356014
An \(AC\) voltage is applied to a resistance \(R\) and an inductor \(L\) in series. If \(R\) and the inductive reactance are both equal to \(3\Omega \) the phase difference between the applied voltage and the current in the circuit is
1 \(\frac{\pi }{4}\)
2 \(\frac{\pi }{6}\)
3 \(\frac{\pi }{2}\)
4 \({\rm{Zero}}\)
Explanation:
As we know that \(\tan \phi = \frac{{{X_L}}}{R} = \frac{{\omega L}}{R}\) \(\tan \phi = \frac{3}{3} \Rightarrow \quad \phi = \frac{\pi }{4}rad\)
PHXII07:ALTERNATING CURRENT
356015
In the \(A.C\). circuit shown, keeping \(‘K’\) pressed,if an iron rod is inserted into the coil, the bulb in the circuit.
1 Gets damaged
2 Glows less brightly
3 Glows with same brightness (as before the rod is inserted)
4 Glows more brightly
Explanation:
As the presence of the iron rod increases the self-inductance and hence the inductive reactance ofthe coil, the net impedance of the circuit increases and that causes the current in the circuit to decrease. \(\therefore \) The bulb glows less brightly.
PHXII07:ALTERNATING CURRENT
356016
An inductance and resistance are connected in series with an \(A.C\) circuit. In this circuit
1 The current and P.d across the resistance lead P.d across the inductance by \(\frac{\pi }{2}\)
2 The current and P.d across the resistance lags behind the P.d across the inductance by angle \(\frac{\pi }{2}\)
3 The current across resistance leads and the P.d across resistance lags behind the P.d across the inductance by \(\frac{\pi }{2}\)
4 The current across resistance lags behind and the P.d across the resistance leads the P.d across the inductance by \(\frac{\pi }{2}\)
Explanation:
Conceptual Question
PHXII07:ALTERNATING CURRENT
356017
A coil of \(200\,\Omega \) resistance and \(1\,H\) inductance is connected to an a.c source of frequency \(\frac{{100}}{\pi }Hz\), phase angle between potential and current will be:
1 \(30^{\circ}\)
2 \(45^{\circ}\)
3 \(60^{\circ}\)
4 \(90^{\circ}\)
Explanation:
\(\cos \phi=\dfrac{R}{Z}=\dfrac{R}{\sqrt{(R)^{2}+\omega^{2} L^{2}}}\) \(\omega=2 \pi v=2 \pi\left(\dfrac{100}{\pi}\right)=200 H z\) \(\Rightarrow \cos \phi=\dfrac{200}{\sqrt{(200)^{2}+(200)^{2} \times(1)^{2}}}\) \(\Rightarrow \cos \phi=\dfrac{1}{\sqrt{2}}=\phi=45^{\circ}\) So, correct option is (2).
356014
An \(AC\) voltage is applied to a resistance \(R\) and an inductor \(L\) in series. If \(R\) and the inductive reactance are both equal to \(3\Omega \) the phase difference between the applied voltage and the current in the circuit is
1 \(\frac{\pi }{4}\)
2 \(\frac{\pi }{6}\)
3 \(\frac{\pi }{2}\)
4 \({\rm{Zero}}\)
Explanation:
As we know that \(\tan \phi = \frac{{{X_L}}}{R} = \frac{{\omega L}}{R}\) \(\tan \phi = \frac{3}{3} \Rightarrow \quad \phi = \frac{\pi }{4}rad\)
PHXII07:ALTERNATING CURRENT
356015
In the \(A.C\). circuit shown, keeping \(‘K’\) pressed,if an iron rod is inserted into the coil, the bulb in the circuit.
1 Gets damaged
2 Glows less brightly
3 Glows with same brightness (as before the rod is inserted)
4 Glows more brightly
Explanation:
As the presence of the iron rod increases the self-inductance and hence the inductive reactance ofthe coil, the net impedance of the circuit increases and that causes the current in the circuit to decrease. \(\therefore \) The bulb glows less brightly.
PHXII07:ALTERNATING CURRENT
356016
An inductance and resistance are connected in series with an \(A.C\) circuit. In this circuit
1 The current and P.d across the resistance lead P.d across the inductance by \(\frac{\pi }{2}\)
2 The current and P.d across the resistance lags behind the P.d across the inductance by angle \(\frac{\pi }{2}\)
3 The current across resistance leads and the P.d across resistance lags behind the P.d across the inductance by \(\frac{\pi }{2}\)
4 The current across resistance lags behind and the P.d across the resistance leads the P.d across the inductance by \(\frac{\pi }{2}\)
Explanation:
Conceptual Question
PHXII07:ALTERNATING CURRENT
356017
A coil of \(200\,\Omega \) resistance and \(1\,H\) inductance is connected to an a.c source of frequency \(\frac{{100}}{\pi }Hz\), phase angle between potential and current will be:
1 \(30^{\circ}\)
2 \(45^{\circ}\)
3 \(60^{\circ}\)
4 \(90^{\circ}\)
Explanation:
\(\cos \phi=\dfrac{R}{Z}=\dfrac{R}{\sqrt{(R)^{2}+\omega^{2} L^{2}}}\) \(\omega=2 \pi v=2 \pi\left(\dfrac{100}{\pi}\right)=200 H z\) \(\Rightarrow \cos \phi=\dfrac{200}{\sqrt{(200)^{2}+(200)^{2} \times(1)^{2}}}\) \(\Rightarrow \cos \phi=\dfrac{1}{\sqrt{2}}=\phi=45^{\circ}\) So, correct option is (2).
