Let's solve each question step by step.

Solution for Each Question:

1. Problem 1: Circuit with Parallel Resistors (20 Ω and 30 Ω) and Series Resistor (15 Ω)

Given:

• Resistors of $20 \, \Omega$ and $30 \, \Omega$ in parallel.
• This parallel combination is connected in series with a $15 \, \Omega$ resistor.
• Current through $15 \, \Omega$ resistor is $3 \, \text{A}$.

Solution Steps:

1. Equivalent Resistance of Parallel Combination (20 Ω and 30 Ω): $\frac{1}{R_{\text{parallel}}} = \frac{1}{20} + \frac{1}{30} = \frac{3 + 2}{60} = \frac{5}{60} = \frac{1}{12} \Rightarrow R_{\text{parallel}} = 12 \, \Omega$

2. Total Resistance of the Circuit: The total resistance $R_{\text{total}}$ is the sum of the parallel resistance and the $15 \, \Omega$ resistor in series: $R_{\text{total}} = 12 \, \Omega + 15 \, \Omega = 27 \, \Omega$

3. Voltage of the Whole Circuit: Using Ohm's Law, $V = I \times R$: $V = 3 \, \text{A} \times 27 \, \Omega = 81 \, \text{V}$

4. Current Through 20 Ω and 30 Ω Resistors: Voltage across the parallel section is $81 \, \text{V} - (3 \, \text{A} \times 15 \, \Omega) = 36 \, \text{V}$.

   • Current through $20 \, \Omega$: $I_{20} = \frac{36}{20} = 1.8 \, \text{A}$
   • Current through $30 \, \Omega$: $I_{30} = \frac{36}{30} = 1.2 \, \text{A}$

5. Total Power: $P = V \times I = 81 \, \text{V} \times 3 \, \text{A} = 243 \, \text{W}$

2. Problem 2: Equivalent Resistance Between Terminals A and B

This problem involves finding the equivalent resistance in a more complex circuit configuration.

We'll analyze each series and parallel grouping:

1. First, calculate resistances in series or parallel combinations if any.
2. Combine them step-by-step toward A and B.

This requires more details, and I'll break it down if needed.

3. Problem 3: Current Through the 2 Ω Resistor

Given a circuit with a $20 \, \text{V}$ supply, and resistors of $16 \, \Omega$, $16 \, \Omega$, $32 \, \Omega$, and $2 \, \Omega$ in a specific configuration.

Steps:

1. Combine parallel and series resistances.
2. Find total resistance and current.

4. Problem 4: Equivalent Resistance Between A and B

This network of resistors requires:

1. Simplifying series and parallel parts.
2. Adding them up toward A-B