The battery shown in the figure is ideal- The values are - -10V-R-5- -L-2H- Initially current in the inductor is zero- The current through the battery at t-2s is

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Standard VIII

Physics

Electric Power

The battery s...

Question

The battery shown in the figure is ideal. The values are $ε = 10 V, R = 5 Ω, L = 2 H$ . Initially current in the inductor is zero. The current through the battery at $t = 2 s$ is

12 A

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A

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3 A

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None of these

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Solution

The correct option is A $12 A$
Potential difference across $R$ and $L$ should always equal to $E$ .

$I_{1} = \frac{E}{R} = \frac{10}{5} = 2 A$

and, $L \frac{d I_{2}}{d t} = E$

$\frac{d I_{2}}{d t} = \frac{E}{L}$

$d I_{2} = 5 d t$

$\int_{0}^{I_{2}} d I_{2} = 5 \int_{0}^{2} d t$

$I_{2} - 0 = 10 A$

$∴$ Current through battery at $t = 2 s$

$= 2 A + 10 A = 12 A$

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The battery shown in the figure is ideal. The values are ε=10V,R=5Ω,L=2H. Initially current in the inductor is zero. The current through the battery at t=2s is

The correct option is A 12APotential difference across R and L should always equal to E.I1=ER=105=2Aand, LdI2dt=EdI2dt=ELdI2=5dt∫I20dI2=5∫20dtI2−0=10A∴ Current through battery at t=2s=2A+10A=12A

The battery shown in the figure is ideal. The values are $ε = 10 V, R = 5 Ω, L = 2 H$ . Initially current in the inductor is zero. The current through the battery at $t = 2 s$ is