Figure shows an L-R circuit. When the switch S is closed, the current through resistor R1, R2 and R3 are I1, I2 and I3 respectively. The value of I1, I2 and I3 at t=0 s is,

Question

Figure shows an L-R circuit. When the switch S is closed, the current through resistor R1, R2 and R3 are I1, I2 and I3 respectively. The value of I1, I2 and I3 at t=0 s is, (A) I1=I2=I3=0 (B) I1=(E/R1), I2=I3=0 (C) I1=0, I2=(E/R2), I3=(E/R3) (D) I1=(E/R1), I2=(E/R2+L1), I3=(E/R3+L3)

Tardigrade Admin · Accepted Answer

At t=0, inductor offers infinite resistance So , I1=(E/R1), I2=I3=0

Q. Figure shows an L-R circuit. When the switch $S$ is closed, the current through resistor $R_{1}, R_{2}$ and $R_{3}$ are $I_{1}, I_{2}$ and $I_{3}$ respectively. The value of $I_{1}, I_{2}$ and $I_{3}$ at $t = 0 s$ is,

$I_{1} = I_{2} = I_{3} = 0$

$I_{1} = \frac{E}{R _{1}}, I_{2} = I_{3} = 0$

$I_{1} = 0, I_{2} = \frac{E}{R _{2}}, I_{3} = \frac{E}{R _{3}}$

$I_{1} = \frac{E}{R _{1}}, I_{2} = \frac{E}{R _{2} + L _{1}}, I_{3} = \frac{E}{R _{3} + L _{3}}$

At $t = 0$ , inductor offers infinite resistance
So $, I_{1} = \frac{E}{R _{1}}, I_{2} = I_{3} = 0$

Q. Figure shows an L-R circuit. When the switch S is closed, the current through resistor R1​,R2​ and R3​ are I1​,I2​ and I3​ respectively. The value of I1​,I2​ and I3​ at t=0s is,

I1​=I2​=I3​=0

I1​=R1​E​,I2​=I3​=0

I1​=0,I2​=R2​E​,I3​=R3​E​

I1​=R1​E​,I2​=R2​+L1​E​,I3​=R3​+L3​E​