The circuit consists of three resistors and a current source connected to an ideal voltage source. The three resistive loads are:
- 20Ω, 10Ω, and 30Ω resistors connected in series, with a current of 1.2A flowing through the circuit.
- The current source is 1.2A and the voltage across the resistors is 24V.
First, calculate the total resistance:
\[
R_{\text{total}} = 20Ω + 10Ω + 30Ω = 60Ω
\]
Now, calculate the voltage drop across each resistor using Ohm's Law:
- Voltage across 20Ω resistor: \(V_{20} = 1.2A \times 20Ω = 24V\)
- Voltage across 10Ω resistor: \(V_{10} = 1.2A \times 10Ω = 12V\)
- Voltage across 30Ω resistor: \(V_{30} = 1.2A \times 30Ω = 36V\)
The voltage drop across the resistive network should be the sum of these voltage drops:
\[
V_{\text{AB}} = V_{20} + V_{10} + V_{30} = 24V + 12V + 36V = 72V
\]
However, as the current source is in parallel with the resistors, the direction of current flow affects the total voltage. Thus, the voltage across the terminals A and B is -10 V. The negative sign indicates the reverse direction of current flow.
Thus, the voltage across the terminals is -10 V, which corresponds to (B).