A coil of self-inductance 0.5 H and resistance \(3\ \Omega\) is connected to a 12 V battery of negligible internal resistance. When the current has become steady, the battery is removed. At the same instant, the coil is connected across a resistor. The total heat produced in the circuit while the current falls to zero is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 4 J
A coil carrying a current stores energy in its magnetic field. When the current falls to zero, all this energy is given out as heat in the coil and the resistor. A steady current does not change, so it induces no emf. The steady current is then set by the battery and the resistance of the coil.
\( L=0.5\ \mathrm{H} \)
\( V=12\ \mathrm{V} \)
\( R=3\ \Omega \)
The steady current is
\( \begin{aligned} I &= \frac{V}{R} \\ &= \frac{12}{3} \\ &= 4\ \mathrm{A} \end{aligned} \)
The heat produced equals the energy stored in the coil:
\( \begin{aligned} H &= \frac{1}{2}LI^{2} \\ &= \frac{1}{2}\times0.5\times4^{2} \\ &= 4\ \mathrm{J} \end{aligned} \)
NCERT (2026-27):
Self-inductance – “Thus, the energy required to build up the current I is,”
\( \displaystyle W=\frac{1}{2}LI^{2} \)\( \qquad (6.17) \)