider with the aid of diagram.

• Voltage Divider: A circuit configuration, typically a series connection of resistors, used to produce an output voltage that is a fraction of its input voltage. The voltage across each resistor is proportional to its resistance.
  • Diagram (conceptual): Two resistors $R_1$ and $R_2$ connected in series to a voltage source $V_{in}$. The output voltage $V_{out}$ is measured across $R_2$.
  • Formula: $V_{out} = V_{in} \frac{R_2}{R_1 + R_2}$
• Current Divider: A circuit configuration, typically a parallel connection of resistors, used to split the input current into multiple branches. The current through each branch is inversely proportional to its resistance.
  • Diagram (conceptual): Two resistors $R_1$ and $R_2$ connected in parallel to a current source $I_{in}$. The current $I_2$ through $R_2$ is a fraction of $I_{in}$.
  • Formula: $I_2 = I_{in} \frac{R_1}{R_1 + R_2}$

2. (a) Parallel Resistors:

• Given:
  • Resistors: $R_1 = 10\, \Omega$, $R_2 = 20\, \Omega$, $R_3 = 30\, \Omega$ (connected in parallel)
  • Supply Voltage: $V_S = 240\, \text{V}$

i. The equivalent resistance.

$$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$$ $$\frac{1}{R_{eq}} = \frac{1}{10\, \Omega} + \frac{1}{20\, \Omega} + \frac{1}{30\, \Omega}$$ $$\frac{1}{R_{eq}} = \frac{6}{60\, \Omega} + \frac{3}{60\, \Omega} + \frac{2}{60\, \Omega} = \frac{11}{60\, \Omega}$$ $$R_{eq} = \frac{60}{11}\, \Omega \approx 5.4545\, \Omega$$

The equivalent resistance is $\boxed{\text{5.4545 } \Omega}$.

ii. The current flowing through the circuit.

$$I_S = \frac{V_S}{R_{eq}}$$ $$I_S = \frac{240\, \text{V}}{60/11\, \Omega} = \frac{240 \times 11}{60}\, \text{A}$$ $$I_S = 4 \times 11\, \text{A} = 44\, \text{A}$$

The current flowing through the circuit is $\boxed{\text{44 A}}$.

iii. The current passing through each resistor.

In a parallel circuit, the voltage across each resistor is equal to the supply voltage ($V_S = 240\, \text{V}$).
Current through $R_1$:

$$I_1 = \frac{V_S}{R_1} = \frac{240\, \text{V}}{10\, \Omega} = 24\, \text{A}$$

Current through $R_2$:

$$I_2 = \frac{V_S}{R_2} = \frac{240\, \text{V}}{20\, \Omega} = 12\, \text{A}$$

Current through $R_3$:

$$I_3 = \frac{V_S}{R_3} = \frac{240\, \text{V}}{30\, \Omega} = 8\, \text{A}$$

The currents passing through each resistor are:
$I_1 = \boxed{\text{24 A}}$
$I_2 = \boxed{\text{12 A}}$
$I_3 = \boxed{\text{8 A}}$

iv. Power dissipated by the $30\, \Omega$ resistor.

For the $30\, \Omega$ resistor ($R_3$), the voltage across it is $V_S = 240\, \text{V}$.

$$P_3 = \frac{V_S^2}{R_3}$$ $$P_3 = \frac{(240\, \text{V})^2}{30\, \Omega} = \frac{57600\, \text{V}^2}{30\, \Omega}$$ $$P_3 = 1920\, \text{W}$$

The power dissipated by the $30\, \Omega$ resistor is $\boxed{\text{1920 W}}$.

2. (b) Household Lighting:

• Given:
  • Number of lights: 22
  • Power per light: $100\, \text{W}$
  • Operating time: 8 hours/day
  • Supply Voltage: $220\, \text{V}$
  • Cost of electricity: Ksh 14.35 per kWh

i. Calculate the total current when all the lights are in operation.

$$I_{total} = \frac{P_{total}}{V}$$ $$I_{total} = \frac{2200\, \text{W}}{220\, \text{V}}$$ $$I_{total} = 10\, \text{A}$$

The total current is $\boxed{\text{10 A}}$.

ii. Determine the annual lighting bill for the house if 1kWh of electricity costs Ksh 14.35.

$$E_{day} = P_{total} \times \text{time per day}$$ $$E_{day} = 2200\, \text{W} \times 8\, \text{h} = 17600\, \text{Wh}$$

Convert to kWh:

$$E_{day} = \frac{17600}{1000}\, \text{kWh} = 17.6\, \text{kWh}$$

$$E_{year} = E_{day} \times 365\, \text{days/year}$$ $$E_{year} = 17.6\, \text{kWh/day} \times 365\, \text{days} = 6424\, \text{kWh}$$

$$\text{Annual Bill} = E_{year} \times \text{Cost per kWh}$$ $$\text{Annual Bill} = 6424\, \text{kWh} \times \text{Ksh 14.35/kWh}$$ $$\text{Annual Bill} = \text{Ksh 92103.40}$$

The annual lighting bill is $\boxed{\text{Ksh 92103.40}}$.

