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A long, fine wire is wound into a coil with inductance 5 mH. The coil is connected across the terminals of a battery, and the current is measured a few seconds after the connection is made. The wire is unwound and wound again into a different coil with L = 10 mH. This second coil is connected across the same battery, and the current is measured in the same way. Compared with the current in the first coil, is the current in the second coil. ( Select all of them that applies)A- Twice as largeB- One-Fourth as largeC- UnchargedD- Half as largeE- Four times as large

Respuesta :

Answer:

unchanged

Explanation:

Let the voltage of the battery be V

Inductance L1 = 5 mH

Inductance L2 = 10 mH

consider resistance R of the circuit (wire, battery).

V = I R + L dI/dt

where, I is the current in the circuit and t is the time.

After few seconds of connection being made, the factor dI/dt is negligible. There is no change in the current flowing through the circuit. when inductor was just attached in the circuit, a current

Answer:

Explanation:

The self inductance of the solenoid depends on the number f turns in the coil. As the battery remains same so the current remains same, but the number of turns changed so that the self inductance is changed.

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