In the circuit shown below, i1 and i2 are the steady-state values of the current through L1 and L2 respectively, then i1 is

Question

In the circuit shown below, i1 and i2 are the steady-state values of the current through L1 and L2 respectively, then i1 is <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-cobh7vq2ywxz.png" /> (A) i1 =(E L2/R (L1 + L2)) (B) i1 =(E L1/R (L1 + L2)) (C) i1 =(E L2/R √L1 L2) (D) i1 =(E L1/R √L1 L2)

Tardigrade Admin · Accepted Answer

Since the potential difference across the inductors is same, we get

L_{1} \frac{d i _{1}}{d t} = L_{2} \frac{d i _{2}}{d t}

L_{1} d i_{1} = L_{2} d i_{2}

L_{1} Δ i_{1} = L_{2} Δ i_{2}

L_{1} i_{1} = L_{2} i_{2}

(Since the initial value of current through both the inductors was zero)

\frac{i _{1}}{i _{2}} = \frac{L _{2}}{L _{1}}

The steady current

i

passing through

R

is

i = \frac{E}{R}

i_{1} = (\frac{L _{2}}{L _{1} + L _{2}}) i = \frac{E L _{2}}{R ( L _{1} + L _{2} )}

Q. In the circuit shown below, $i_{1}$ and $i_{2}$ are the steady-state values of the current through $L_{1}$ and $L_{2}$ respectively, then $i_{1}$ is

$i_{1 =} \frac{E L _{2}}{R ( L _{1} + L _{2} )}$

$i_{1 =} \frac{E L _{1}}{R ( L _{1} + L _{2} )}$

$i_{1 =} \frac{E L _{2}}{R L _{1} L _{2}}$

$i_{1 =} \frac{E L _{1}}{R L _{1} L _{2}}$

Q. In the circuit shown below, i1​ and i2​ are the steady-state values of the current through L1​ and L2​ respectively, then i1​ is

i1=​R(L1​+L2​)EL2​​

i1=​R(L1​+L2​)EL1​​

i1=​RL1​L2​​EL2​​

i1=​RL1​L2​​EL1​​

Q. In the circuit shown below, $i_{1}$ and $i_{2}$ are the steady-state values of the current through $L_{1}$ and $L_{2}$ respectively, then $i_{1}$ is

$i_{1 =} \frac{E L _{2}}{R ( L _{1} + L _{2} )}$

$i_{1 =} \frac{E L _{1}}{R ( L _{1} + L _{2} )}$

$i_{1 =} \frac{E L _{2}}{R L _{1} L _{2}}$

$i_{1 =} \frac{E L _{1}}{R L _{1} L _{2}}$