LR, RC and LCR Circuits - NEET Physics Chapterwise MCQs & PYQs

NEET LR, RC and LCR Circuits MCQs & PYQs

Practice NEET LR, RC and LCR Circuits Questions

Question 41:

easy

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.

 

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.

💬 Discuss this Question

Question 42:

easy

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.

 

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.

💬 Discuss this Question

Question 43:

easy

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.

 

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.

💬 Discuss this Question

Question 44:

easy

Assertion (A): In ac supply we cannot feel any fluctuations of current in bulbs.

Reason (R): House hold ac supply has very low frequency.

 

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.

💬 Discuss this Question

Question 45:

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.

 

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).

💬 Discuss this Question

Question 46:

easy

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.

 

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).

💬 Discuss this Question

Question 47:

easy

Assertion (A): In series RL circuit voltage leads the current.

Reason (R): In series \( \text{LCR} \) circuit current may lead the voltage.

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).

💬 Discuss this Question

Question 48:

easy

Consider the following statements and choose the correct option.

Statement A: Capacitive reactance is inversely proportional to frequency.

Statement B: Inductive reactance is directly proportional to frequency.

 

The formulae are \(X_C = \frac{1}{2\pi f C}\) (inversely proportional) and \(X_L = 2\pi f L\) (directly proportional). Hence, both statements are correct.

💬 Discuss this Question

Question 49:

moderate

An a.c. voltage given by relation, \(V = 300\sin(100t)\text{ volt}\) is connected with resistor having resistance \(60\ \Omega\) and inductor of inductance \(L\). If peak current in the circuit is \(3\text{ A}\), the value of \(L\) will be (where \(t\) denotes the time in s)

Short Solution: 

Peak voltage \(V_0 = 300\text{ V}\), peak current \(I_0 = 3\text{ A}\), so impedance \(Z = V_0/I_0 = 100\ \Omega\). Using \(Z = \sqrt{R^2 + (\omega L)^2}\), we have \(100 = \sqrt{60^2 + (100L)^2}\), which gives \(100L = 80 \implies L = 0.8\text{ H}\).

Detailed Solution: 

Understanding the Given Problem

We are given a standard alternating current (AC) circuit containing a resistor and an unknown inductor connected in series.
Given Data:
  • AC Voltage Equation: $V = 300\sin(100t)\text{ V}$
  • Resistance ($R$): $60\ \Omega$
  • Peak Current ($I_0$): $3\text{ A}$
From the standard AC voltage equation $V = V_0\sin(\omega t)$, we can extract two crucial values by comparing it to our given equation:
  • Peak Voltage ($V_0$): $300\text{ V}$
  • Angular Frequency ($\omega$): $100\text{ rad/s}$

Step 1: Calculate the Total Impedance ($Z$)

In an AC circuit, the total opposition to the current flow is called impedance ($Z$). It can be found using the peak voltage and peak current, similar to Ohm's law:
$$Z = \frac{V_0}{I_0}$$
Substitute the known values:
$$Z = \frac{300\text{ V}}{3\text{ A}} = 100\ \Omega$$

Step 2: Use the L-R Circuit Impedance Formula

For a series L-R circuit, the impedance is the vector sum of the resistance ($R$) and the inductive reactance ($X_L$). The formula is:
$$Z = \sqrt{R^2 + X_L^2}$$
We also know that inductive reactance is given by $X_L = \omega L$. Substituting this into the impedance formula gives:
$$Z = \sqrt{R^2 + (\omega L)^2}$$

Step 3: Solve for the Inductance ($L$)

Now, substitute the values we have gathered ($Z = 100\ \Omega$, $R = 60\ \Omega$, $\omega = 100\text{ rad/s}$) into the equation:
$$100 = \sqrt{60^2 + (100L)^2}$$
Square both sides of the equation to remove the square root:
$$(100)^2 = 60^2 + (100L)^2$$
$$10000 = 3600 + (100L)^2$$
Subtract $3600$ from both sides:
$$6400 = (100L)^2$$
Take the square root of both sides:
$$80 = 100L$$
Finally, solve for $L$:
$$L = \frac{80}{100} = 0.8\text{ H}$$

Final Answer

The correct value of the inductance $L$ is 0.8 H, which corresponds to Option 3.
💬 Discuss this Question

Question 50:

easy

Consider a series \(LCR\) circuit in which reactance and resistance are \(100\ \Omega\) each. When the circuit is connected to ac source \(220\text{ V}\), \(50\text{ Hz}\), then current drawn from the source is

Here, \(R = 100\ \Omega\) and net reactance \(X = 100\ \Omega\). Total impedance is \(Z = \sqrt{R^2 + X^2} = 100\sqrt{2}\ \Omega\). The current \(I = \frac{V}{Z} = \frac{220}{100\sqrt{2}} = 1.1\sqrt{2}\text{ A}\).

💬 Discuss this Question