154962
A $20 \mathrm{~V} \mathrm{AC}$ is applied to a circuit consisting of a resistor and a coil with negligible resistance. If the voltage across the resistor is $12 \mathrm{~V}$, the voltage across the coil is -
1 $16 \mathrm{~V}$
2 $10 \mathrm{~V}$
3 $8 \mathrm{~V}$
4 $6 \mathrm{~V}$
Explanation:
A Given, voltage (V) $=20 \mathrm{~V}$ Voltage across resistor $\left(\mathrm{V}_{\mathrm{R}}\right)=12 \mathrm{~V}$ In L-R combination, voltage is given as $\mathrm{V}=\sqrt{\mathrm{V}_{\mathrm{R}}^{2}+\mathrm{V}_{\mathrm{L}}^{2}}$ Squaring on both side, $\mathrm{V}^{2} =\mathrm{V}_{\mathrm{R}}^{2}+\mathrm{V}_{\mathrm{L}}^{2}$ $\mathrm{~V}_{\mathrm{L}}^{2} =\mathrm{V}^{2}-\mathrm{V}_{\mathrm{R}}^{2}$ $\mathrm{~V}_{\mathrm{L}}^{2} =20^{2}-12^{2}$ $\mathrm{~V}_{\mathrm{L}}^{2} =400-144$ $\mathrm{~V}_{\mathrm{L}}^{2} =256$ $\mathrm{~V}_{\mathrm{L}} =\sqrt{256}$ $\mathrm{~V}_{\mathrm{L}} =16 \mathrm{~V}$ Hence, the voltage across the coil is $16 \mathrm{~V}$.
AP EAMCET-20.08.2021
Alternating Current
154963
An alternating current is given by $i=$ $2 \sin \omega t+6 \cos \omega t$. The rms current in amperes is
154964
A resistance of $20 \Omega$ is connected to an alternating current source of $110 \mathrm{~V}$. If the frequency of the $\mathrm{AC}$ source is $50 \mathrm{~Hz}$, then the time taken by the current to change from its maximum value to the rms value is
1 $4 \mathrm{~ms}$
2 $2.5 \mathrm{~s}$
3 $2 \mathrm{~s}$
4 $2.5 \mathrm{~ms}$
Explanation:
D Given, Resistance $(\mathrm{R})=20 \Omega$ Voltage $(\mathrm{V})=110 \mathrm{~V}$ Frequency $(\mathrm{f})=50 \mathrm{~Hz}$ We know that, $i=i_{0} \cos 2 \pi f t$ $i_{\text {rms }}=i_{0} / \sqrt{2}$ On comparing time to move maximum value to r.m.s value $t =\frac{1}{8 f}$ $t =\frac{1}{8 \times 50}$ $t =2.5 \times 10^{-3} s$ $t =2.5 \mathrm{~ms}$
TS EAMCET 04.08.2021
Alternating Current
154965
In the given circuit the peak voltage across $C, L$ and $R$ are $30 \mathrm{~V}, 110 \mathrm{~V}$ and $60 \mathrm{~V}$ respectively. The rms value of the applied voltage is
1 $100 \mathrm{~V}$
2 $200 \mathrm{~V}$
3 $70.7 \mathrm{~V}$
4 $141 \mathrm{~V}$
Explanation:
C Given, Voltage across resistor $\left(\mathrm{V}_{\mathrm{R}}\right)=60 \mathrm{~V}$ Voltage across capacitor $\left(\mathrm{V}_{\mathrm{C}}\right)=30 \mathrm{~V}$ Voltage across inductor $\left(\mathrm{V}_{\mathrm{L}}\right)=110 \mathrm{~V}$ Resultant voltage in LCR circuit, $\mathrm{V}_{0}=\sqrt{\mathrm{V}_{\mathrm{R}}^{2}+\left(\mathrm{V}_{\mathrm{L}}-\mathrm{V}_{\mathrm{C}}\right)^{2}}$ $\mathrm{~V}_{0}=\sqrt{(60)^{2}+(110-30)^{2}}$ $\mathrm{~V}_{0}=\sqrt{3600+(80)^{2}}$ $\mathrm{~V}_{0}=\sqrt{3600+6400}$ $\mathrm{~V}_{0}=\sqrt{10,000}$ $\mathrm{~V}_{0}=100$ Now, rms voltage, $\mathrm{V}_{\text {rms }} =\frac{\mathrm{V}_{0}}{\sqrt{2}}=\frac{100}{\sqrt{2}}=\frac{50 \times \sqrt{2} \times \sqrt{2}}{\sqrt{2}}$ $\mathrm{~V}_{\text {rms }} =50 \sqrt{2}$ $\mathrm{~V}_{\text {rms }} =50 \times 1.414=70.7 \mathrm{~V}$
