Question

A current of 2 mA was passed through an unknown resistor which dissipated a power of 4.4 W. Dissipated power when an ideal power supply of 11 V is connected across it is

A. 11×10⁻⁵W
B. 11×10⁻³W
C. 11×10⁻⁴W
D. 11×10⁵W

Answer 1

Using the formula P = V²/R, the power dissipation of a resistor when an ideal power supply of 11 V is applied across it is found to be 110µW, much lower than the provided incorrect answer of 11×10⁵W.

Step-by-step explanation:

To determine the power dissipated when a resistor has a voltage of 11 V applied across it, we use the formula P = IV, where P is power in watts (W), I is current in amperes (A), and V is voltage in volts (V). Given that the power dissipated at a 2 mA (0.002 A) current is 4.4 W, we can find the resistance using the power formula P = I²R, which rearranges to R = P/I². From the given information, we have:

R = 4.4 W / (0.002 A)²

R = 4.4 W / 0.000004 A² = 1,100,000 Ω (or 1.1 MΩ)

Now, applying the 11 V across this resistor, the power dissipated would be calculated using P = V²/R:

P = (11 V)² / 1,100,000 Ω

P = 121 V² / 1,100,000 Ω

P = 0.00011 W or 110µW

Therefore, the power dissipated when an ideal power supply of 11 V is connected across the resistor is 110µW, which is substantially lower than the initial 4.4 W with a 2 mA current. We do not reach the disproportionate value of 11×10⁵W as the question seems to erroneously suggest.