A coil of self inductance 7.5 mH and resistance of 30 $\Omega$ is placed across the terminals of a 12-V battery of negligible internal resistance.The current in this circuit after 50 $\mu$s is approximately

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The current in the circuit is given by:$$I = \frac{V}{R} * (1 - e^(-Rt/L))$$where:* V is the voltage of the battery (12 V)* R is the resistance of the circuit (30 Ω)* t is the time (50 μs = 50 * 10^-6 s)* L is the inductance of the coil (7.5 mH = 7.5 * 10^-3 H)Plugging in the values, we get:$$I = \frac{12}{30} * (1 - e^(-30 * 50 * 10^-6 / 7.5 * 10^-3)) = 0.073 A = 73 mA$$Therefore, the current in the circuit after 50 μs is approximately 73 mA.

A Coil of Self Inductance 7 Ω\OmegaΩ Is Placed Across the Terminals of a 12-V Battery of 30

A Coil of Self Inductance 7 $\Omega$ Is Placed Across the Terminals of a 12-V Battery of 30