3. (a) Series Capacitors:

• Given:
  • Capacitors: $C_1 = 6\, \mu\text{F}$, $C_2 = 12\, \mu\text{F}$ (connected in series)
  • Supply Voltage: $V_S = 120\, \text{V}$

$$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}$$ $$\frac{1}{C_{eq}} = \frac{1}{6\, \mu\text{F}} + \frac{1}{12\, \mu\text{F}}$$ $$\frac{1}{C_{eq}} = \frac{2}{12\, \mu\text{F}} + \frac{1}{12\, \mu\text{F}} = \frac{3}{12\, \mu\text{F}} = \frac{1}{4\, \mu\text{F}}$$ $$C_{eq} = 4\, \mu\text{F}$$

In a series capacitor circuit, the charge on each capacitor is the same and equal to the total charge.

$$Q_{total} = C_{eq} \times V_S$$ $$Q_{total} = (4 \times 10^{-6}\, \text{F}) \times 120\, \text{V} = 480 \times 10^{-6}\, \text{C} = 480\, \mu\text{C}$$

So, the charge on each capacitor is:
$Q_1 = \boxed{\text{480 } \mu\text{C}}$
$Q_2 = \boxed{\text{480 } \mu\text{C}}$

$$V_1 = \frac{Q_1}{C_1} = \frac{480\, \mu\text{C}}{6\, \mu\text{F}} = 80\, \text{V}$$ $$V_2 = \frac{Q_2}{C_2} = \frac{480\, \mu\text{C}}{12\, \mu\text{F}} = 40\, \text{V}$$

The voltage across each capacitor is:
$V_1 = \boxed{\text{80 V}}$
$V_2 = \boxed{\text{40 V}}$

$$E = \frac{1}{2} C V^2$$

For $C_1$:

$$E_1 = \frac{1}{2} (6 \times 10^{-6}\, \text{F}) (80\, \text{V})^2 = \frac{1}{2} \times 6 \times 10^{-6} \times 6400\, \text{J}$$ $$E_1 = 19200 \times 10^{-6}\, \text{J} = 0.0192\, \text{J}$$

For $C_2$:

$$E_2 = \frac{1}{2} (12 \times 10^{-6}\, \text{F}) (40\, \text{V})^2 = \frac{1}{2} \times 12 \times 10^{-6} \times 1600\, \text{J}$$ $$E_2 = 9600 \times 10^{-6}\, \text{J} = 0.0096\, \text{J}$$

The energy stored in each capacitor is:
$E_1 = \boxed{\text{0.0192 J}}$
$E_2 = \boxed{\text{0.0096 J}}$

3. (b) RC Circuit (Charging Capacitor):

• Given:
  • Capacitor: $C = 10\, \mu\text{F} = 10 \times 10^{-6}\, \text{F}$
  • Resistor: $R = 1\, \text{k}\Omega = 1 \times 10^3\, \Omega$
  • Supply Voltage: $V_S = 100\, \text{V}$

i. The time constant.

$$\tau = RC$$ $$\tau = (1 \times 10^3\, \Omega) \times (10 \times 10^{-6}\, \text{F})$$ $$\tau = 10 \times 10^{-3}\, \text{s} = 0.01\, \text{s}$$

The time constant is $\boxed{\text{0.01 s}}$.

ii. The initial charging current.

At the instant the switch is closed ($t=0$), the capacitor acts as a short circuit, so the current is limited only by the resistor.

$$I_0 = \frac{V_S}{R}$$ $$I_0 = \frac{100\, \text{V}}{1 \times 10^3\, \Omega}$$ $$I_0 = 0.1\, \text{A}$$

The initial charging current is $\boxed{\text{0.1 A}}$.

iii. The voltage on the capacitor after $10\, \text{ms}$.

The voltage across a charging capacitor is given by $V_C(t) = V_S (1 - e^{-t/\tau})$.
Given $t = 10\, \text{ms} = 0.01\, \text{s}$ and $\tau = 0.01\, \text{s}$.

$$V_C(0.01\, \text{s}) = 100\, \text{V} (1 - e^{-0.01\, \text{s}/0.01\, \text{s}})$$ $$V_C(0.01\, \text{s}) = 100\, \text{V} (1 - e^{-1})$$ $$V_C(0.01\, \text{s}) = 100\, \text{V} (1 - 0.36788)$$ $$V_C(0.01\, \text{s}) = 100\, \text{V} (0.63212)$$ $$V_C(0.01\, \text{s}) \approx 63.21\, \text{V}$$

The voltage on the capacitor after $10\, \text{ms}$ is $\boxed{\text{63.21 V}}$.

iv. The time taken for the capacitor to charge to 90% of its final value.

The final voltage is $V_S = 100\, \text{V}$. So, 90% of the final value is $0.9 \times 100\, \text{V} = 90\, \text{V}$.

$$V_C(t) = V_S (1 - e^{-t/\tau})$$ $$90\, \text{V} = 100\, \text{V} (1 - e^{-t/\tau})$$

$$0.9 = 1 - e^{-t/\tau}$$ $$e^{-t/\tau} = 1 - 0.9$$ $$e^{-t/\tau} = 0.1$$

Take the natural logarithm of both sides:

$$-t/\tau = \ln(0.1)$$ $$-t/\tau \approx -2.302585$$ $$t = \tau \times 2.302585$$

Using $\tau = 0.01\, \text{s}$:

$$t = 0.01\, \text{s} \times 2.302585$$ $$t \approx 0.023026\, \text{s}$$

The time taken is $\boxed{\text{0.023026 s}}$.

2 zimekamilika. Zimebaki 3 leo — tuma inayofuata.