154998
A circuit element is connected to an alternating voltage supply of variable frequency. When the frequency is increased, current in the circuit is found to remain same. The circuit element is
1 Capacitor only
2 Resistor only
3 Inductor only
4 Either a capacitor or an inductor
Explanation:
B Resistance is not dependent upon frequency of AC source. So the resistance of the circuit remains same and current will be same as it depends on the amplitude of AC source only. Read the following passage and answer question No. 45: AC supply represented by $\mathrm{V}=141.4 \sin (100 \pi \mathrm{t})$ is applied to a series L-C-R circuit of impedance $50 \Omega$ where resistance has a value $40 \Omega$. The circuit is behaving capacitive.
COMEDK 2011
Alternating Current
155002
The peak value of alternating voltage is 846 volts. Its root mean square value is
1 $\frac{846}{\sqrt{2}}$ volts
2 846 volts
3 $846 \sqrt{2}$ Volts
4 zero
Explanation:
A Given, Peak voltage, $\mathrm{V}_{0}=846 \mathrm{~V}$ The rms value of voltage is, $I_{\mathrm{rms}}=\frac{\mathrm{V}_{0}}{\sqrt{2}}=\frac{846}{\sqrt{2}} \mathrm{~V}$
COMEDK 2016
Alternating Current
155008
RMS value of $\mathrm{AC}$ is of the peak value.
1 $7 \%$
2 $7.7 \%$
3 $70 \%$
4 $70.7 \%$
Explanation:
D We know that, $\mathrm{I}_{\mathrm{rms}}=\frac{\mathrm{I}_{0}}{\sqrt{2}}=0.707 \mathrm{I}_{0}$ $\mathrm{I}_{0}=\text { Peak value }$ $\therefore$ It is $70.7 \%$ of peak value.
VITEEE-2006
Alternating Current
155012
The time taken by an alternating current of 50 $\mathrm{Hz}$ in reaching from zero to its maximum value will be
1 $0.5 \mathrm{~s}$
2 $0.005 \mathrm{~s}$
3 $0.05 \mathrm{~s}$
4 $5 \mathrm{~s}$
Explanation:
B Given, $\mathrm{f}=50 \mathrm{~Hz}$ Time required to reach from zero to maximum is quarter time period. $\mathrm{T}=\frac{1}{\mathrm{f}}=\frac{1}{50}=0.02 \mathrm{sec}$ Time required $=\frac{\mathrm{T}}{4}=\frac{0.02}{4}=0.005 \mathrm{~s}$
154998
A circuit element is connected to an alternating voltage supply of variable frequency. When the frequency is increased, current in the circuit is found to remain same. The circuit element is
1 Capacitor only
2 Resistor only
3 Inductor only
4 Either a capacitor or an inductor
Explanation:
B Resistance is not dependent upon frequency of AC source. So the resistance of the circuit remains same and current will be same as it depends on the amplitude of AC source only. Read the following passage and answer question No. 45: AC supply represented by $\mathrm{V}=141.4 \sin (100 \pi \mathrm{t})$ is applied to a series L-C-R circuit of impedance $50 \Omega$ where resistance has a value $40 \Omega$. The circuit is behaving capacitive.
COMEDK 2011
Alternating Current
155002
The peak value of alternating voltage is 846 volts. Its root mean square value is
1 $\frac{846}{\sqrt{2}}$ volts
2 846 volts
3 $846 \sqrt{2}$ Volts
4 zero
Explanation:
A Given, Peak voltage, $\mathrm{V}_{0}=846 \mathrm{~V}$ The rms value of voltage is, $I_{\mathrm{rms}}=\frac{\mathrm{V}_{0}}{\sqrt{2}}=\frac{846}{\sqrt{2}} \mathrm{~V}$
COMEDK 2016
Alternating Current
155008
RMS value of $\mathrm{AC}$ is of the peak value.
1 $7 \%$
2 $7.7 \%$
3 $70 \%$
4 $70.7 \%$
Explanation:
D We know that, $\mathrm{I}_{\mathrm{rms}}=\frac{\mathrm{I}_{0}}{\sqrt{2}}=0.707 \mathrm{I}_{0}$ $\mathrm{I}_{0}=\text { Peak value }$ $\therefore$ It is $70.7 \%$ of peak value.
VITEEE-2006
Alternating Current
155012
The time taken by an alternating current of 50 $\mathrm{Hz}$ in reaching from zero to its maximum value will be
1 $0.5 \mathrm{~s}$
2 $0.005 \mathrm{~s}$
3 $0.05 \mathrm{~s}$
4 $5 \mathrm{~s}$
Explanation:
B Given, $\mathrm{f}=50 \mathrm{~Hz}$ Time required to reach from zero to maximum is quarter time period. $\mathrm{T}=\frac{1}{\mathrm{f}}=\frac{1}{50}=0.02 \mathrm{sec}$ Time required $=\frac{\mathrm{T}}{4}=\frac{0.02}{4}=0.005 \mathrm{~s}$
154998
A circuit element is connected to an alternating voltage supply of variable frequency. When the frequency is increased, current in the circuit is found to remain same. The circuit element is
1 Capacitor only
2 Resistor only
3 Inductor only
4 Either a capacitor or an inductor
Explanation:
B Resistance is not dependent upon frequency of AC source. So the resistance of the circuit remains same and current will be same as it depends on the amplitude of AC source only. Read the following passage and answer question No. 45: AC supply represented by $\mathrm{V}=141.4 \sin (100 \pi \mathrm{t})$ is applied to a series L-C-R circuit of impedance $50 \Omega$ where resistance has a value $40 \Omega$. The circuit is behaving capacitive.
