Question

An AC source is connected to an inductance of 100 mH, a capacitance of 100 μF and a resistance of 120 Ω as shown in the figure. The time in which the resistance having a thermal capacity 2 J/°C will get heated by 16°C is ___ s.

Fig. Circuit

Answer 1

The correct answer is 15
Given : L = 100 × 10–3H
C = 100 × 10–6 F
R = 120 Ω
ω L = 10 Ω
$\frac{1}{ωC} = \frac{1}{10^4 × 10^{-6}} = 100Ω$
$⇒ X_C – X_L = 90 Ω$
$⇒ Z = \sqrt{90² + 120²} = 150Ω$
$⇒ I_{rms} = \frac{20}{150} = \frac{2}{15}A$
For heat resistance by 16°C heat required = 32 J
$⇒ ( \frac{2}{15} )² × 120 × t = 32$

$t = \frac{32 ×15 × 15}{4 × 120 }$
= 15
∴ Resistance will get heated by 16°C is 15s