154962
A $20 \mathrm{~V} \mathrm{AC}$ is applied to a circuit consisting of a resistor and a coil with negligible resistance. If the voltage across the resistor is $12 \mathrm{~V}$, the voltage across the coil is -
1 $16 \mathrm{~V}$
2 $10 \mathrm{~V}$
3 $8 \mathrm{~V}$
4 $6 \mathrm{~V}$
Explanation:
A Given, voltage (V) $=20 \mathrm{~V}$ Voltage across resistor $\left(\mathrm{V}_{\mathrm{R}}\right)=12 \mathrm{~V}$ In L-R combination, voltage is given as $\mathrm{V}=\sqrt{\mathrm{V}_{\mathrm{R}}^{2}+\mathrm{V}_{\mathrm{L}}^{2}}$ Squaring on both side, $\mathrm{V}^{2} =\mathrm{V}_{\mathrm{R}}^{2}+\mathrm{V}_{\mathrm{L}}^{2}$ $\mathrm{~V}_{\mathrm{L}}^{2} =\mathrm{V}^{2}-\mathrm{V}_{\mathrm{R}}^{2}$ $\mathrm{~V}_{\mathrm{L}}^{2} =20^{2}-12^{2}$ $\mathrm{~V}_{\mathrm{L}}^{2} =400-144$ $\mathrm{~V}_{\mathrm{L}}^{2} =256$ $\mathrm{~V}_{\mathrm{L}} =\sqrt{256}$ $\mathrm{~V}_{\mathrm{L}} =16 \mathrm{~V}$ Hence, the voltage across the coil is $16 \mathrm{~V}$.
AP EAMCET-20.08.2021
Alternating Current
154963
An alternating current is given by $i=$ $2 \sin \omega t+6 \cos \omega t$. The rms current in amperes is
154964
A resistance of $20 \Omega$ is connected to an alternating current source of $110 \mathrm{~V}$. If the frequency of the $\mathrm{AC}$ source is $50 \mathrm{~Hz}$, then the time taken by the current to change from its maximum value to the rms value is
1 $4 \mathrm{~ms}$
2 $2.5 \mathrm{~s}$
3 $2 \mathrm{~s}$
4 $2.5 \mathrm{~ms}$
Explanation:
D Given, Resistance $(\mathrm{R})=20 \Omega$ Voltage $(\mathrm{V})=110 \mathrm{~V}$ Frequency $(\mathrm{f})=50 \mathrm{~Hz}$ We know that, $i=i_{0} \cos 2 \pi f t$ $i_{\text {rms }}=i_{0} / \sqrt{2}$ On comparing time to move maximum value to r.m.s value $t =\frac{1}{8 f}$ $t =\frac{1}{8 \times 50}$ $t =2.5 \times 10^{-3} s$ $t =2.5 \mathrm{~ms}$
TS EAMCET 04.08.2021
Alternating Current
154965
In the given circuit the peak voltage across $C, L$ and $R$ are $30 \mathrm{~V}, 110 \mathrm{~V}$ and $60 \mathrm{~V}$ respectively. The rms value of the applied voltage is
1 $100 \mathrm{~V}$
2 $200 \mathrm{~V}$
3 $70.7 \mathrm{~V}$
4 $141 \mathrm{~V}$
Explanation:
C Given, Voltage across resistor $\left(\mathrm{V}_{\mathrm{R}}\right)=60 \mathrm{~V}$ Voltage across capacitor $\left(\mathrm{V}_{\mathrm{C}}\right)=30 \mathrm{~V}$ Voltage across inductor $\left(\mathrm{V}_{\mathrm{L}}\right)=110 \mathrm{~V}$ Resultant voltage in LCR circuit, $\mathrm{V}_{0}=\sqrt{\mathrm{V}_{\mathrm{R}}^{2}+\left(\mathrm{V}_{\mathrm{L}}-\mathrm{V}_{\mathrm{C}}\right)^{2}}$ $\mathrm{~V}_{0}=\sqrt{(60)^{2}+(110-30)^{2}}$ $\mathrm{~V}_{0}=\sqrt{3600+(80)^{2}}$ $\mathrm{~V}_{0}=\sqrt{3600+6400}$ $\mathrm{~V}_{0}=\sqrt{10,000}$ $\mathrm{~V}_{0}=100$ Now, rms voltage, $\mathrm{V}_{\text {rms }} =\frac{\mathrm{V}_{0}}{\sqrt{2}}=\frac{100}{\sqrt{2}}=\frac{50 \times \sqrt{2} \times \sqrt{2}}{\sqrt{2}}$ $\mathrm{~V}_{\text {rms }} =50 \sqrt{2}$ $\mathrm{~V}_{\text {rms }} =50 \times 1.414=70.7 \mathrm{~V}$
