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Domestic Electric Circuits

In the given ...

Question

In the given circuit find the ratio of $i_{1}$ to $i_{2}$ . Where $i_{1}$ is the initial (at $t = 0)$ current, and $i_{2}$ is steady state (at $t = \infty)$ current through the battery.

1.0

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0.8

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Solution

The correct option is B $0.8$
Inductor is open circuited at $t = 0$ and short circuited at $t = \infty$
Hence, $I_{1} = \frac{10 V}{10 R} = 1$ A
$I_{2} = \frac{10 V}{8 R} = \frac{10}{8}$ A
So, $\frac{I_{1}}{I_{2}} = \frac{1}{\frac{10}{8}} = \frac{8}{10} = 0.8$

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In the given circuit find the ratio of i1 to i2. Where i1 is the initial (at t=0) current, and i2 is steady state (at t=∞) current through the battery.

The correct option is B 0.8Inductor is open circuited at t=0 and short circuited at t=∞Hence, I1=10V10R=1AI2=10V8R=108ASo, I1I2=1108=810=0.8

In the given circuit find the ratio of $i_{1}$ to $i_{2}$ . Where $i_{1}$ is the initial (at $t = 0)$ current, and $i_{2}$ is steady state (at $t = \infty)$ current through the battery.

The correct option is B $0.8$
Inductor is open circuited at $t = 0$ and short circuited at $t = \infty$
Hence, $I_{1} = \frac{10 V}{10 R} = 1$ A
$I_{2} = \frac{10 V}{8 R} = \frac{10}{8}$ A
So, $\frac{I_{1}}{I_{2}} = \frac{1}{\frac{10}{8}} = \frac{8}{10} = 0.8$