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RLC Circuit

Refer to the ...

Question

Refer to the circuit diagram and the corresponding graphs. The current rises when key $K$ is pressed. With $R = R_{1}$ and $L = L_{1}$ the rise of current is shown by curve (1), while curve (2) shows the rise of current when $R = R_{2}$ and $L = L_{2}$ . The maximum current is same for both curves, then :

A

R_{1} = R_{2}, L_{1} = L_{2}

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B

R_{1} > R_{2}, L_{1} = L_{2}

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C

R_{1} > R_{2}, L_{1} < L_{2}

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D

R_{1} = R_{2}, L_{1} < L_{2}

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Solution

The correct option is D $R_{1} = R_{2}, L_{1} < L_{2}$
Consider the expression for the current rising exponentially in the $L R$ circuit. The time constant is $(L / R)$ . In this case, the curve (1) is rising faster than curve (2) indicating that $(L_{1} / R_{1}) < (L_{2} / R_{2})$ . However, in both the cases the maximum current is the same and equal to $(V / R_{1})$ or $(V / R_{2})$ . Which means $R_{1} = R_{2}$ .

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Similar questions

Q. Refer to the circuit diagram and the corresponding graphs. The current rises when key

K

is pressed. With

R = R_{1}

L = L_{1}

the rise of current is shown by curve (1), while curve (2) shows the rise of current when

R = R_{2}

L = L_{2}

. The maximum current is same for both curves, then :

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R_{2}

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Q. Two loops P and Q are made from a same uniform wire. The radii of P and Q are

r_{1}

r_{2}

respectively and their moments of inertia are

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l_{2}

respectively. If

\frac{l_{2}}{l_{1}}

\frac{r_{2}}{r_{1}}

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E_{1} = 4.0 V, R_{1} = 2 Ω, E_{2} = 6.0 V, R_{2} = 4 Ω

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, graph between peak current and frequency (

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The correct order for

R_{1}, R_{2} a n d R_{3}

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