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).
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.
1. Both statements A and B are correct
2. Both statements A and B are incorrect
3. Statement A is correct and B is incorrect
4. Statement A is incorrect and B is correct
View Answer
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.
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
1. \(\frac{\pi}{2\omega}\)
2. \(\frac{\pi}{4\omega}\)
3. \(\frac{\pi}{\omega}\)
4. \(\frac{2\pi}{\omega}\)
View Answer
The RMS value is \(i = \frac{i_0}{\sqrt{2}}\). Thus, \(\frac{i_0}{\sqrt{2}} = i_0 sin \omega t ⇒ \omega t = \frac{\pi}{4} ⇒ t = \frac{\pi}{4\omega}\).
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)
1. 0.6 H
2. 0.4 H
3. 0.8 H
4. 1 H
View Answer
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 \(100L = 80 \implies L = 0.8\text{ H}\).
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
1. \(2.2\sqrt{2}\text{ A}\)
2. \(1.1\sqrt{2}\text{ A}\)
3. \(3.3\sqrt{2}\text{ A}\)
4. \(2.2\text{ A}\)
View Answer
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 \(I = \frac{V}{Z} = \frac{220}{100\sqrt{2}} = 1.1\sqrt{2}\text{ A}\).