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A 680 Ω load resistor, RL, is connected across a constant current source of 1.2 A. The internal source resistance, RS, is 12 kΩ. The load current, RL, is
Answer & Solution
Correct Answer:
Option
D
The load current is determined by analyzing the circuit and applying Ohm's law and the voltage-divider principle.
Given:
Load resistance, RL = 680 Ω
Internal source resistance, RS = 12 kΩ = 12,000 Ω
Constant current source, IS = 1.2 A
Step 1: Determine the total voltage of the circuit
The total voltage generated by the current source is:
Vtotal = IS × RS = 1.2 A × 12,000 Ω = 14,400 V
Step 2: Calculate the equivalent resistance
The load resistor and internal source resistance form a voltage divider. The voltage across RL is calculated as:
VL = Vtotal × (RL / (RL + RS))
VL = 14,400 × (680 / (680 + 12,000))
VL = 14,400 × (680 / 12,680) = 774 V
Step 3: Find the load current
The load current is:
IL = VL / RL = 774 / 680 = 1.14 A
Given:
Load resistance, RL = 680 Ω
Internal source resistance, RS = 12 kΩ = 12,000 Ω
Constant current source, IS = 1.2 A
Step 1: Determine the total voltage of the circuit
The total voltage generated by the current source is:
Vtotal = IS × RS = 1.2 A × 12,000 Ω = 14,400 V
Step 2: Calculate the equivalent resistance
The load resistor and internal source resistance form a voltage divider. The voltage across RL is calculated as:
VL = Vtotal × (RL / (RL + RS))
VL = 14,400 × (680 / (680 + 12,000))
VL = 14,400 × (680 / 12,680) = 774 V
Step 3: Find the load current
The load current is:
IL = VL / RL = 774 / 680 = 1.14 A
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LoginIs= source current
Rs= source resistance
Rl= load resistance
IL = (Is*Rs)/(Rs + Rl)