# The Voltage and Current of a Discharged Inductor at the Instant the Switch is Closed

The figure shows an LR circuit with a switch and a 240-volt battery. At the instant the switch is closed the current in the circuit and the potential difference between points a and b, Vab, are

Choices :

a.     0 A,    0 V
b.     0 A, -240 V
c.     0 A, +240 V
d. 0.024 A,    0 V
e. 0.024 A, +240 V


Answer : c. 0 A, +240 V

My Questions

• When the battery discharging, the voltage will drop continuously, the voltage across a and b at the start instant should be equal to the original voltage of the battery 240V. However, how can we determine whether Vab is +240V or -240V?

• Moreover, if the potential different at the start instant is 240V, why should the current equals 0A?

• There will be zero current as the inductance needs time to load with $L\frac{\mathrm{d}i}{\mathrm{d}t}$, in other words the coil needs to fill up. Create a loop and write an equation to determine the sign of the voltage(most straightforward method). Jun 20 '14 at 10:15

the voltage through a and b at the start instant should be equal to the original voltage of the battery 240V.

To be sure, it's the voltage across a and b, not through - voltage is across, current is through.

However, how can we determine whether Vab is +240V or -240V?

Kirchoff's voltage law. The KVL equation for this series circuit, assuming the switch is closed, can be written as:

$$V_{ab} = 240V - i\cdot10\mathrm k \Omega$$

and, initially, $i = 0$.

if the potential different at the start instant is 240V, why should the current equals 0A?

(1) To satisfy the above KVL equation.

(2) we also have $V_{ab} = L \frac{di}{dt} = 2.5mH \frac{di}{dt}$.

So the voltage across the inductor is proportional to the rate of change of current, not the instantaneous current.

When the switch is closed, the instantaneous current is initially zero since the current must be continuous (else the derivative in the above equation doesn't exist).

However, $\frac{di}{dt}$ is not continuous; the rate of change 'jumps' from zero, before the switch is closed, to $\frac{240V}{2.5mH}$ the instant the switch is closed.