274641
If an LCR series circuit is connected to an ac source, then at resonance the voltage across
1 $\text{R}$ is zero
2 $\text{R}$ equals the applied voltage
3 $\text{C}$ is zero
4 L equals the applied voltage
Explanation:
(b) In series RLC circuit,
Voltage, $\text{V}=\sqrt{\text{V}_{\text{R}}^{2}+{{\left( {{\text{V}}_{\text{L}}}-{{\text{V}}_{\text{C}}} \right)}^{2}}}$
And, at resonance, ${{\text{V}}_{\text{L}}}={{\text{V}}_{\text{C}}}$
Hence, $V={{V}_{R}}$
NCERT Page-245 / N-189
AC (NCERT)
274642
In a series resonant circuit, having $\text{L},\text{C}$ and $\text{R}$ as its elements, the resonant current is $i$. The power dissipated in circuit at resonance is
1 $\frac{{{i}^{2}}R}{\left( \omega L-1/\omega C \right)}$
2 zero
3 ${{i}^{2}}\omega \text{L}$
4 ${{\text{i}}^{2}}\text{R}$
Whereas $\omega $ is angular resonant frequency
Explanation:
(d) At resonance $\omega \text{L}=1/\omega \text{C}$
and $i=E/R$, So power dissipated in circuit is $P={{i}^{2}}R$.
NCERT Page-245 / N-189
AC (NCERT)
274643
At resonant frequency the current amplitude in series $LCR$ circuit is
1 maximum
2 minimum (
3 zero
4 infinity
Explanation:
(a)
NCERT Page-245 / N-189
AC (NCERT)
274644
An A. C. of frequency $f$ is flowing in a circuit containing a resistance $R$ and capacitance $C$ in series. The impedance of the circuit is equal to
274641
If an LCR series circuit is connected to an ac source, then at resonance the voltage across
1 $\text{R}$ is zero
2 $\text{R}$ equals the applied voltage
3 $\text{C}$ is zero
4 L equals the applied voltage
Explanation:
(b) In series RLC circuit,
Voltage, $\text{V}=\sqrt{\text{V}_{\text{R}}^{2}+{{\left( {{\text{V}}_{\text{L}}}-{{\text{V}}_{\text{C}}} \right)}^{2}}}$
And, at resonance, ${{\text{V}}_{\text{L}}}={{\text{V}}_{\text{C}}}$
Hence, $V={{V}_{R}}$
NCERT Page-245 / N-189
AC (NCERT)
274642
In a series resonant circuit, having $\text{L},\text{C}$ and $\text{R}$ as its elements, the resonant current is $i$. The power dissipated in circuit at resonance is
1 $\frac{{{i}^{2}}R}{\left( \omega L-1/\omega C \right)}$
2 zero
3 ${{i}^{2}}\omega \text{L}$
4 ${{\text{i}}^{2}}\text{R}$
Whereas $\omega $ is angular resonant frequency
Explanation:
(d) At resonance $\omega \text{L}=1/\omega \text{C}$
and $i=E/R$, So power dissipated in circuit is $P={{i}^{2}}R$.
NCERT Page-245 / N-189
AC (NCERT)
274643
At resonant frequency the current amplitude in series $LCR$ circuit is
1 maximum
2 minimum (
3 zero
4 infinity
Explanation:
(a)
NCERT Page-245 / N-189
AC (NCERT)
274644
An A. C. of frequency $f$ is flowing in a circuit containing a resistance $R$ and capacitance $C$ in series. The impedance of the circuit is equal to
274641
If an LCR series circuit is connected to an ac source, then at resonance the voltage across
1 $\text{R}$ is zero
2 $\text{R}$ equals the applied voltage
3 $\text{C}$ is zero
4 L equals the applied voltage
Explanation:
(b) In series RLC circuit,
Voltage, $\text{V}=\sqrt{\text{V}_{\text{R}}^{2}+{{\left( {{\text{V}}_{\text{L}}}-{{\text{V}}_{\text{C}}} \right)}^{2}}}$
And, at resonance, ${{\text{V}}_{\text{L}}}={{\text{V}}_{\text{C}}}$
Hence, $V={{V}_{R}}$
NCERT Page-245 / N-189
AC (NCERT)
274642
In a series resonant circuit, having $\text{L},\text{C}$ and $\text{R}$ as its elements, the resonant current is $i$. The power dissipated in circuit at resonance is
1 $\frac{{{i}^{2}}R}{\left( \omega L-1/\omega C \right)}$
2 zero
3 ${{i}^{2}}\omega \text{L}$
4 ${{\text{i}}^{2}}\text{R}$
Whereas $\omega $ is angular resonant frequency
Explanation:
(d) At resonance $\omega \text{L}=1/\omega \text{C}$
and $i=E/R$, So power dissipated in circuit is $P={{i}^{2}}R$.
NCERT Page-245 / N-189
AC (NCERT)
274643
At resonant frequency the current amplitude in series $LCR$ circuit is
1 maximum
2 minimum (
3 zero
4 infinity
Explanation:
(a)
NCERT Page-245 / N-189
AC (NCERT)
274644
An A. C. of frequency $f$ is flowing in a circuit containing a resistance $R$ and capacitance $C$ in series. The impedance of the circuit is equal to
274641
If an LCR series circuit is connected to an ac source, then at resonance the voltage across
1 $\text{R}$ is zero
2 $\text{R}$ equals the applied voltage
3 $\text{C}$ is zero
4 L equals the applied voltage
Explanation:
(b) In series RLC circuit,
Voltage, $\text{V}=\sqrt{\text{V}_{\text{R}}^{2}+{{\left( {{\text{V}}_{\text{L}}}-{{\text{V}}_{\text{C}}} \right)}^{2}}}$
And, at resonance, ${{\text{V}}_{\text{L}}}={{\text{V}}_{\text{C}}}$
Hence, $V={{V}_{R}}$
NCERT Page-245 / N-189
AC (NCERT)
274642
In a series resonant circuit, having $\text{L},\text{C}$ and $\text{R}$ as its elements, the resonant current is $i$. The power dissipated in circuit at resonance is
1 $\frac{{{i}^{2}}R}{\left( \omega L-1/\omega C \right)}$
2 zero
3 ${{i}^{2}}\omega \text{L}$
4 ${{\text{i}}^{2}}\text{R}$
Whereas $\omega $ is angular resonant frequency
Explanation:
(d) At resonance $\omega \text{L}=1/\omega \text{C}$
and $i=E/R$, So power dissipated in circuit is $P={{i}^{2}}R$.
NCERT Page-245 / N-189
AC (NCERT)
274643
At resonant frequency the current amplitude in series $LCR$ circuit is
1 maximum
2 minimum (
3 zero
4 infinity
Explanation:
(a)
NCERT Page-245 / N-189
AC (NCERT)
274644
An A. C. of frequency $f$ is flowing in a circuit containing a resistance $R$ and capacitance $C$ in series. The impedance of the circuit is equal to
