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An inductor with an inductance of 2.30H and a resistance of 8.00 Ω isconnected to the terminals of a battery with an emf of 6.00 V andnegligible internal resistance.

Find the initial rate of increase ofcurrent in the circuit.

User Page David
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Final answer:

The initial rate of increase of current in an inductor-resistor circuit connected to a 6.00 V battery is calculated using inductive reactance and Ohm's Law, resulting in an initial current increase rate of approximately 2.609 A/s.

Step-by-step explanation:

The initial rate of increase of current in the circuit, when an inductor with an inductance of 2.30H and a resistance of 8.00 Ω is connected to a battery of emf 6.00 V, can be found using the formula for inductive reactance and Ohm's Law. At the moment the circuit is closed, the current is initially zero and the entire battery voltage appears across the inductor.

The voltage across an inductor (V_L) is given by V_L = L*(di/dt), where L is the inductance and di/dt is the rate of change of current. Since the resistance (R) is also present, the total voltage (V) is split between the inductive reactance and the resistance. As per Ohm's Law, V = iR + L*(di/dt). Initially, i = 0, so V = L*(di/dt). Therefore, the initial rate of increase of current di/dt = V/L = 6.00V / 2.30H. The initial current increase rate is thus approximately 2.609 A/s.

User Joarleymoraes
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