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Flux as the Flow

In the circui...

Question

In the circuit shown emf $E = 10 V,$ Resistance $R = 1 Ω$ and coil inductances $L_{1} = 1 H a n d L_{2} = 2 H$ are known. The internal resistance of the source and the coil are negligible.
The steady state currents in the coils $L_{1} a n d L_{2}$ after switch S was closed:

i_{1} = \frac{10}{3} A, i_{2} = \frac{20}{3} A

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i_{1} = i_{2} = \frac{10}{3} A

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i_{1} = \frac{20}{3} A, i_{2} = \frac{10}{3} A

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i_{1} = i_{2} \frac{20}{3} A

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Solution

The correct option is D $i_{1} = \frac{20}{3} A, i_{2} = \frac{10}{3} A$
$i_{1} \times 1 = i_{2} \times 2$
and $i_{1} + i_{2} = \frac{10}{1} = 10$
$3 (i_{2}) = 10$
$i_{2} = \frac{10}{3}$
$i_{1} = \frac{20}{3}$

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In the circuit shown emf E=10V, Resistance R=1Ω and coil inductances L1=1H and L2=2H are known. The internal resistance of the source and the coil are negligible. The steady state currents in the coils L1 and L2 after switch S was closed:

The correct option is D i1=203A,i2=103Ai1×1=i2×2and i1+i2=101=103(i2)=10i2=103i1=203

In the circuit shown emf $E = 10 V,$ Resistance $R = 1 Ω$ and coil inductances $L_{1} = 1 H a n d L_{2} = 2 H$ are known. The internal resistance of the source and the coil are negligible.
The steady state currents in the coils $L_{1} a n d L_{2}$ after switch S was closed:

The correct option is D $i_{1} = \frac{20}{3} A, i_{2} = \frac{10}{3} A$
$i_{1} \times 1 = i_{2} \times 2$
and $i_{1} + i_{2} = \frac{10}{1} = 10$
$3 (i_{2}) = 10$
$i_{2} = \frac{10}{3}$
$i_{1} = \frac{20}{3}$