Sankalp NEET Full Test-4 Question-29 Solution

Question: 29: Two resistors of resistance, 100 \Omega and 200 \Omega are connected in parallel in an electrical circuit. The ratio of the thermal energy developed in 100 \Omega to that in 200 \Omega in a given time is

(1) 1: 4

(2) 4: 1

(3) 1: 2

(4) 2: 1

Answer: Option (4)

Explanation:

When resistors are connected in parallel, the potential difference across each resistor is the same.

The thermal energy produced in a resistor in time t is given by

H = Pt

where P is the power dissipated in the resistor.

Power in a resistor in terms of potential difference is

P = \frac{V^2}{R}

Since both resistors have the same potential difference, the thermal energy developed is inversely proportional to resistance.

Thus, the ratio of thermal energies is

\frac{H_{100}}{H_{200}} = \frac{V^2/100}{V^2/200} \frac{H_{100}}{H_{200}} = \frac{200}{100} = 2

Therefore, the ratio of thermal energy developed in 100 \Omega

to that in 200 \Omega is

2 : 1

Hence, the correct answer is Option (4).