Question

A coil of self-inductance 2⋅5H and resistance of $20\Omega $ is connected to a battery of emf 120V having internal resistance of $5\Omega $. Find the current in the circuit in steady state.

Answer 1

Hint: An ideal battery has zero resistance, but a perfectly ideal thing does not exist in the real world. In this question, we have a battery with internal resistance, this means that the battery is based on practical or real-world application. Almost all the batteries we have including the Inverter battery and the smartphone battery all have some internal resistance which is really small.

Complete step-by-step solution:
In this question, we have been given that
Self Inductance of coil (L) = 2.5 H
Resistance (R) = $20\Omega $
Internal Resistance ${{R}_{i}}=5\Omega $
EMF of battery (e) = 120 V
Here, we have two resistance, one is internal resistance and the other one is connected externally
To find the total resistance, we can add them up as the external resistance must be connected in series with internal resistance. We can not connect the external resistance with the internal resistance of the battery in parallel combination.
Therefore, Net Resistance (${{R}_{net}}$) = $20\Omega +5\Omega $
${{R}_{net}}=25\Omega $
Now, the current in the steady state can be given as
$I=\dfrac{emf}{{{R}_{net}}}$
Putting the values in the above equation, we get
$I=\dfrac{120}{25}$
$I=4.8A$
Therefore, we can say that the current in the steady state will be equal to 4.8 Ampere

Note: Self-inductance is defined because the phenomenon during which a change in current in a circuit produces an induced electro-motive-force within the same circuit. Self-inductance is an impact that's noticed when one coil experiences the effect of inductance. Under the consequences of self-inductance and changes in current induce an EMF or electro-motive force therein same wire or coil, producing what's often termed a back-EMF