154962
A $20 \mathrm{~V} \mathrm{AC}$ is applied to a circuit consisting of a resistor and a coil with negligible resistance. If the voltage across the resistor is $12 \mathrm{~V}$, the voltage across the coil is -
1 $16 \mathrm{~V}$
2 $10 \mathrm{~V}$
3 $8 \mathrm{~V}$
4 $6 \mathrm{~V}$
Explanation:
A Given, voltage (V) $=20 \mathrm{~V}$ Voltage across resistor $\left(\mathrm{V}_{\mathrm{R}}\right)=12 \mathrm{~V}$ In L-R combination, voltage is given as $\mathrm{V}=\sqrt{\mathrm{V}_{\mathrm{R}}^{2}+\mathrm{V}_{\mathrm{L}}^{2}}$ Squaring on both side, $\mathrm{V}^{2} =\mathrm{V}_{\mathrm{R}}^{2}+\mathrm{V}_{\mathrm{L}}^{2}$ $\mathrm{~V}_{\mathrm{L}}^{2} =\mathrm{V}^{2}-\mathrm{V}_{\mathrm{R}}^{2}$ $\mathrm{~V}_{\mathrm{L}}^{2} =20^{2}-12^{2}$ $\mathrm{~V}_{\mathrm{L}}^{2} =400-144$ $\mathrm{~V}_{\mathrm{L}}^{2} =256$ $\mathrm{~V}_{\mathrm{L}} =\sqrt{256}$ $\mathrm{~V}_{\mathrm{L}} =16 \mathrm{~V}$ Hence, the voltage across the coil is $16 \mathrm{~V}$.
AP EAMCET-20.08.2021
Alternating Current
154963
An alternating current is given by $i=$ $2 \sin \omega t+6 \cos \omega t$. The rms current in amperes is
154964
A resistance of $20 \Omega$ is connected to an alternating current source of $110 \mathrm{~V}$. If the frequency of the $\mathrm{AC}$ source is $50 \mathrm{~Hz}$, then the time taken by the current to change from its maximum value to the rms value is
1 $4 \mathrm{~ms}$
2 $2.5 \mathrm{~s}$
3 $2 \mathrm{~s}$
4 $2.5 \mathrm{~ms}$
Explanation:
D Given, Resistance $(\mathrm{R})=20 \Omega$ Voltage $(\mathrm{V})=110 \mathrm{~V}$ Frequency $(\mathrm{f})=50 \mathrm{~Hz}$ We know that, $i=i_{0} \cos 2 \pi f t$ $i_{\text {rms }}=i_{0} / \sqrt{2}$ On comparing time to move maximum value to r.m.s value $t =\frac{1}{8 f}$ $t =\frac{1}{8 \times 50}$ $t =2.5 \times 10^{-3} s$ $t =2.5 \mathrm{~ms}$
TS EAMCET 04.08.2021
Alternating Current
154965
In the given circuit the peak voltage across $C, L$ and $R$ are $30 \mathrm{~V}, 110 \mathrm{~V}$ and $60 \mathrm{~V}$ respectively. The rms value of the applied voltage is
1 $100 \mathrm{~V}$
2 $200 \mathrm{~V}$
3 $70.7 \mathrm{~V}$
4 $141 \mathrm{~V}$
Explanation:
C Given, Voltage across resistor $\left(\mathrm{V}_{\mathrm{R}}\right)=60 \mathrm{~V}$ Voltage across capacitor $\left(\mathrm{V}_{\mathrm{C}}\right)=30 \mathrm{~V}$ Voltage across inductor $\left(\mathrm{V}_{\mathrm{L}}\right)=110 \mathrm{~V}$ Resultant voltage in LCR circuit, $\mathrm{V}_{0}=\sqrt{\mathrm{V}_{\mathrm{R}}^{2}+\left(\mathrm{V}_{\mathrm{L}}-\mathrm{V}_{\mathrm{C}}\right)^{2}}$ $\mathrm{~V}_{0}=\sqrt{(60)^{2}+(110-30)^{2}}$ $\mathrm{~V}_{0}=\sqrt{3600+(80)^{2}}$ $\mathrm{~V}_{0}=\sqrt{3600+6400}$ $\mathrm{~V}_{0}=\sqrt{10,000}$ $\mathrm{~V}_{0}=100$ Now, rms voltage, $\mathrm{V}_{\text {rms }} =\frac{\mathrm{V}_{0}}{\sqrt{2}}=\frac{100}{\sqrt{2}}=\frac{50 \times \sqrt{2} \times \sqrt{2}}{\sqrt{2}}$ $\mathrm{~V}_{\text {rms }} =50 \sqrt{2}$ $\mathrm{~V}_{\text {rms }} =50 \times 1.414=70.7 \mathrm{~V}$
