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51.

In an `LR` circuit as shows in Fig.when the swtich is closed, how much time will it take for the current to grow to a value `n` times the maximum value of current (where `n lt 1)`?

Answer» We know that `I = (epsilon)/(R ) (1 - e^(-t//tau))`
`I = n(epsilon)/(R )` (given)
`n = (epsilon)/(R ) = (epsilon)/(R ) (1 - e^(-t//tau))`
or `e^(-t//tau) = 1 -n`
or `t = tau` In `((1)/(1 - n))`
Discussion
52.

A simple `LR` circuit is connected to a battery at time `t = 0`. The energy stored in the inductor reaches half its maximum value at timeA. `(R )/(L)` In `[(sqrt(2))/(sqrt(2) - 1)]`B. `(L )/(R)` In `[(sqrt(2)-1)/(sqrt(2))]`C. `(L )/(R)` In `[(sqrt(2))/(sqrt(2) - 1)]`D. `(R)/(L)` In `[(sqrt(2)-1)/(sqrt(2))]`

Answer» Correct Answer - C
`U_(max) = (1)/(2) LI_(0)^(2)`, `U = (U_(max))/(2)`
`rArr (1)/(2) LI^(2) = (1)/(2) [(1)/(2) LI_(0)^(2)] rArr I_(0)^(2)[1 - e^(-t//tau)]^(2) = (I_(0)^(2))/(2)`
`rArr e^(-t//tau) = 1 - (1)/(sqrt(2)) = (sqrt(2) - 1)/(sqrt(2))`
`rArr -(t)/(tau) = In ((sqrt(2) - 1)/(sqrt(2))) rArr t = tau In ((sqrt(2))/(sqrt(2) - 1))`
Discussion
53.

Two resistors of `10 Omega` and `20 Omega` and an ideal inductor of `10 H` are connected to a `2 V` battery as shows in Fig. key `K` is inserted at time `t = 0`. The initial `(t = 0)` and final `(t rarr oo)` currents through the battery are A. `(1)/(15) A`,`(1)/(10) A`B. `(1)/(10) A`,`(1)/(15) A`C. `(2)/(15) A`,`(1)/(10) A`D. `(1)/(15) A`,`(2)/(25) A`

Answer» Correct Answer - A
At `t = 0`,i.e., when the key is just pressed, no current exiests inside the inductor. So `10 Omega` and `20 Omega` resistance are in series and a net resistance of `(10 + 20) = 30 Omega` exists across the circuit.
Hence, `I_(1) = (2)/(30) = (1)/(15) A`
As `t rarr oo`, the current in the inductor grows to attain a mximum value, i.e., the entrie current passes through the inductor and no current passes through `10 Omega` resistor.
Hence, `I_(2) = (2)/(20) = (1)/(10) A`
Discussion
54.

The potential difference across a `2-H` inductor as a funtion of time is shows in Fig. At `t = 0`, current is zero. Choose the corrent stetement. A. Current at `t = 2 s` is `5 A`B. Current at `t = 2 s` is `10 A`C. Current versus time graph across the inductor will be Fig. D. Current versus time graph across the inductor will be Fig.

Answer» Correct Answer - A::C
`V = L(dI)/(dt) rArr underset(0) overset(1) intdI = (1)/(L) underset(0) overset(1)int Vdt`
`rArr I = (1)/(L)` [Area under `nu -t` graph from `t = 0` to `t = t`]
At `t = 2 s`, `I = (1)/(2) [(1)/(2) xx 2xx 10] = 5 A`
For `t = 0` to `2 s, V = 5t`
`I = (1)/(L) underset(0) overset(t) int 5t dt = (1)/(2) [(5t^(2))/(2)]_(0)^(t) = (5t^(2))/(4)`
For `t = 2` to `4 s, V = - 5t + 20`
`I = (1)/(2) underset(0) overset(t)int (-5t + 20)dt rArr I = (-5t^(2))/(4) + 10t`
Hence the correct graph is (c )`.
Discussion
55.

Two parallel resistanceless rails are connected by an inductor of inductance `L` at one end as shows in Fig. A magnetic field `B` exists in the space which is perendicular to the plane of the rails. Now a conductor of length `l` and mass `m` is placed transverse on the rail and given an inpulse `J` toward the rightward direction. Then choose the correct option (S). A. Velocity of the conductor is half of the initial velocity after a displaacement of the conductor `d = sqrt((3J^(2)L)/(4B^(2)l^(2)m))`B. Current flowing through the inductor at the instant when velocity of the conductor is half of the initial velocity is `i = sqrt((3J^(2))/(4Lm))`C. Velocity of the conductor is half of the initial velocity after a displacement of the conductor `d = sqrt((3J^(2)L)/(B^(2)l^(2)m))`D. Current flowing through the inductor at the instant when velocity of the conductor is half of the initial velocity is `i = sqrt((3j^(2))/(mL))`

Answer» Correct Answer - A::B
`L(di)/(dt) = B nu l rArr int di = (Bl)/(L) int nu dt rArr I = (Bl)/(L) x` (i)
`F = ma rArr - iBl = m nu (d nu)/(dx)`
`rArr -(B^(2)l^(2)x)/(L) = mnu (d nu)/(dx) rArr -(B^(2)l^(2))/(L) underset(0) overset(d) int x dx = underset(nu _(0)) overset(nu_(0)//2)int nu dnu`
`rArr -(B^(2)l^(2)d^(2))/(L) = (- 3 nu_(0)^(2))/(8), nu_(0) = (J)/(m)`
`rArr d = sqrt((3J^(2)L)/(4B^(2)l^(2)m))`
Put `x = d` in (i), `i = (Bl)/(L) sqrt((3J^(2)L)/(4B^(2)l^(2)m)) = sqrt((3J^(2))/(4Lm))`
Discussion