155245
The value of alternating emf $E$ in the given circuit will be :
1 $220 \mathrm{~V}$
2 $140 \mathrm{~V}$
3 $100 \mathrm{~V}$
4 $20 \mathrm{~V}$
Explanation:
C Given that, $\mathrm{V}_{\mathrm{R}}=80 \mathrm{~V}, \mathrm{~V}_{\mathrm{C}}=100 \mathrm{~V}, \mathrm{~V}_{\mathrm{L}}=40$ $\mathrm{V}$ $\text { In series }(E) =\sqrt{V_{R}^{2}+\left(V_{L}-V_{C}\right)^{2}}$ $E =\sqrt{(80)^{2}+(40-100)^{2}}$ $E =\sqrt{(80)^{2}+(-60)^{2}}$ $E =\sqrt{10000}$ $E =100 \mathrm{~V}$
Karnataka CET-2008
Alternating Current
155246
In a LCR circuit the potential difference between the terminals of the inductance is 60 $\mathrm{V}$, between the terminals of the capacitor is 30 $V$ and that between the terminals of resistance is $40 \mathrm{~V}$. The supply voltage will be equal to :
1 $130 \mathrm{~V}$
2 $10 \mathrm{~V}$
3 $50 \mathrm{~V}$
4 $70 \mathrm{~V}$
Explanation:
C Given, \(\mathrm{V}_{\mathrm{L}}=60 \mathrm{~V}, \mathrm{~V}_{\mathrm{C}}=30 \mathrm{~V}, \mathrm{~V}_{\mathrm{R}}=40 \mathrm{~V}\) We know that, \(\text { Supply voltage }(\mathrm{V})=\sqrt{\mathrm{V}_{\mathrm{R}}^2+\left(\mathrm{V}_{\mathrm{L}}-\mathrm{V}_{\mathrm{C}}\right)^2}\) \(\mathrm{V} =\sqrt{(40)^2+(60-30)^2}\) \(\mathrm{~V}=\sqrt{(40)^2+(30)^2}\) \(\mathrm{~V}=\sqrt{2500}\) \(\mathrm{~V}=50 \text { Volt }\)
Karnataka CET-2004
Alternating Current
155247
In the circuit shown in the figure, the A.C source gives a voltage $V=20 \cos (2000t)$. Neglecting source resistance, the voltmeter and ammeter reading will be :
155248
In the circuit shown in the figure, neglecting source resistance, the voltmeter and ammeter readings will respectively be :
1 $150 \mathrm{~V}, 8 \mathrm{~A}$
2 $0 \mathrm{~V}, 8 \mathrm{~A}$
3 $0 \mathrm{~V}, 3 \mathrm{~A}$
4 $150 \mathrm{~V}, 3 \mathrm{~A}$
Explanation:
B Given that, $\mathrm{X}_{\mathrm{C}}=\mathrm{X}_{\mathrm{L}}=25 \Omega, \mathrm{V}=240$ Volt, $\mathrm{R}$ $=30 \Omega$ So, potential between $\mathrm{X}_{\mathrm{L}}$ and $\mathrm{X}_{\mathrm{C}}$ is zero Then, $\quad(Z)=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}$ $Z=\sqrt{(30)^{2}+(25-25)^{2}}$ $Z=30 \Omega$ Then, current $(\mathrm{I})=\frac{\mathrm{V}}{\mathrm{Z}}=\frac{240}{30}$ $\mathrm{I}=8 \text { Amp. }$
Karnataka CET-2001
Alternating Current
155249
An LCR series circuit is under resonance. If $I_{m}$ is current amplitude, $V_{m}$ is voltage amplitude, $R$ is the resistance, $Z$ is the impedance, $X_{L}$ is the inductive reactance and $X_{C}$ is the capacitive reactance then,
1 $I_{m}=\frac{V_{m}}{Z}$
2 $I_{m}=\frac{V_{m}}{X_{L}}$
3 $I_{m}=\frac{V_{m}}{X_{C}}$
4 $I_{m}=\frac{V_{m}}{R}$
Explanation:
D Given that, Impedance of the circuit $(Z)=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}$ At resonance, $\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}$ Then, $Z=\sqrt{R^{2}+\left(X_{C}-X_{C}\right)^{2}}$ $\mathrm{Z}=\mathrm{R}$ $\therefore \quad \mathrm{I}_{\mathrm{m}}=\frac{\mathrm{V}_{\mathrm{m}}}{\mathrm{Z}}$ $\mathrm{I}_{\mathrm{m}}=\frac{\mathrm{V}_{\mathrm{m}}}{\mathrm{R}}$
155245
