A. State Kirchhoff's law of voltage and law of current. ii. Using Kirchhoff's law, calculate the current in each branch of the circuit.

$$140 \text{ V} - (I_1+I_2) \times 20 \Omega - I_2 \times 6 \Omega = 0$$ $$140 - 20I_1 - 20I_2 - 6I_2 = 0$$ $$140 - 20I_1 - 26I_2 = 0$$ $$20I_1 + 26I_2 = 140 \quad (1)$$

$$-I_1 \times 5 \Omega - 90 \text{ V} + I_2 \times 6 \Omega = 0$$ $$-5I_1 + 6I_2 = 90 \quad (2)$$

From equation (2), multiply by 4:

$$4 \times (-5I_1 + 6I_2) = 4 \times 90$$ $$-20I_1 + 24I_2 = 360 \quad (3)$$

Add equation (1) and equation (3):

$$(20I_1 + 26I_2) + (-20I_1 + 24I_2) = 140 + 360$$ $$50I_2 = 500$$ $$I_2 = \frac{500}{50} = 10 \text{ A}$$

$$-5I_1 + 6(10) = 90$$ $$-5I_1 + 60 = 90$$ $$-5I_1 = 90 - 60$$ $$-5I_1 = 30$$ $$I_1 = \frac{30}{-5} = -6 \text{ A}$$

• Current in branch AB (through $20 \Omega$ resistor): $I_{AB} = I_1 + I_2 = -6 \text{ A} + 10 \text{ A} = 4 \text{ A}$.
• Current in branch BC (through $5 \Omega$ resistor): $I_{BC} = I_1 = -6 \text{ A}$. A negative sign indicates the current flows in the opposite direction to the assumed $I_1$ (i.e., from C to B).
• Current in branch BE (through $6 \Omega$ resistor): $I_{BE} = I_2 = 10 \text{ A}$.

The currents in each branch are:

• Current through the $20 \Omega$ resistor (branch AB): $\boxed{\text{4 A (from A to B)}}$
• Current through the $5 \Omega$ resistor (branch BC): $\boxed{\text{6 A (from C to B)}}$
• Current through the $6 \Omega$ resistor (branch BE): $\boxed{\text{10 A (from B to E)}}$
• Current through the 140 V source (branch FA): $\boxed{\text{4 A (from F to A)}}$
• Current through the 90 V source (branch CD): $\boxed{\text{6 A (from D to C)}}$

9.b.i: Determine resistor R4

Given: Total voltage $V_{total} = 48 \text{ V}$, $R_1 = 16 \Omega$, $R_2 = 20 \Omega$, $R_3 = 32 \Omega$. Power dissipated by $R_1$, $P_1 = 81 \text{ W}$.

The power dissipated by $R_1$ is given by $P_1 = \frac{V_{parallel}^2}{R_1}$.

$$V_{parallel}^2 = P_1 \times R_1 = 81 \text{ W} \times 16 \Omega = 1296 \text{ V}^2$$ $$V_{parallel} = \sqrt{1296} = 36 \text{ V}$$

Since $R_4$ is in series with the parallel combination, the voltage across $R_4$ is:

$$V_4 = V_{total} - V_{parallel} = 48 \text{ V} - 36 \text{ V} = 12 \text{ V}$$

$$\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$$ $$\frac{1}{R_p} = \frac{1}{16 \Omega} + \frac{1}{20 \Omega} + \frac{1}{32 \Omega}$$

To find a common denominator (LCM of 16, 20, 32 is 160):

$$\frac{1}{R_p} = \frac{10}{160} + \frac{8}{160} + \frac{5}{160} = \frac{10+8+5}{160} = \frac{23}{160} \Omega^{-1}$$ $$R_p = \frac{160}{23} \Omega$$

This current flows through the parallel combination and $R_4$.

$$I_{total} = \frac{V_{parallel}}{R_p} = \frac{36 \text{ V}}{\frac{160}{23} \Omega} = \frac{36 \times 23}{160} \text{ A} = \frac{828}{160} \text{ A} = \frac{207}{40} \text{ A}$$

$$R_4 = \frac{V_4}{I_{total}} = \frac{12 \text{ V}}{\frac{207}{40} \text{ A}} = \frac{12 \times 40}{207} \Omega = \frac{480}{207} \Omega$$

Simplifying the fraction by dividing numerator and denominator by 3:

$$R_4 = \frac{160}{69} \Omega$$

The value of resistor R4 is $\boxed{\frac{160}{69} \Omega \approx 2.32 \Omega}$.

9.b.ii: Determine total resistance of the circuit

The total resistance of the circuit is the sum of the equivalent parallel resistance ($R_p$) and $R_4$.

$$R_{total} = R_p + R_4$$ $$R_{total} = \frac{160}{23} \Omega + \frac{160}{69} \Omega$$

To add these fractions, find a common denominator (69):

$$R_{total} = \frac{160 \times 3}{23 \times 3} \Omega + \frac{160}{69} \Omega = \frac{480}{69} \Omega + \frac{160}{69} \Omega$$ $$R_{total} = \frac{480+160}{69} \Omega = \frac{640}{69} \Omega$$

The total resistance of the circuit is $\boxed{\frac{640}{69} \Omega \approx 9.28 \Omega}$.