Let's work through the problems one by one:

Problem 1:

Determine the voltage $V_{ab}$ from the given circuit

The circuit consists of three resistors and two current sources:

• Resistor $R_1 = 6 \, \Omega$
• Resistor $R_2 = 4 \, \Omega$
• Resistor $R_3 = 10 \, \Omega$
• Voltage source $V = 10 \, V$
• Current source $I_1 = 3 \, A$
• Current source $I_2 = 5 \, A$

To find the voltage $V_{ab}$, we can use Kirchhoff's Voltage Law (KVL). Let's solve step-by-step:

• The left side of the circuit with the $6 \, \Omega$ resistor and $3 \, A$ current source is in series, so the voltage across the $6 \, \Omega$ resistor is: $V_{R1} = I_1 \times R_1 = 3 \times 6 = 18 \, V$

• The right side with the $4 \, \Omega$ resistor and $5 \, A$ current source will have a voltage across it of: $V_{R2} = I_2 \times R_2 = 5 \times 4 = 20 \, V$

The total voltage $V_{ab}$ across the circuit is determined by considering the net effect of these two voltages along with the 10 V source in the middle.

Problem 2:

Determine the values of $i$ and $V$ in the second circuit

The circuit contains:

• Resistor $9 \, \Omega$
• Resistor $3 \, \Omega$
• Resistor $12 \, \Omega$
• Voltage source $V = 7 \, V$

To find the current $i$, we can calculate the equivalent resistance of the circuit:

• The $3 \, \Omega$ and $12 \, \Omega$ resistors are in parallel. The equivalent resistance $R_{\text{eq}}$ is given by: $\frac{1}{R_{\text{eq}}} = \frac{1}{3} + \frac{1}{12} = \frac{4}{12} + \frac{1}{12} = \frac{5}{12}$ So, $R_{\text{eq}} = \frac{12}{5} = 2.4 \, \Omega$

• Now, add the $9 \, \Omega$ resistor in series: $R_{\text{total}} = 9 + 2.4 = 11.4 \, \Omega$

• Using Ohm's law, the current $i$ through the circuit is: $i = \frac{V}{R_{\text{total}}} = \frac{7}{11.4} \approx 0.614 \, A$

Problem 3:

Determine the current $i$ in the third circuit

This circuit contains:

• A $20 \, V$ voltage source
• Resistors: $2 \, \Omega$, $6 \, \Omega$, $2 \, \Omega$, $8 \, \Omega$, and $4 \, \Omega$

To solve for the current $i$, we need to reduce the resistances step by step:

• The two $2 \, \Omega$ resistors are in series, so their equivalent resistance is $4 \, \Omega$.
• The $4 \, \Omega$ is in parallel with the $8 \, \Omega$ resistor: $\frac{1}{R_{\text{eq}}} = \frac{1}{4} + \frac{1}{8} = \frac{2}{8} + \frac{1}{8} = \frac{3}{8}$ So, $R_{\text{eq}} = \frac{8}{3} \approx 2.67 \, \Omega$

Now the total resistance is the combination of these with the $6 \, \Omega$ resistor in series: $R_{\text{total}} = 6 + 2.67 = 8.67 \, \Omega$

Using Ohm’s law, the total current is: $i = \frac{V}{R_{\text{total}}} = \frac{20}{8.67} \approx 2.31 \, A$

Would you like further details or have any other questions?

Here are five related questions for further practice:

1. How would the result for $V_{ab}$ change if the current sources were reversed?
2. What would happen to the total current in Problem 2 if one of the resistors was doubled?
3. How do you calculate power dissipated by each resistor in the circuits?
4. Can you explain the difference between series and parallel circuits in terms of total resistance?
5. How would the results change if an additional resistor was added in series to Problem 3's circuit?

Tip: Always double-check whether resistors are in series or parallel as this affects how you calculate total resistance.