The value of alternating emf $E$ in the given circuit will be :
1 $220 \mathrm{~V}$
2 $140 \mathrm{~V}$
3 $100 \mathrm{~V}$
4 $20 \mathrm{~V}$
Explanation:
C Given that, $\mathrm{V}_{\mathrm{R}}=80 \mathrm{~V}, \mathrm{~V}_{\mathrm{C}}=100 \mathrm{~V}, \mathrm{~V}_{\mathrm{L}}=40$ $\mathrm{V}$ $\text { In series }(E) =\sqrt{V_{R}^{2}+\left(V_{L}-V_{C}\right)^{2}}$ $E =\sqrt{(80)^{2}+(40-100)^{2}}$ $E =\sqrt{(80)^{2}+(-60)^{2}}$ $E =\sqrt{10000}$ $E =100 \mathrm{~V}$
Karnataka CET-2008
Alternating Current
155246
In a LCR circuit the potential difference between the terminals of the inductance is 60 $\mathrm{V}$, between the terminals of the capacitor is 30 $V$ and that between the terminals of resistance is $40 \mathrm{~V}$. The supply voltage will be equal to :
1 $130 \mathrm{~V}$
2 $10 \mathrm{~V}$
3 $50 \mathrm{~V}$
4 $70 \mathrm{~V}$
Explanation:
C Given, \(\mathrm{V}_{\mathrm{L}}=60 \mathrm{~V}, \mathrm{~V}_{\mathrm{C}}=30 \mathrm{~V}, \mathrm{~V}_{\mathrm{R}}=40 \mathrm{~V}\) We know that, \(\text { Supply voltage }(\mathrm{V})=\sqrt{\mathrm{V}_{\mathrm{R}}^2+\left(\mathrm{V}_{\mathrm{L}}-\mathrm{V}_{\mathrm{C}}\right)^2}\) \(\mathrm{V} =\sqrt{(40)^2+(60-30)^2}\) \(\mathrm{~V}=\sqrt{(40)^2+(30)^2}\) \(\mathrm{~V}=\sqrt{2500}\) \(\mathrm{~V}=50 \text { Volt }\)
Karnataka CET-2004
Alternating Current
155247
In the circuit shown in the figure, the A.C source gives a voltage $V=20 \cos (2000t)$. Neglecting source resistance, the voltmeter and ammeter reading will be :
155248
In the circuit shown in the figure, neglecting source resistance, the voltmeter and ammeter readings will respectively be :
1 $150 \mathrm{~V}, 8 \mathrm{~A}$
2 $0 \mathrm{~V}, 8 \mathrm{~A}$
3 $0 \mathrm{~V}, 3 \mathrm{~A}$
4 $150 \mathrm{~V}, 3 \mathrm{~A}$
Explanation:
B Given that, $\mathrm{X}_{\mathrm{C}}=\mathrm{X}_{\mathrm{L}}=25 \Omega, \mathrm{V}=240$ Volt, $\mathrm{R}$ $=30 \Omega$ So, potential between $\mathrm{X}_{\mathrm{L}}$ and $\mathrm{X}_{\mathrm{C}}$ is zero Then, $\quad(Z)=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}$ $Z=\sqrt{(30)^{2}+(25-25)^{2}}$ $Z=30 \Omega$ Then, current $(\mathrm{I})=\frac{\mathrm{V}}{\mathrm{Z}}=\frac{240}{30}$ $\mathrm{I}=8 \text { Amp. }$
Karnataka CET-2001
Alternating Current
155249
An LCR series circuit is under resonance. If $I_{m}$ is current amplitude, $V_{m}$ is voltage amplitude, $R$ is the resistance, $Z$ is the impedance, $X_{L}$ is the inductive reactance and $X_{C}$ is the capacitive reactance then,
1 $I_{m}=\frac{V_{m}}{Z}$
2 $I_{m}=\frac{V_{m}}{X_{L}}$
3 $I_{m}=\frac{V_{m}}{X_{C}}$
4 $I_{m}=\frac{V_{m}}{R}$
Explanation:
D Given that, Impedance of the circuit $(Z)=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}$ At resonance, $\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}$ Then, $Z=\sqrt{R^{2}+\left(X_{C}-X_{C}\right)^{2}}$ $\mathrm{Z}=\mathrm{R}$ $\therefore \quad \mathrm{I}_{\mathrm{m}}=\frac{\mathrm{V}_{\mathrm{m}}}{\mathrm{Z}}$ $\mathrm{I}_{\mathrm{m}}=\frac{\mathrm{V}_{\mathrm{m}}}{\mathrm{R}}$
155245
