In an electrical circuit four identical bulbs are connected in parallel to each other to a battery of 10 V (negligible internal resistance). When all the four bulbs glow, a current of 4 A is recorded. Then the power dissipated in the circuit and the resistance of each bulb are ___________ and ____________, respectively.

Correct option is B

Correct Answer: B 40 W, 10 Ω

Explanation:
The problem involves four identical bulbs connected in parallel, powered by a 10 V battery. The total current drawn by the circuit is 4 A. Let us determine the total power dissipated and the resistance of each bulb.

1. Total Power Dissipated in the Circuit:
   The power dissipated in the circuit is given by the formula:

P=V×I

Substituting V=10 , V and I=4A:

P=10×4=40 W

1. Resistance of Each Bulb:
   Since the bulbs are identical and connected in parallel, the current is evenly divided among the bulbs.
   Current through each bulb:

​$$I_{\text{per bulb}} = \frac{\text{Total Current}}{\text{Number of Bulbs}} = \frac{4}{4} = 1 \, \text{A}$$​​

Using Ohm's Law, the resistance of each bulb can be calculated as:

​$$R = \frac{V}{I}$$​​

Substituting V=10 V and I=1A:

$$R = \frac{10}{1} = 10$$Ω​ 

Thus, each bulb has a resistance of 10 Ω, and the total power dissipated in the circuit is 40 W.

Information Booster:

·         In a parallel circuit, the total current is the sum of the currents through each branch.

·         The voltage across all components in a parallel circuit is the same.

·         Identical resistors in parallel divide the current equally.

·         Power dissipation can be calculated using P=VI, P=I²R, or P=V²/R.

·         The equivalent resistance of a parallel circuit is always less than the smallest individual resistance.