Alternating Current - NEET Physics Chapterwise MCQs & PYQs

NEET Alternating Current MCQs & PYQs

Practice NEET Alternating Current Questions

Question 71:

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).

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Question 72:

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).

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Question 73:

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).

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Question 74:

moderate

The maximum power is dissipated for an ac in a/an

The average power dissipated in an AC circuit is \(P_{\text{avg}} = V_{\text{rms}} I_{\text{rms}} cos\phi\). For a purely resistive circuit, the phase difference \(\phi = 0\), hence \(cos\phi = 1\) (maximum), resulting in maximum power dissipation.

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Question 75:

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.

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Question 76:

moderate

An ac current flowing in a circuit is given by \(i = i_0 sin \omega t\). The minimum time taken to reach zero to rms value of current is

The RMS value is \(i = \frac{i_0}{\sqrt{2}}\). Thus, \(\frac{i_0}{\sqrt{2}} = i_0 sin \omega t ⇒ \omega t = \frac{\pi}{4} ⇒ t = \frac{\pi}{4\omega}\).

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Question 77:

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.
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Question 78:

easy

Which of the following AC circuit will not dissipate energy? (Symbols have their usual meaning)

In a purely reactive circuit like a series LC circuit, the average power dissipated is zero because the phase difference between the current and voltage is \( 90^\circ \), making the power factor \( \cos\phi = 0 \).

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Question 79:

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 \(I = \frac{V}{Z} = \frac{220}{100\sqrt{2}} = 1.1\sqrt{2}\text{ A}\).

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Question 80:

easy

1. The peak voltage of the ac source is equal to (2022)

The root-mean-square (rms) voltage for a sinusoidal alternating voltage is given by $V_{rms} = \frac{V_0}{\sqrt{2}}$. Rearranging for peak voltage $V_0$, we get $V_0 = \sqrt{2} V_{rms}$. Hence, the peak voltage is $\sqrt{2}$ times the rms value of the ac source.

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