An e.m.f. of 15 volt is applied in a circuit containing 5 henry inductance and 10 ohm resistance. The ratio of the currents at time t = $\infty$ and at t = 1 second is

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Solution

At $t = \infty$ ,
Resistance offered by inductor is zero,So $I_{1} = \frac{15}{10} = 1.5 A$

At $t = 1 s e c o n d$ ,

Current $(I_{2}) = \frac{V}{R} (1 - e^{- R t / L})$

$\Rightarrow I_{2} = 1.5 (1 - \frac{1}{e^{2}}) = 1.365 A$

So, $I_{1} : I_{2} = 1.5 : 1.365$

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An e.m.f. of 15 volt is applied in a circuit containing 5 henry inductance and 10 ohm resistance. The ratio of the currents at time t = ∞ and at t = 1 second is

At t=∞,Resistance offered by inductor is zero,So I1=1510=1.5AAt t=1second,Current (I2)=VR(1−e−Rt/L)⇒I2=1.5(1−1e2)=1.365ASo, I1:I2=1.5:1.365

An e.m.f. of 15 volt is applied in a circuit containing 5 henry inductance and 10 ohm resistance. The ratio of the currents at time t = $\infty$ and at t = 1 second is

At $t = \infty$ ,
Resistance offered by inductor is zero,So $I_{1} = \frac{15}{10} = 1.5 A$

At $t = 1 s e c o n d$ ,

Current $(I_{2}) = \frac{V}{R} (1 - e^{- R t / L})$

$\Rightarrow I_{2} = 1.5 (1 - \frac{1}{e^{2}}) = 1.365 A$

So, $I_{1} : I_{2} = 1.5 : 1.365$