JEE MAIN - Physics (2021 - 16th March Evening Shift - No. 18)
The energy dissipated by a resistor is 10 mJ in 1 s when an electric current of 2 mA flows through it. The resistance is ___________$$\Omega$$. (Round off to the Nearest Integer)
উত্তর
2500
ব্যাখ্যা
Given, energy dissipated by a resistor, H = 10 mJ = 10 $$\times$$ 10$$-$$3 J
Time, t = 1 s
Electric current, I = 2 mA = 2 $$\times$$ 10$$-$$3 A
Resistance, R = ?
According to Joule's law of heating,
H = I2Rt
$$ \Rightarrow R = {H \over {{I^2}T}}$$ ....... (i)
Substituting the given values in Eq. (i), we get
$$R = {{10 \times {{10}^{ - 3}}} \over {{{(2 \times {{10}^{ - 3}})}^2} \times 1}}$$
$$ \Rightarrow R = {{{{10}^{ - 2}}} \over {4 \times {{10}^{ - 6}}}} \Rightarrow R = 0.25 \times {10^4}$$
$$ \Rightarrow R = 2500\,\Omega $$