362968
A galvanometer may be converted into ammeter or a voltmeter.The resistance of the device so obtained will be the largest in case of
1 Ammeter of range 1\(A\)
2 Ammeter of range 10\(A\)
3 Voltmeter of range 1\(V\)
4 Voltmeter of range 10\(V\)
Explanation:
A voltmeter having more range has the highest resistance.
PHXII04:MOVING CHARGES AND MAGNETISM
362969
Three voltmeters, all having different internal resistances are joined as shown in figure. When some potential difference is applied across \(A\) and \(B\), their readings are \({V_1},{\rm{ }}{V_2}\) and \({V_3}\). Choose the correct options.
1 \({V_1} \ne {V_3} - {V_2}\)
2 \({V_1} + {V_2} = {V_3}\)
3 \({V_1} = {V_2}\)
4 \({V_1} + {V_2} > {V_2}\)
Explanation:
Using ohm's law across points \(A\) and \(B\) \(V=i R\) \(V_{A B}=I_{1}\left(R_{1}+R_{2}\right)\) \(V_{A B}=I_{2} R_{3}\) \({I_1}\left( {{R_1} + {R_2}} \right) = {I_2}{R_3}\) \( \Rightarrow {V_1} + {V_2} = {V_3}\)
JEE - 2024
PHXII04:MOVING CHARGES AND MAGNETISM
362970
A voltmeter having resistance of \(50 \times {10^3}ohm\) is used to measure the voltage in a circuit. To increase the range of measurement 3 times the additional series resistance required is
1 \({10^5}ohm\)
2 \(9 \times {10^6}ohm\)
3 \(150k.ohm\)
4 \(900k.ohm\)
Explanation:
Given that \({R_v} = 50 \times {10^3}ohm\) The range of a voltmeter is \(\mathrm{V}=\mathrm{i}_{\mathrm{g}} \mathrm{R}_{\mathrm{v}}\) As the range is to be made three times \(3 \mathrm{~V}=\mathrm{i}_{\mathrm{g}}\left(\mathrm{R}_{\mathrm{v}}+\mathrm{S}\right)\) From both the equations we get \(\mathrm{S}=10^{5} \Omega\)
PHXII04:MOVING CHARGES AND MAGNETISM
362971
A galvanometer has a resistance of \(50\,\Omega \) and it allows maximum current of \(5\,mA.\) It can be converted into voltmeter to measure upto \(100\,V\) by connecting it in series to a resistor of resistance
1 \(20050\,\Omega \)
2 \(19500\,\Omega \)
3 \(5975\,\Omega \)
4 \(19950\,\Omega \)
Explanation:
Given, \({R_G} = 50\,\Omega ,I = 5\;mA,\;V = 100\;V\) Let the resistance is \(R\). \(\left(R+R_{G}\right) I=V\) \((R+50) \times 5 \times 10^{-3}=100\) \(R = 20000 - 50 = 19950\,\Omega \)
NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXII04:MOVING CHARGES AND MAGNETISM
362968
A galvanometer may be converted into ammeter or a voltmeter.The resistance of the device so obtained will be the largest in case of
1 Ammeter of range 1\(A\)
2 Ammeter of range 10\(A\)
3 Voltmeter of range 1\(V\)
4 Voltmeter of range 10\(V\)
Explanation:
A voltmeter having more range has the highest resistance.
PHXII04:MOVING CHARGES AND MAGNETISM
362969
Three voltmeters, all having different internal resistances are joined as shown in figure. When some potential difference is applied across \(A\) and \(B\), their readings are \({V_1},{\rm{ }}{V_2}\) and \({V_3}\). Choose the correct options.
1 \({V_1} \ne {V_3} - {V_2}\)
2 \({V_1} + {V_2} = {V_3}\)
3 \({V_1} = {V_2}\)
4 \({V_1} + {V_2} > {V_2}\)
Explanation:
Using ohm's law across points \(A\) and \(B\) \(V=i R\) \(V_{A B}=I_{1}\left(R_{1}+R_{2}\right)\) \(V_{A B}=I_{2} R_{3}\) \({I_1}\left( {{R_1} + {R_2}} \right) = {I_2}{R_3}\) \( \Rightarrow {V_1} + {V_2} = {V_3}\)
JEE - 2024
PHXII04:MOVING CHARGES AND MAGNETISM
362970
A voltmeter having resistance of \(50 \times {10^3}ohm\) is used to measure the voltage in a circuit. To increase the range of measurement 3 times the additional series resistance required is
1 \({10^5}ohm\)
2 \(9 \times {10^6}ohm\)
3 \(150k.ohm\)
4 \(900k.ohm\)
Explanation:
Given that \({R_v} = 50 \times {10^3}ohm\) The range of a voltmeter is \(\mathrm{V}=\mathrm{i}_{\mathrm{g}} \mathrm{R}_{\mathrm{v}}\) As the range is to be made three times \(3 \mathrm{~V}=\mathrm{i}_{\mathrm{g}}\left(\mathrm{R}_{\mathrm{v}}+\mathrm{S}\right)\) From both the equations we get \(\mathrm{S}=10^{5} \Omega\)
PHXII04:MOVING CHARGES AND MAGNETISM
362971
A galvanometer has a resistance of \(50\,\Omega \) and it allows maximum current of \(5\,mA.\) It can be converted into voltmeter to measure upto \(100\,V\) by connecting it in series to a resistor of resistance
1 \(20050\,\Omega \)
2 \(19500\,\Omega \)
3 \(5975\,\Omega \)
4 \(19950\,\Omega \)
Explanation:
Given, \({R_G} = 50\,\Omega ,I = 5\;mA,\;V = 100\;V\) Let the resistance is \(R\). \(\left(R+R_{G}\right) I=V\) \((R+50) \times 5 \times 10^{-3}=100\) \(R = 20000 - 50 = 19950\,\Omega \)
362968
A galvanometer may be converted into ammeter or a voltmeter.The resistance of the device so obtained will be the largest in case of
1 Ammeter of range 1\(A\)
2 Ammeter of range 10\(A\)
3 Voltmeter of range 1\(V\)
4 Voltmeter of range 10\(V\)
Explanation:
A voltmeter having more range has the highest resistance.
