Ncert Exercises & Solutions

In the L-C-R circuit, shown in the figure the AC driving voltage is $V = V_{m} sin ωt$ $V = V_{m} \sin ωt$ .

(a) Write down the equation of motion for q(t).

(b) At $t = t_{0}$ $t = t_{0}$ , the voltage source stops and R is short-circuited. Now write down how much energy is stored in each of L and C.

Hint: Use Kirchoff's law.

Step 1: (a) Consider the R-L-C circuit shown in the adjacent diagram.

Given,

V = V_{m} sin ωt

$V = V_{m} \sin ωt$

Let current at any instant be i.

Applying KVL in the given circuit;

i R + L \frac{d i}{d t} + \frac{q}{C} - V_{m} sin ω t = 0 . . . (i)

$i R + L \frac{d i}{d t} + \frac{q}{C} - V_{m} \sin ω t = 0 . . . (i)$

Now, we can write,

i = \frac{d q}{d t} \Rightarrow \frac{d i}{d t} = \frac{d^{2} q}{d t^{2}}

$i = \frac{d q}{d t} \Rightarrow \frac{d i}{d t} = \frac{d^{2} q}{d t^{2}}$

From Eq. (i),

\frac{d q}{d t} R + L \frac{d^{2} q}{d t^{2}} + \frac{q}{C} = V_{m} sin ω t

$\frac{d q}{d t} R + L \frac{d^{2} q}{d t^{2}} + \frac{q}{C} = V_{m} \sin ω t$

\Rightarrow L \frac{d^{2} q}{d t^{2}} + R \frac{d q}{d t} + \frac{q}{C} = V_{m} sin ω t

$\Rightarrow L \frac{d^{2} q}{d t^{2}} + R \frac{d q}{d t} + \frac{q}{C} = V_{m} \sin ω t$

This is the required equation of variation (motion) of charge.

(b) Step 2:

L e t q = q_{m} sin (ω t + ϕ) = - q_{m} cos (ω t + ϕ)

$L e t q = q_{m} \sin (ω t + ϕ) = - q_{m} \cos (ω t + ϕ)$

i = i_{m} sin (ω t + ϕ) = q_{m} ω sin (ω t + ϕ)

$i = i_{m} \sin (ω t + ϕ) = q_{m} ω \sin (ω t + ϕ)$

i_{m} = \frac{V_{m}}{Z} = \frac{V_{m}}{\sqrt{R^{2} + {(X_{C} - X_{L})}^{2}}}

$i_{m} = \frac{V_{m}}{Z} = \frac{V_{m}}{\sqrt{R^{2} + {(X_{C} - X_{L})}^{2}}}$

ϕ = {tan}^{- 1} (\frac{X_{C} - X_{L}}{R})

$ϕ = \tan^{- 1} (\frac{X_{C} - X_{L}}{R})$

Step 3: When R is short-circuited at

t = t_{0}

$t = t_{0}$ , the energy is stored in L and C.

U_{L} = \frac{1}{2} {Li}^{2} = \frac{1}{2} L {[\frac{V_{m}}{\sqrt{{(R^{2} + X_{C} - X_{L})}^{2}}}]}^{2} {sin}^{2} ({ωt}_{0} + ϕ)

$U_{L} = \frac{1}{2} {Li}^{2} = \frac{1}{2} L {[\frac{V_{m}}{\sqrt{{(R^{2} + X_{C} - X_{L})}^{2}}}]}^{2} \sin^{2} ({ωt}_{0} + ϕ)$

and U_{C} = \frac{1}{2} \times \frac{q^{2}}{C} = \frac{1}{2 C} [q^{2} m {cos}^{2} ({ωt}_{0} + ϕ)]

$and U_{C} = \frac{1}{2} \times \frac{q^{2}}{C} = \frac{1}{2 C} [q^{2} m \cos^{2} ({ωt}_{0} + ϕ)]$

$= \frac{1}{2 C} \times {(\frac{i_{m}}{ω})}^{2} \cos^{2} ({ωt}_{0} + φ) [∵ i_{m} = q_{m} ω]$

$= \frac{1}{2 C} {[\frac{V_{m}}{\sqrt{R^{2} + {(X_{C} - X_{L})}^{2}}}]}^{2} \frac{\cos^{2} ({ωt}_{0} + ϕ)}{ω^{2}}$

$= \frac{1}{2 {Cω}^{2}} {[\frac{V_{m}}{\sqrt{R^{2} + {(X_{C} - X_{L})}^{2}}}]}^{2} \cos^{2} ({ωt}_{0} + ϕ)$

(c) Step 4: When R is short-circuited, the circuit becomes an L-C oscillator. The capacitor will go on discharging and all energy will go to L and back and forth. Hene, there is an oscillation of energy from electrostatic to magnetic and magnetic to electrostatic.

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