COMEDK 2011
Alternating Current
155002
The peak value of alternating voltage is 846 volts. Its root mean square value is
1 $\frac{846}{\sqrt{2}}$ volts
2 846 volts
3 $846 \sqrt{2}$ Volts
4 zero
Explanation:
A Given, Peak voltage, $\mathrm{V}_{0}=846 \mathrm{~V}$ The rms value of voltage is, $I_{\mathrm{rms}}=\frac{\mathrm{V}_{0}}{\sqrt{2}}=\frac{846}{\sqrt{2}} \mathrm{~V}$
COMEDK 2016
Alternating Current
155008
RMS value of $\mathrm{AC}$ is of the peak value.
1 $7 \%$
2 $7.7 \%$
3 $70 \%$
4 $70.7 \%$
Explanation:
D We know that, $\mathrm{I}_{\mathrm{rms}}=\frac{\mathrm{I}_{0}}{\sqrt{2}}=0.707 \mathrm{I}_{0}$ $\mathrm{I}_{0}=\text { Peak value }$ $\therefore$ It is $70.7 \%$ of peak value.
VITEEE-2006
Alternating Current
155012
The time taken by an alternating current of 50 $\mathrm{Hz}$ in reaching from zero to its maximum value will be
1 $0.5 \mathrm{~s}$
2 $0.005 \mathrm{~s}$
3 $0.05 \mathrm{~s}$
4 $5 \mathrm{~s}$
Explanation:
B Given, $\mathrm{f}=50 \mathrm{~Hz}$ Time required to reach from zero to maximum is quarter time period. $\mathrm{T}=\frac{1}{\mathrm{f}}=\frac{1}{50}=0.02 \mathrm{sec}$ Time required $=\frac{\mathrm{T}}{4}=\frac{0.02}{4}=0.005 \mathrm{~s}$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Alternating Current
154998
A circuit element is connected to an alternating voltage supply of variable frequency. When the frequency is increased, current in the circuit is found to remain same. The circuit element is
1 Capacitor only
2 Resistor only
3 Inductor only
4 Either a capacitor or an inductor
Explanation:
B Resistance is not dependent upon frequency of AC source. So the resistance of the circuit remains same and current will be same as it depends on the amplitude of AC source only. Read the following passage and answer question No. 45: AC supply represented by $\mathrm{V}=141.4 \sin (100 \pi \mathrm{t})$ is applied to a series L-C-R circuit of impedance $50 \Omega$ where resistance has a value $40 \Omega$. The circuit is behaving capacitive.
COMEDK 2011
Alternating Current
155002
The peak value of alternating voltage is 846 volts. Its root mean square value is
1 $\frac{846}{\sqrt{2}}$ volts
2 846 volts
3 $846 \sqrt{2}$ Volts
4 zero
Explanation:
A Given, Peak voltage, $\mathrm{V}_{0}=846 \mathrm{~V}$ The rms value of voltage is, $I_{\mathrm{rms}}=\frac{\mathrm{V}_{0}}{\sqrt{2}}=\frac{846}{\sqrt{2}} \mathrm{~V}$
COMEDK 2016
Alternating Current
155008
RMS value of $\mathrm{AC}$ is of the peak value.
1 $7 \%$
2 $7.7 \%$
3 $70 \%$
4 $70.7 \%$
Explanation:
D We know that, $\mathrm{I}_{\mathrm{rms}}=\frac{\mathrm{I}_{0}}{\sqrt{2}}=0.707 \mathrm{I}_{0}$ $\mathrm{I}_{0}=\text { Peak value }$ $\therefore$ It is $70.7 \%$ of peak value.
VITEEE-2006
Alternating Current
155012
The time taken by an alternating current of 50 $\mathrm{Hz}$ in reaching from zero to its maximum value will be
1 $0.5 \mathrm{~s}$
2 $0.005 \mathrm{~s}$
3 $0.05 \mathrm{~s}$
4 $5 \mathrm{~s}$
Explanation:
B Given, $\mathrm{f}=50 \mathrm{~Hz}$ Time required to reach from zero to maximum is quarter time period. $\mathrm{T}=\frac{1}{\mathrm{f}}=\frac{1}{50}=0.02 \mathrm{sec}$ Time required $=\frac{\mathrm{T}}{4}=\frac{0.02}{4}=0.005 \mathrm{~s}$