The value of alternating emf $E$ in the given circuit will be :
1 $220 \mathrm{~V}$
2 $140 \mathrm{~V}$
3 $100 \mathrm{~V}$
4 $20 \mathrm{~V}$
Explanation:
C Given that, $\mathrm{V}_{\mathrm{R}}=80 \mathrm{~V}, \mathrm{~V}_{\mathrm{C}}=100 \mathrm{~V}, \mathrm{~V}_{\mathrm{L}}=40$ $\mathrm{V}$ $\text { In series }(E) =\sqrt{V_{R}^{2}+\left(V_{L}-V_{C}\right)^{2}}$ $E =\sqrt{(80)^{2}+(40-100)^{2}}$ $E =\sqrt{(80)^{2}+(-60)^{2}}$ $E =\sqrt{10000}$ $E =100 \mathrm{~V}$
Karnataka CET-2008
Alternating Current
155246
In a LCR circuit the potential difference between the terminals of the inductance is 60 $\mathrm{V}$, between the terminals of the capacitor is 30 $V$ and that between the terminals of resistance is $40 \mathrm{~V}$. The supply voltage will be equal to :
1 $130 \mathrm{~V}$
2 $10 \mathrm{~V}$
3 $50 \mathrm{~V}$
4 $70 \mathrm{~V}$
Explanation:
C Given, \(\mathrm{V}_{\mathrm{L}}=60 \mathrm{~V}, \mathrm{~V}_{\mathrm{C}}=30 \mathrm{~V}, \mathrm{~V}_{\mathrm{R}}=40 \mathrm{~V}\) We know that, \(\text { Supply voltage }(\mathrm{V})=\sqrt{\mathrm{V}_{\mathrm{R}}^2+\left(\mathrm{V}_{\mathrm{L}}-\mathrm{V}_{\mathrm{C}}\right)^2}\) \(\mathrm{V} =\sqrt{(40)^2+(60-30)^2}\) \(\mathrm{~V}=\sqrt{(40)^2+(30)^2}\) \(\mathrm{~V}=\sqrt{2500}\) \(\mathrm{~V}=50 \text { Volt }\)
Karnataka CET-2004
Alternating Current
155247
In the circuit shown in the figure, the A.C source gives a voltage $V=20 \cos (2000t)$. Neglecting source resistance, the voltmeter and ammeter reading will be :
155248
In the circuit shown in the figure, neglecting source resistance, the voltmeter and ammeter readings will respectively be :
1 $150 \mathrm{~V}, 8 \mathrm{~A}$
2 $0 \mathrm{~V}, 8 \mathrm{~A}$
3 $0 \mathrm{~V}, 3 \mathrm{~A}$
4 $150 \mathrm{~V}, 3 \mathrm{~A}$
Explanation:
B Given that, $\mathrm{X}_{\mathrm{C}}=\mathrm{X}_{\mathrm{L}}=25 \Omega, \mathrm{V}=240$ Volt, $\mathrm{R}$ $=30 \Omega$ So, potential between $\mathrm{X}_{\mathrm{L}}$ and $\mathrm{X}_{\mathrm{C}}$ is zero Then, $\quad(Z)=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}$ $Z=\sqrt{(30)^{2}+(25-25)^{2}}$ $Z=30 \Omega$ Then, current $(\mathrm{I})=\frac{\mathrm{V}}{\mathrm{Z}}=\frac{240}{30}$ $\mathrm{I}=8 \text { Amp. }$
Karnataka CET-2001
Alternating Current
155249
An LCR series circuit is under resonance. If $I_{m}$ is current amplitude, $V_{m}$ is voltage amplitude, $R$ is the resistance, $Z$ is the impedance, $X_{L}$ is the inductive reactance and $X_{C}$ is the capacitive reactance then,
1 $I_{m}=\frac{V_{m}}{Z}$
2 $I_{m}=\frac{V_{m}}{X_{L}}$
3 $I_{m}=\frac{V_{m}}{X_{C}}$
4 $I_{m}=\frac{V_{m}}{R}$
Explanation:
D Given that, Impedance of the circuit $(Z)=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}$ At resonance, $\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}$ Then, $Z=\sqrt{R^{2}+\left(X_{C}-X_{C}\right)^{2}}$ $\mathrm{Z}=\mathrm{R}$ $\therefore \quad \mathrm{I}_{\mathrm{m}}=\frac{\mathrm{V}_{\mathrm{m}}}{\mathrm{Z}}$ $\mathrm{I}_{\mathrm{m}}=\frac{\mathrm{V}_{\mathrm{m}}}{\mathrm{R}}$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Alternating Current
