A direct current of 4 ~A and an alternating current of peak value 4 ~A flow through resistance of 3\, \Omega and 2\,\Omega respectively. The ratio of heat produced in the two resistances in same interval of time will be :

Alternating Current PYQ — JEE Main 2022 · 27 Jul · Shift 1 · Q58 with Solution

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Correct answer: B

Heat produced in a resistor

In a time interval $t$ , heat produced is $H = I^2 R t$ where $I$ is the effective current through the resistor.
For the direct current case

Given:
- $I_{dc} = 4\,\text{A}$
- $R_1 = 3\,\Omega$
So, $H_{dc} = I_{dc}^2 R_1 t = 4^2 \cdot 3 \cdot t = 16 \cdot 3\, t = 48t$
For the alternating current case

Given peak current: $I_0 = 4\,\text{A}$

RMS current is $I_{rms} = \frac{I_0}{\sqrt{2}} = \frac{4}{\sqrt{2}}$

Resistance: $R_2 = 2\,\Omega$

Hence, $H_{ac} = I_{rms}^2 R_2 t = \left(\frac{4}{\sqrt{2}}\right)^2 \cdot 2 \cdot t$

Now, $\left(\frac{4}{\sqrt{2}}\right)^2 = \frac{16}{2} = 8$

Therefore, $H_{ac} = 8 \cdot 2 \cdot t = 16t$
Find the ratio

$H_{dc} : H_{ac} = 48t : 16t = 3:1$
Option check
- A: $3:2$ ✗
- B: $3:1$ ✓
- C: $3:4$ ✗
- D: $4:3$ ✗

Therefore, the correct answer is B.