Answer :
The inductor voltage is initially 6 volts when the switch is closed.
When the switch in an RL circuit is closed, the inductor opposes the sudden change in current by inducing a voltage across its terminals. This is due to the property of self-inductance. The induced voltage follows Lenz's law, which states that the direction of the induced voltage opposes the change in current that produced it.
At the instant the switch is closed, the current through the circuit is initially zero. Since the current is changing from zero to a non-zero value, the inductor induces a voltage that tries to counteract this change. This induced voltage is given by the equation:
V = L * di/dt
Where:
V is the induced voltage,
L is the inductance of the inductor,
di/dt is the rate of change of current with respect to time.
However, at the instant the switch is closed, the rate of change of current (di/dt) is at its maximum value, which causes the induced voltage to be at its maximum as well. In this case, since the current is increasing from zero to a positive value, the induced voltage will be in the same direction as the battery voltage.
Therefore, the inductor voltage is initially equal to the battery voltage, which is 6 volts, when the switch is closed.
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Final Answer:
The inductor voltage at the instant the switch is closed is 6 volts.
Explanation:
When an inductor, a 1 kΩ resistor, and a switch are connected in series across a 6 V battery, the inductor resists changes in current flow due to its inherent property of inductance. As the switch is closed, current begins to flow through the circuit.
However, at the instant of closure, the inductor resists the sudden change in current and consequently tries to maintain the status quo. This results in a transient situation where the inductor voltage is momentarily zero.
This phenomenon is governed by the equation \(V_L = -L \frac{di}{dt}\), where \(V_L\) is the inductor voltage, \(L\) is the inductance, \(di\) is the change in current, and \(dt\) is the change in time.
Since the switch closure results in an instantaneous change in current (\(di\) becomes very large) and the time interval (\(dt\)) approaches zero, the inductor voltage (\(V_L\)) at that exact moment is effectively zero.
As time progresses, the inductor's behavior changes, and its voltage will eventually reach a steady-state value as the current stabilizes.
But at the moment the switch is closed, the inductor voltage is indeed zero due to the sudden change in current and the associated transient behavior.
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