155245
The value of alternating emf $E$ in the given circuit will be :
1 $220 \mathrm{~V}$
2 $140 \mathrm{~V}$
3 $100 \mathrm{~V}$
4 $20 \mathrm{~V}$
Explanation:
C Given that, $\mathrm{V}_{\mathrm{R}}=80 \mathrm{~V}, \mathrm{~V}_{\mathrm{C}}=100 \mathrm{~V}, \mathrm{~V}_{\mathrm{L}}=40$ $\mathrm{V}$ $\text { In series }(E) =\sqrt{V_{R}^{2}+\left(V_{L}-V_{C}\right)^{2}}$ $E =\sqrt{(80)^{2}+(40-100)^{2}}$ $E =\sqrt{(80)^{2}+(-60)^{2}}$ $E =\sqrt{10000}$ $E =100 \mathrm{~V}$
Karnataka CET-2008
Alternating Current
155246
In a LCR circuit the potential difference between the terminals of the inductance is 60 $\mathrm{V}$, between the terminals of the capacitor is 30 $V$ and that between the terminals of resistance is $40 \mathrm{~V}$. The supply voltage will be equal to :
1 $130 \mathrm{~V}$
2 $10 \mathrm{~V}$
3 $50 \mathrm{~V}$
4 $70 \mathrm{~V}$
Explanation:
C Given, \(\mathrm{V}_{\mathrm{L}}=60 \mathrm{~V}, \mathrm{~V}_{\mathrm{C}}=30 \mathrm{~V}, \mathrm{~V}_{\mathrm{R}}=40 \mathrm{~V}\) We know that, \(\text { Supply voltage }(\mathrm{V})=\sqrt{\mathrm{V}_{\mathrm{R}}^2+\left(\mathrm{V}_{\mathrm{L}}-\mathrm{V}_{\mathrm{C}}\right)^2}\) \(\mathrm{V} =\sqrt{(40)^2+(60-30)^2}\) \(\mathrm{~V}=\sqrt{(40)^2+(30)^2}\) \(\mathrm{~V}=\sqrt{2500}\) \(\mathrm{~V}=50 \text { Volt }\)
Karnataka CET-2004
Alternating Current
155247
In the circuit shown in the figure, the A.C source gives a voltage $V=20 \cos (2000t)$. Neglecting source resistance, the voltmeter and ammeter reading will be :
155248
In the circuit shown in the figure, neglecting source resistance, the voltmeter and ammeter readings will respectively be :
1 $150 \mathrm{~V}, 8 \mathrm{~A}$
2 $0 \mathrm{~V}, 8 \mathrm{~A}$
3 $0 \mathrm{~V}, 3 \mathrm{~A}$
4 $150 \mathrm{~V}, 3 \mathrm{~A}$
Explanation:
B Given that, $\mathrm{X}_{\mathrm{C}}=\mathrm{X}_{\mathrm{L}}=25 \Omega, \mathrm{V}=240$ Volt, $\mathrm{R}$ $=30 \Omega$ So, potential between $\mathrm{X}_{\mathrm{L}}$ and $\mathrm{X}_{\mathrm{C}}$ is zero Then, $\quad(Z)=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}$ $Z=\sqrt{(30)^{2}+(25-25)^{2}}$ $Z=30 \Omega$ Then, current $(\mathrm{I})=\frac{\mathrm{V}}{\mathrm{Z}}=\frac{240}{30}$ $\mathrm{I}=8 \text { Amp. }$
Karnataka CET-2001
Alternating Current
155249
An LCR series circuit is under resonance. If $I_{m}$ is current amplitude, $V_{m}$ is voltage amplitude, $R$ is the resistance, $Z$ is the impedance, $X_{L}$ is the inductive reactance and $X_{C}$ is the capacitive reactance then,
1 $I_{m}=\frac{V_{m}}{Z}$
2 $I_{m}=\frac{V_{m}}{X_{L}}$
3 $I_{m}=\frac{V_{m}}{X_{C}}$
4 $I_{m}=\frac{V_{m}}{R}$
Explanation:
D Given that, Impedance of the circuit $(Z)=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}$ At resonance, $\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}$ Then, $Z=\sqrt{R^{2}+\left(X_{C}-X_{C}\right)^{2}}$ $\mathrm{Z}=\mathrm{R}$ $\therefore \quad \mathrm{I}_{\mathrm{m}}=\frac{\mathrm{V}_{\mathrm{m}}}{\mathrm{Z}}$ $\mathrm{I}_{\mathrm{m}}=\frac{\mathrm{V}_{\mathrm{m}}}{\mathrm{R}}$
