Final answer:
The power dissipated in the lamp with a resistance of 2.1Ω connected to a 4 V battery is approximately 7.6 watts.
Step-by-step explanation:
The power dissipated in a lamp when connected to a battery with specified open circuit voltage and short circuit current can be calculated using Ohm's Law and the power formula P = IV, where P is power, I is current, and V is voltage. Given that the lamp has a resistance R=2.1Ω, and the battery has an open circuit voltage of Voc=4 V, we must first calculate the current flowing through the lamp using Ohm's Law, which states I = V/R.
The current through the lamp is I = Voc/R = 4 V / 2.1Ω ≈ 1.90476 A. The power dissipated in the lamp can then be found by using the formula P = I²R, yielding P = (1.90476 A)² * 2.1Ω ≈ 7.6 W. Thus, the power dissipated in the lamp is approximately 7.6 watts.