In the given circuit, let i1 be the current drawn from battery at time t=0 and i2 be steady current at t=∞, then the ratio (i1/i2) is

Question

In the given circuit, let i1 be the current drawn from battery at time t=0 and i2 be steady current at t=∞, then the ratio (i1/i2) is (A) 1.0 (B) 0.8 (C) 1.2 (D) 1.5

Tardigrade Admin · Accepted Answer

At t=0, inductor behaves as open circuit, so i1=(10/10)=1 A At t=∞, inductor behave as short circuit, so i2=(10/8)=(5/4) A Hence, (i1/i2)=(1/5 / 4) =(4/5)=0.8

Q. In the given circuit, let $i_{1}$ be the current drawn from battery at time $t = 0$ and $i_{2}$ be steady current at $t = \infty$ , then the ratio $\frac{i _{1}}{i _{2}}$ is

1.0

0.8

1.2

1.5

At $t = 0$ , inductor behaves as open circuit, so
$i_{1} = \frac{10}{10} = 1 A$
At $t = \infty$ , inductor behave as short circuit, so
$i_{2} = \frac{10}{8} = \frac{5}{4} A$
Hence, $\frac{i _{1}}{i _{2}} = \frac{1}{5/4}$
$= \frac{4}{5} = 0.8$

Q. In the given circuit, let i1​ be the current drawn from battery at time t=0 and i2​ be steady current at t=∞, then the ratio i2​i1​​ is

1.0

0.8

1.2

1.5

At t=0, inductor behaves as open circuit, so i1​=1010​=1A At t=∞, inductor behave as short circuit, so i2​=810​=45​A Hence, i2​i1​​=5/41​ =54​=0.8

Q. In the given circuit, let $i_{1}$ be the current drawn from battery at time $t = 0$ and $i_{2}$ be steady current at $t = \infty$ , then the ratio $\frac{i _{1}}{i _{2}}$ is

At $t = 0$ , inductor behaves as open circuit, so
$i_{1} = \frac{10}{10} = 1 A$
At $t = \infty$ , inductor behave as short circuit, so
$i_{2} = \frac{10}{8} = \frac{5}{4} A$
Hence, $\frac{i _{1}}{i _{2}} = \frac{1}{5/4}$
$= \frac{4}{5} = 0.8$