Assertion (A): If the resistance of a series resonant LCR circuit is decreased, then the peak current versus frequency graph will be taller and narrower.
Reason (R): If the resistance of a series resonant LCR circuit decreased, then its resonance will be unaffected.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer
When resistance \(R\) in a series \(LCR\) circuit decreases, the peak current \(I_{max} = V/R\) at resonance increases, making the peak taller. The quality factor \(Q = (\omega_0 L)/R\) increases, leading to a narrower resonance curve. The resonant frequency \(omega_0 = 1/\sqrt{LC}\) remains unchanged, but the overall resonance behavior (sharpness, peak current) is affected. Therefore, (A) is true and (R) is false.
Assertion (A): The impedance of series L-C-R circuit can be greater, equal or less than the resistance.
Reason (R): The minimum impedance of series LCR circuit depends over angular frequency of applied emf.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer
The impedance of a series \(LCR\) circuit is given by \(Z = \sqrt{R^2 + (X_L - X_C)^2}\). Since \((X_L - X_C)^2\) is always non-negative, \(Z\) is always greater than or equal to \(R\). Thus, (A) is false. The minimum impedance occurs at resonance, where \(Z_{min} = R\). This minimum value depends only on \(R\) and not on the angular frequency \(omega\). Thus, (R) is also false.
Assertion (A): When frequency is greater than resonance frequency in a series LCR circuit, it will be an inductive circuit.
Reason (R): Resultant Voltage Will lead the current.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer
When frequency \(f > f_r\) in a series LCR circuit, the inductive reactance \(X_L\) is greater than the capacitive reactance \(X_C\). This makes the circuit inductive. In an inductive circuit, the resultant voltage leads the current. Hence, both assertion and reason are true, and the reason correctly explains the assertion.
Assertion (A): In ac supply we cannot feel any fluctuations of current in bulbs.
Reason (R): House hold ac supply has very low frequency.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer
Our eyes cannot detect flickers above \(10-15 \text{ Hz}\). Household AC supply (\(50/60 \text{ Hz}\)) changes too rapidly for us to perceive fluctuations in bulbs. Thus, Assertion (A) is true. However, \(50/60 \text{ Hz}\) is not considered a "very low frequency", thus Reason (R) is false.
Assertion (A): An electric lamp is connected in series with a long solenoid of copper with air core and then connected to \( \text{ac} \) source. If an iron rod is inserted in solenoid, the lamp will become dim.
Reason (R): If an iron rod is inserted in solenoid, the inductance of solenoid increases.
1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer
Inserting an iron rod (ferromagnetic material) into a solenoid significantly increases its inductance \( L \) (R is true). In an \( \text{ac} \) circuit, this increases inductive reactance \( X_L =\omega L \), which in turn increases the total impedance \( Z \) of the circuit. Higher \( Z \) leads to lower current \( I = V/Z \), making the lamp dim (A is true). (R) provides the correct explanation for (A).
Assertion (A): For an electric lamp connected in series with a variable capacitor and \( \text{ac} \) source, its brightness increases with increase of capacitance.
Reason (R): Capacitive reactance decreases with increase in capacitance of capacitor.
1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer
Capacitive reactance is given by \( X_C = frac{1}{omega C} \). As capacitance \( C \) increases, \( X_C \) decreases (R is true). A decrease in \( X_C \) leads to a decrease in the total circuit impedance \( Z \). With constant voltage \( V \), a lower \( Z \) results in higher current \( I = V/Z \), thus increasing the lamp's brightness (A is true). (R) correctly explains (A).
Assertion (A): In series RL circuit voltage leads the current.
Reason (R): In series \( \text{LCR} \) circuit current may lead the voltage.
1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer
In a series \( text{RL} \) circuit, the inductive reactance \( X_L \) causes the voltage to lead the current, so (A) is true. In a series \( text{LCR} \) circuit, if \( X_C > X_L \), the circuit is capacitive, and current leads the voltage, so (R) is true. Both statements are true, but (R) is about a different circuit type and does not explain (A).