Question

An inductor with an inductance of 3.00 H and a resistor with a resistance of 14.0 Ω are connected to the terminals of a battery with an emf of 10.0 V and negligible internal resistance.
Find the initial rate of increase of the current in the circuit.

Answer 1

Answer:

Initial rate of increase of the current in the circuit is 3.33A/s

Explanation:

The total voltage V in the circuit is expressed according to ohm's law which states that the current I flowing through a metallic conductor at constant temperature is directly proportional to the potential difference across its ends.

Since there are two elements (inductor and resistor), the total voltage across them is;

V = VL + VR

VL = L(dI/dt)

VR = IR

V = L(dI/dt) + IR

Note that at the instant the switch is closed, the source voltage will be equal to the voltage on the inductor making the voltage across the resistor to be zero. The equation above then becomes;

V = L(dI/dt)+0

V = L(dI/dt)

dI/dt = V/L

Given V = 10Volts and L = 3.0Henry

dI/dt = 10/3

dI/dt = 3.33A/s

Therefore, the initial rate of increase of the current in the circuit is 3.33A/s

Answer 2

Answer:

3.33 A/s

Explanation:

At the instant that the switch is closed, the voltage across the inductor is equal to the emf of the battery.

So, the initial voltage across the inductor is equal to the Emf of the battery.

And when this happens, there is no flow of current across the resistor, at this initial instant.

Ohm's law, stated for inductors is given as

V = L (dI/dt)

V = voltage across the inductor = 10.0 V

L = inductance of the inductor = 3.00 H

(dI/dt) = Rate of change of current in the circuit.

10 = 3 × (dI/dt)

(dI/dt) = (10/3) = 3.33 A/s

Hope this Helps!!!