356014
An \(AC\) voltage is applied to a resistance \(R\) and an inductor \(L\) in series. If \(R\) and the inductive reactance are both equal to \(3\Omega \) the phase difference between the applied voltage and the current in the circuit is
1 \(\frac{\pi }{4}\)
2 \(\frac{\pi }{6}\)
3 \(\frac{\pi }{2}\)
4 \({\rm{Zero}}\)
Explanation:
As we know that \(\tan \phi = \frac{{{X_L}}}{R} = \frac{{\omega L}}{R}\) \(\tan \phi = \frac{3}{3} \Rightarrow \quad \phi = \frac{\pi }{4}rad\)
PHXII07:ALTERNATING CURRENT
356015
In the \(A.C\). circuit shown, keeping \(‘K’\) pressed,if an iron rod is inserted into the coil, the bulb in the circuit.
1 Gets damaged
2 Glows less brightly
3 Glows with same brightness (as before the rod is inserted)
4 Glows more brightly
Explanation:
As the presence of the iron rod increases the self-inductance and hence the inductive reactance ofthe coil, the net impedance of the circuit increases and that causes the current in the circuit to decrease. \(\therefore \) The bulb glows less brightly.
PHXII07:ALTERNATING CURRENT
356016
An inductance and resistance are connected in series with an \(A.C\) circuit. In this circuit
1 The current and P.d across the resistance lead P.d across the inductance by \(\frac{\pi }{2}\)
2 The current and P.d across the resistance lags behind the P.d across the inductance by angle \(\frac{\pi }{2}\)
3 The current across resistance leads and the P.d across resistance lags behind the P.d across the inductance by \(\frac{\pi }{2}\)
4 The current across resistance lags behind and the P.d across the resistance leads the P.d across the inductance by \(\frac{\pi }{2}\)
Explanation:
Conceptual Question
PHXII07:ALTERNATING CURRENT
356017
A coil of \(200\,\Omega \) resistance and \(1\,H\) inductance is connected to an a.c source of frequency \(\frac{{100}}{\pi }Hz\), phase angle between potential and current will be:
1 \(30^{\circ}\)
2 \(45^{\circ}\)
3 \(60^{\circ}\)
4 \(90^{\circ}\)
Explanation:
\(\cos \phi=\dfrac{R}{Z}=\dfrac{R}{\sqrt{(R)^{2}+\omega^{2} L^{2}}}\) \(\omega=2 \pi v=2 \pi\left(\dfrac{100}{\pi}\right)=200 H z\) \(\Rightarrow \cos \phi=\dfrac{200}{\sqrt{(200)^{2}+(200)^{2} \times(1)^{2}}}\) \(\Rightarrow \cos \phi=\dfrac{1}{\sqrt{2}}=\phi=45^{\circ}\) So, correct option is (2).
356014
An \(AC\) voltage is applied to a resistance \(R\) and an inductor \(L\) in series. If \(R\) and the inductive reactance are both equal to \(3\Omega \) the phase difference between the applied voltage and the current in the circuit is
1 \(\frac{\pi }{4}\)
2 \(\frac{\pi }{6}\)
3 \(\frac{\pi }{2}\)
4 \({\rm{Zero}}\)
Explanation:
As we know that \(\tan \phi = \frac{{{X_L}}}{R} = \frac{{\omega L}}{R}\) \(\tan \phi = \frac{3}{3} \Rightarrow \quad \phi = \frac{\pi }{4}rad\)
PHXII07:ALTERNATING CURRENT
356015
In the \(A.C\). circuit shown, keeping \(‘K’\) pressed,if an iron rod is inserted into the coil, the bulb in the circuit.
1 Gets damaged
2 Glows less brightly
3 Glows with same brightness (as before the rod is inserted)
4 Glows more brightly
Explanation:
As the presence of the iron rod increases the self-inductance and hence the inductive reactance ofthe coil, the net impedance of the circuit increases and that causes the current in the circuit to decrease. \(\therefore \) The bulb glows less brightly.
PHXII07:ALTERNATING CURRENT
356016
An inductance and resistance are connected in series with an \(A.C\) circuit. In this circuit
1 The current and P.d across the resistance lead P.d across the inductance by \(\frac{\pi }{2}\)
2 The current and P.d across the resistance lags behind the P.d across the inductance by angle \(\frac{\pi }{2}\)
3 The current across resistance leads and the P.d across resistance lags behind the P.d across the inductance by \(\frac{\pi }{2}\)
4 The current across resistance lags behind and the P.d across the resistance leads the P.d across the inductance by \(\frac{\pi }{2}\)
Explanation:
Conceptual Question
PHXII07:ALTERNATING CURRENT
356017
A coil of \(200\,\Omega \) resistance and \(1\,H\) inductance is connected to an a.c source of frequency \(\frac{{100}}{\pi }Hz\), phase angle between potential and current will be:
1 \(30^{\circ}\)
2 \(45^{\circ}\)
3 \(60^{\circ}\)
4 \(90^{\circ}\)
Explanation:
\(\cos \phi=\dfrac{R}{Z}=\dfrac{R}{\sqrt{(R)^{2}+\omega^{2} L^{2}}}\) \(\omega=2 \pi v=2 \pi\left(\dfrac{100}{\pi}\right)=200 H z\) \(\Rightarrow \cos \phi=\dfrac{200}{\sqrt{(200)^{2}+(200)^{2} \times(1)^{2}}}\) \(\Rightarrow \cos \phi=\dfrac{1}{\sqrt{2}}=\phi=45^{\circ}\) So, correct option is (2).
