- Home
- Standard 12
- Physics
Two resistors of $10\, \Omega$ and $20\, \Omega$ and an ideal inductor of $10\, H$ are connected to a $2\, V$ battery as shown. The key $K$ is inserted at time $t = 0$. The initial $(t = 0)$ and final ($t \rightarrow \infty$ ) currents through battery are
$\frac{1}{{15}} \,A, \frac{1}{{10}} \,A$
$\frac{1}{{10}} \,A, \frac{1}{{15}} \,A$
$\frac{2}{{15}} \,A, \frac{1}{{10}} \,A$
$\frac{1}{{15}} \,A, \frac{2}{{25}} \,A$
Solution
At $t=0,$ i.e., when the key is just pressed, no current exiests inside the inductor. So $10 \Omega$ and $20 \Omega$ resistance are in series and a net resistance of $(10+20)=30 \Omega$ exists across the circuit.
Hence, $I_{1}=\frac{2}{30}=\frac{1}{15} A$
As $t \rightarrow \infty,$ the current in the inductor grows to attain a mximum value, i.e., the entrie current passes through the inductor and no current passes through $10 \Omega$ resistor.
Hence, $I_{2}=\frac{2}{20}=\frac{1}{10} A$