155245
The value of alternating emf $E$ in the given circuit will be :
1 $220 \mathrm{~V}$
2 $140 \mathrm{~V}$
3 $100 \mathrm{~V}$
4 $20 \mathrm{~V}$
Explanation:
C Given that, $\mathrm{V}_{\mathrm{R}}=80 \mathrm{~V}, \mathrm{~V}_{\mathrm{C}}=100 \mathrm{~V}, \mathrm{~V}_{\mathrm{L}}=40$ $\mathrm{V}$ $\text { In series }(E) =\sqrt{V_{R}^{2}+\left(V_{L}-V_{C}\right)^{2}}$ $E =\sqrt{(80)^{2}+(40-100)^{2}}$ $E =\sqrt{(80)^{2}+(-60)^{2}}$ $E =\sqrt{10000}$ $E =100 \mathrm{~V}$
Karnataka CET-2008
Alternating Current
155246
In a LCR circuit the potential difference between the terminals of the inductance is 60 $\mathrm{V}$, between the terminals of the capacitor is 30 $V$ and that between the terminals of resistance is $40 \mathrm{~V}$. The supply voltage will be equal to :
1 $130 \mathrm{~V}$
2 $10 \mathrm{~V}$
3 $50 \mathrm{~V}$
4 $70 \mathrm{~V}$
Explanation:
C Given, \(\mathrm{V}_{\mathrm{L}}=60 \mathrm{~V}, \mathrm{~V}_{\mathrm{C}}=30 \mathrm{~V}, \mathrm{~V}_{\mathrm{R}}=40 \mathrm{~V}\) We know that, \(\text { Supply voltage }(\mathrm{V})=\sqrt{\mathrm{V}_{\mathrm{R}}^2+\left(\mathrm{V}_{\mathrm{L}}-\mathrm{V}_{\mathrm{C}}\right)^2}\) \(\mathrm{V} =\sqrt{(40)^2+(60-30)^2}\) \(\mathrm{~V}=\sqrt{(40)^2+(30)^2}\) \(\mathrm{~V}=\sqrt{2500}\) \(\mathrm{~V}=50 \text { Volt }\)
Karnataka CET-2004
Alternating Current
155247
In the circuit shown in the figure, the A.C source gives a voltage $V=20 \cos (2000t)$. Neglecting source resistance, the voltmeter and ammeter reading will be :
155248
In the circuit shown in the figure, neglecting source resistance, the voltmeter and ammeter readings will respectively be :
1 $150 \mathrm{~V}, 8 \mathrm{~A}$
2 $0 \mathrm{~V}, 8 \mathrm{~A}$
3 $0 \mathrm{~V}, 3 \mathrm{~A}$
4 $150 \mathrm{~V}, 3 \mathrm{~A}$
Explanation:
B Given that, $\mathrm{X}_{\mathrm{C}}=\mathrm{X}_{\mathrm{L}}=25 \Omega, \mathrm{V}=240$ Volt, $\mathrm{R}$ $=30 \Omega$ So, potential between $\mathrm{X}_{\mathrm{L}}$ and $\mathrm{X}_{\mathrm{C}}$ is zero Then, $\quad(Z)=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}$ $Z=\sqrt{(30)^{2}+(25-25)^{2}}$ $Z=30 \Omega$ Then, current $(\mathrm{I})=\frac{\mathrm{V}}{\mathrm{Z}}=\frac{240}{30}$ $\mathrm{I}=8 \text { Amp. }$
Karnataka CET-2001
Alternating Current
155249
An LCR series circuit is under resonance. If $I_{m}$ is current amplitude, $V_{m}$ is voltage amplitude, $R$ is the resistance, $Z$ is the impedance, $X_{L}$ is the inductive reactance and $X_{C}$ is the capacitive reactance then,
1 $I_{m}=\frac{V_{m}}{Z}$
2 $I_{m}=\frac{V_{m}}{X_{L}}$
3 $I_{m}=\frac{V_{m}}{X_{C}}$
4 $I_{m}=\frac{V_{m}}{R}$
Explanation:
D Given that, Impedance of the circuit $(Z)=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}$ At resonance, $\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}$ Then, $Z=\sqrt{R^{2}+\left(X_{C}-X_{C}\right)^{2}}$ $\mathrm{Z}=\mathrm{R}$ $\therefore \quad \mathrm{I}_{\mathrm{m}}=\frac{\mathrm{V}_{\mathrm{m}}}{\mathrm{Z}}$ $\mathrm{I}_{\mathrm{m}}=\frac{\mathrm{V}_{\mathrm{m}}}{\mathrm{R}}$
