An alternating $L - C - R$ circuit is shown in figure, then match the column : (Here symbols have their usual meaning)

Column-I Column-II

$(A) .$ If $ω L_{1} - \frac{1}{C_{1} ω} = R_{1}$ , and $ω L_{2} - \frac{1}{ω C_{2}} = R_{2}$ $(1) .$ $I_{1}$ and $I_{2}$ are in same Phase.

$(B) .$ If $ω L_{1} - \frac{1}{C_{1} ω} = R_{1}$ , and $\frac{1}{ω C_{2}} - ω L_{2} = R_{2}$ $(2) .$ $I = I_{1} + I_{2}$ where , $I_{1}, I_{2}, I_{3}$ are RMS value of currents.

$(C) .$ If capacitor $C_{1}$ and inductor $L_{2}$ are removed from circuit and $ω L_{1} = R_{1}; \frac{1}{ω C_{2}} = R_{2}$ $(3) .$ Magnitude of phase difference between $I_{1}$ and $I_{2}$ is $\frac{π}{2}$

$(D) .$ If capacitors $C_{1}$ and $C_{2}$ are both removed from the circuit and $ω L_{1} = R_{1}; ω L_{2} = R_{2}$ $(4) .$ $I = \sqrt{{(I_{1})}_{1}^{2} + {(I_{2})}_{2}^{2}}$ where, $I_{1}, I_{2}, I_{3}$ are RMS value of currents.

$(5) .$ $I_{1}$ is lags and $I_{2}$ is leads from source voltage.


$(A) .$ If $ω L_{1} - \frac{1}{C_{1} ω} = R_{1}$ , and $ω L_{2} - \frac{1}{ω C_{2}} = R_{2}$	$(1) .$ $I_{1}$ and $I_{2}$ are in same Phase.
$(B) .$ If $ω L_{1} - \frac{1}{C_{1} ω} = R_{1}$ , and $\frac{1}{ω C_{2}} - ω L_{2} = R_{2}$	$(2) .$ $I = I_{1} + I_{2}$ where , $I_{1}, I_{2}, I_{3}$ are RMS value of currents.
$(C) .$ If capacitor $C_{1}$ and inductor $L_{2}$ are removed from circuit and $ω L_{1} = R_{1}; \frac{1}{ω C_{2}} = R_{2}$	$(3) .$ Magnitude of phase difference between $I_{1}$ and $I_{2}$ is $\frac{π}{2}$
$(D) .$ If capacitors $C_{1}$ and $C_{2}$ are both removed from the circuit and $ω L_{1} = R_{1}; ω L_{2} = R_{2}$	$(4) .$ $I = \sqrt{{(I_{1})}_{1}^{2} + {(I_{2})}_{2}^{2}}$ where, $I_{1}, I_{2}, I_{3}$ are RMS value of currents.
$(5) .$ $I_{1}$ is lags and $I_{2}$ is leads from source voltage.

(A) \to (1, 2); (B) \to (3, 4, 5); (C) \to (3, 4, 5); (D) \to (1, 2)

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(A) \to (1, 4); (B) \to (3, 5); (C) \to (3, 5); (D) \to (1, 2)

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(A) \to (1, 4); (B) \to (3, 4); (C) \to (3, 4); (D) \to (1, 4)

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(A) \to (1, 2); (B) \to (3, 4, 5); (C) \to (3, 4, 5); (D) \to (1, 4)

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Solution

The correct option is A $(A) \to (1, 2); (B) \to (3, 4, 5); (C) \to (3, 4, 5); (D) \to (1, 2)$
From $A$ :

It is clear that, $X_{L 1} - X_{C 1} = R_{1}$ and $X_{L 2} - X_{C 2} = R_{2}$

Thus , Phase angle of first $LCR$ branch is $ϕ_{1} = {tan}^{- 1} (\frac{X_{C 1} - X_{L 1}}{R_{1}})$

$\Rightarrow ϕ_{1} = {tan}^{- 1} (- 1) = - \frac{π}{4}$

Phase angle of Second $LCR$ branch is $ϕ_{2} = {tan}^{- 1} (\frac{X_{C 2} - X_{L 2}}{R_{2}})$

$\Rightarrow ϕ_{2} = {tan}^{- 1} (- 1) = - \frac{π}{4}$

Thus, value of current in both braches are in same phase.

Since, the phase difference is zero

Using, $I = \sqrt{{(I_{1})}_{1}^{2} + {(I_{1})}_{1}^{2} + 2 I_{1} I_{2} cos ϕ}$

We can say that, $I = I_{1} + I_{2}$

From $B$ :

Phase angle of first $LCR$ branch is $ϕ_{1} = {tan}^{- 1} (\frac{X_{C 1} - X_{L 1}}{R_{1}})$

$\Rightarrow ϕ_{1} = {tan}^{- 1} (- 1) = - \frac{π}{4}$

Phase angle of Second $LCR$ branch is $ϕ_{2} = {tan}^{- 1} (\frac{X_{C 2} - X_{L 2}}{R_{2}})$

$\Rightarrow ϕ_{2} = {tan}^{- 1} (1) = \frac{π}{4}$

Thus, $| ϕ_{1} - ϕ_{2} | = \frac{π}{2}$

So, clearly we can say that, $I_{1}$ is lagging and $I_{2}$ is leading from sourrce voltage.

Using, $I = \sqrt{{(I_{1})}_{1}^{2} + {(I_{1})}_{1}^{2} + 2 I_{1} I_{2} cos ϕ}$

We can say that, $I = \sqrt{{(I_{1})}_{1}^{2} + {(I_{2})}_{2}^{2}}$

From $C$ :

When $C_{1}$ is removed first $LCR$ branch becomes an $LR$ branch and when $L_{2}$ is removed second $LCR$ branch becomes a $CR$ branch.

Phase angle of $LR$ branch is $ϕ_{1} = {tan}^{- 1} (\frac{- X_{L 1}}{R_{1}}) = - \frac{π}{4}$

Phase angle of $CR$ branch is $ϕ_{2} = {tan}^{- 1} (\frac{X_{C 2}}{R_{2}}) = \frac{π}{4}$

Thus, $| ϕ_{1} - ϕ_{2} | = \frac{π}{2}$

So, clearly we can say that, $I_{1}$ is lagging and $I_{2}$ is leading from sourrce voltage.

Using, $I = \sqrt{{(I_{1})}_{1}^{2} + {(I_{1})}_{1}^{2} + 2 I_{1} I_{2} cos ϕ}$

We can say that, $I = \sqrt{{(I_{1})}_{1}^{2} + {(I_{2})}_{2}^{2}}$

From $D$ :

When both $C_{1} & C_{2}$ are removed from the circuit,First and second $LCR$ branches become $LR$ branches.

Phase angle of first $LR$ branch is $ϕ_{1} = {tan}^{- 1} (\frac{- X_{L 1}}{R_{1}}) = - \frac{π}{4}$

Phase angle of second $LR$ branch is $ϕ_{2} = {tan}^{- 1} (\frac{- X_{L 2}}{R_{2}}) = - \frac{π}{4}$

So, the current in both the branches have same phase.

Since, the phase difference is zero

Using, $I = \sqrt{{(I_{1})}_{1}^{2} + {(I_{1})}_{1}^{2} + 2 I_{1} I_{2} cos ϕ}$

We can say that, $I = I_{1} + I_{2}$

Hence, option (a) is the correct alternative.

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