Physics NCERT Exemplar Solutions Class 12th Chapter Seven Alternating Current Overview Class 12th Physics Physics NCERT Exemplar Solutions Class 11th Chapter Five

For a series LCR circuit, I vs curve is shown:

Choose the most appropriate answer from the options given below:

Option 1 - <p>To the left of  <span class="mathml" contenteditable="false"> <math> <mrow> <mi>ω</mi> </mrow> </math> </span>the circuit is mainly capacitive.</p>

Option 2 - <p>To the left of <span class="mathml" contenteditable="false"> <math> <mrow> <mi>ω</mi> </mrow> </math> </span> the circuit is mainly inductive.</p>

Option 3 - <p>At <span class="mathml" contenteditable="false"> <math> <mrow> <mi>ω</mi> </mrow> </math> </span> impedance of the circuit is equal to the resistance of the circuit.</p>

Option 4 - <p>At <span class="mathml" contenteditable="false"> <math> <mrow> <mi>ω</mi> </mrow> </math> </span>, impedance of the circuit is 0.</p>

3 Views|Posted 6 months ago

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Alok Kumar Singh | Contributor-Level 10

6 months ago

Correct Option - 3

Detailed Solution:

At w_r impedance of the circuit is equal to the resistance of the circuit.

Left of w_r, circuit is mainly capacitive

Right of w_r, the circuit is mainly Inductive.

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Power factor of series LCR circuit at resonance is

At resonance V_L = V_C

_{$\Rightarrow c o s ? = \frac{V_{R}}{\sqrt[]{V_{R}^{2} + {(V_{L} - V_{C})}^{2}}} = 1$}

An electric bulb and a capacitor are connected in series with an AC source. On increasing the frequency of the source, the brightness of the bulb:

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E = 100V

Figure shows a circuit that contains four identical resistors with resistance R = $2.0 Ω,$ two identical inductors with inductance L = 2.0mH and an ideal battery with emf E = 9V. The current ‘i’ just after switch ‘S’ is closed will be:

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Physics NCERT Exemplar Solutions Class 12th Chapter Seven 2025

Physics NCERT Exemplar Solutions Class 12th Chapter Seven 2025

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