Question

If the open circuit voltage of a circuit containing ideal sources and resistors is measured at 10 , while the current through the short circuit across the circuit is 600 , what would be the power absorbed by a 60 resistor placed across the terminals

Answer 1

Answer:

Power absorbed by the 60 ohms resistor = 1.02 W

Explanation:

Open Circuit Voltage, $V_{oc} = 10 V$

Short circuit current, $I_{sc} = 600 mA = 0.6A$

Thevenin Resistance, $R_{th} = V_{oc} /I_{sc}$

$R_{th} = 10/0.6\\R_{th} = 16.66 ohms$

$R_{l} = 60 ohms$

Current flowing through the terminals, $I = V_{oc} /(R_{th} + R_{l} )$

$I = 10/(60 + 16.66)\\I = 10/76.66\\I = 0.13 A$

Power absorbed by the 60 ohms resistor

$P = I^{2} R_{l} \\P = 0.13^{2} * 60\\P = 1.02 W$