Question

Light bulbs are typically rated by their power dissipation when operated at a given voltage.

Answer 1

(b) At 24 W and 12 V, the bulbs can operate with the largest current of 2 A.

Answer:

Option B.

Explanation:

Power consumed in light bulbs are measured in terms of voltage and current. As it is known that power dissipated in an electric circuit is directly proportional to the square of current and also to resistance.

Power = I²R

As by Ohm's law, V = IR, then Power = VI

Here, V is the voltage and I is the current flowing through the bulb.

So here the power and voltage is given and we have to determine the largest current. So from the formula it can stated that current flowing through the bulb will be directly proportional to the power dissipated by the circuit and inversely proportional to the voltage at which the circuit is operated.

So for the given options, the current value is as follows:

(a) I = P/V = 3.2/6=0.5333 A

(b) I = P/V = 24/12 = 2 A

(c) I= P/V = 16/18 =0.9 A

(d) I = P/V = 32/24 = 1.3 A

Thus, the option B is having the maximum voltage at which the bulb can be operated. So at 24 W and 12 V, the bulbs can operate with the largest current of 2 A.