154962
A $20 \mathrm{~V} \mathrm{AC}$ is applied to a circuit consisting of a resistor and a coil with negligible resistance. If the voltage across the resistor is $12 \mathrm{~V}$, the voltage across the coil is -
1 $16 \mathrm{~V}$
2 $10 \mathrm{~V}$
3 $8 \mathrm{~V}$
4 $6 \mathrm{~V}$
Explanation:
A Given, voltage (V) $=20 \mathrm{~V}$ Voltage across resistor $\left(\mathrm{V}_{\mathrm{R}}\right)=12 \mathrm{~V}$ In L-R combination, voltage is given as $\mathrm{V}=\sqrt{\mathrm{V}_{\mathrm{R}}^{2}+\mathrm{V}_{\mathrm{L}}^{2}}$ Squaring on both side, $\mathrm{V}^{2} =\mathrm{V}_{\mathrm{R}}^{2}+\mathrm{V}_{\mathrm{L}}^{2}$ $\mathrm{~V}_{\mathrm{L}}^{2} =\mathrm{V}^{2}-\mathrm{V}_{\mathrm{R}}^{2}$ $\mathrm{~V}_{\mathrm{L}}^{2} =20^{2}-12^{2}$ $\mathrm{~V}_{\mathrm{L}}^{2} =400-144$ $\mathrm{~V}_{\mathrm{L}}^{2} =256$ $\mathrm{~V}_{\mathrm{L}} =\sqrt{256}$ $\mathrm{~V}_{\mathrm{L}} =16 \mathrm{~V}$ Hence, the voltage across the coil is $16 \mathrm{~V}$.
AP EAMCET-20.08.2021
Alternating Current
154963
An alternating current is given by $i=$ $2 \sin \omega t+6 \cos \omega t$. The rms current in amperes is
154964
A resistance of $20 \Omega$ is connected to an alternating current source of $110 \mathrm{~V}$. If the frequency of the $\mathrm{AC}$ source is $50 \mathrm{~Hz}$, then the time taken by the current to change from its maximum value to the rms value is
1 $4 \mathrm{~ms}$
2 $2.5 \mathrm{~s}$
3 $2 \mathrm{~s}$
4 $2.5 \mathrm{~ms}$
Explanation:
D Given, Resistance $(\mathrm{R})=20 \Omega$ Voltage $(\mathrm{V})=110 \mathrm{~V}$ Frequency $(\mathrm{f})=50 \mathrm{~Hz}$ We know that, $i=i_{0} \cos 2 \pi f t$ $i_{\text {rms }}=i_{0} / \sqrt{2}$ On comparing time to move maximum value to r.m.s value $t =\frac{1}{8 f}$ $t =\frac{1}{8 \times 50}$ $t =2.5 \times 10^{-3} s$ $t =2.5 \mathrm{~ms}$
TS EAMCET 04.08.2021
Alternating Current
154965
In the given circuit the peak voltage across $C, L$ and $R$ are $30 \mathrm{~V}, 110 \mathrm{~V}$ and $60 \mathrm{~V}$ respectively. The rms value of the applied voltage is
1 $100 \mathrm{~V}$
2 $200 \mathrm{~V}$
3 $70.7 \mathrm{~V}$
4 $141 \mathrm{~V}$
Explanation:
C Given, Voltage across resistor $\left(\mathrm{V}_{\mathrm{R}}\right)=60 \mathrm{~V}$ Voltage across capacitor $\left(\mathrm{V}_{\mathrm{C}}\right)=30 \mathrm{~V}$ Voltage across inductor $\left(\mathrm{V}_{\mathrm{L}}\right)=110 \mathrm{~V}$ Resultant voltage in LCR circuit, $\mathrm{V}_{0}=\sqrt{\mathrm{V}_{\mathrm{R}}^{2}+\left(\mathrm{V}_{\mathrm{L}}-\mathrm{V}_{\mathrm{C}}\right)^{2}}$ $\mathrm{~V}_{0}=\sqrt{(60)^{2}+(110-30)^{2}}$ $\mathrm{~V}_{0}=\sqrt{3600+(80)^{2}}$ $\mathrm{~V}_{0}=\sqrt{3600+6400}$ $\mathrm{~V}_{0}=\sqrt{10,000}$ $\mathrm{~V}_{0}=100$ Now, rms voltage, $\mathrm{V}_{\text {rms }} =\frac{\mathrm{V}_{0}}{\sqrt{2}}=\frac{100}{\sqrt{2}}=\frac{50 \times \sqrt{2} \times \sqrt{2}}{\sqrt{2}}$ $\mathrm{~V}_{\text {rms }} =50 \sqrt{2}$ $\mathrm{~V}_{\text {rms }} =50 \times 1.414=70.7 \mathrm{~V}$
