Three resistive loads are connected to ideal voltage and current sources as shown in the circuit below. The voltage $V_{AB}$ across the terminals A and B is equal to \(\underline{\hspace{1cm}}\) V.
Three resistive loads are connected to ideal voltage and current sources as shown in the circuit below. The voltage $V_{AB}$ across the terminals A and B is equal to \(\underline{\hspace{1cm}}\) V.
Show Hint
- +10
- -10
- -6
- +6
The Correct Option is B
Solution and Explanation
- 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).
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