PHXII04:MOVING CHARGES AND MAGNETISM
362969
Three voltmeters, all having different internal resistances are joined as shown in figure. When some potential difference is applied across \(A\) and \(B\), their readings are \({V_1},{\rm{ }}{V_2}\) and \({V_3}\). Choose the correct options.
1 \({V_1} \ne {V_3} - {V_2}\)
2 \({V_1} + {V_2} = {V_3}\)
3 \({V_1} = {V_2}\)
4 \({V_1} + {V_2} > {V_2}\)
Explanation:
Using ohm's law across points \(A\) and \(B\) \(V=i R\) \(V_{A B}=I_{1}\left(R_{1}+R_{2}\right)\) \(V_{A B}=I_{2} R_{3}\) \({I_1}\left( {{R_1} + {R_2}} \right) = {I_2}{R_3}\) \( \Rightarrow {V_1} + {V_2} = {V_3}\)
JEE - 2024
PHXII04:MOVING CHARGES AND MAGNETISM
362970
A voltmeter having resistance of \(50 \times {10^3}ohm\) is used to measure the voltage in a circuit. To increase the range of measurement 3 times the additional series resistance required is
1 \({10^5}ohm\)
2 \(9 \times {10^6}ohm\)
3 \(150k.ohm\)
4 \(900k.ohm\)
Explanation:
Given that \({R_v} = 50 \times {10^3}ohm\) The range of a voltmeter is \(\mathrm{V}=\mathrm{i}_{\mathrm{g}} \mathrm{R}_{\mathrm{v}}\) As the range is to be made three times \(3 \mathrm{~V}=\mathrm{i}_{\mathrm{g}}\left(\mathrm{R}_{\mathrm{v}}+\mathrm{S}\right)\) From both the equations we get \(\mathrm{S}=10^{5} \Omega\)
PHXII04:MOVING CHARGES AND MAGNETISM
362971
A galvanometer has a resistance of \(50\,\Omega \) and it allows maximum current of \(5\,mA.\) It can be converted into voltmeter to measure upto \(100\,V\) by connecting it in series to a resistor of resistance
1 \(20050\,\Omega \)
2 \(19500\,\Omega \)
3 \(5975\,\Omega \)
4 \(19950\,\Omega \)
Explanation:
Given, \({R_G} = 50\,\Omega ,I = 5\;mA,\;V = 100\;V\) Let the resistance is \(R\). \(\left(R+R_{G}\right) I=V\) \((R+50) \times 5 \times 10^{-3}=100\) \(R = 20000 - 50 = 19950\,\Omega \)
362968
A galvanometer may be converted into ammeter or a voltmeter.The resistance of the device so obtained will be the largest in case of
1 Ammeter of range 1\(A\)
2 Ammeter of range 10\(A\)
3 Voltmeter of range 1\(V\)
4 Voltmeter of range 10\(V\)
Explanation:
A voltmeter having more range has the highest resistance.
PHXII04:MOVING CHARGES AND MAGNETISM
362969
Three voltmeters, all having different internal resistances are joined as shown in figure. When some potential difference is applied across \(A\) and \(B\), their readings are \({V_1},{\rm{ }}{V_2}\) and \({V_3}\). Choose the correct options.
1 \({V_1} \ne {V_3} - {V_2}\)
2 \({V_1} + {V_2} = {V_3}\)
3 \({V_1} = {V_2}\)
4 \({V_1} + {V_2} > {V_2}\)
Explanation:
Using ohm's law across points \(A\) and \(B\) \(V=i R\) \(V_{A B}=I_{1}\left(R_{1}+R_{2}\right)\) \(V_{A B}=I_{2} R_{3}\) \({I_1}\left( {{R_1} + {R_2}} \right) = {I_2}{R_3}\) \( \Rightarrow {V_1} + {V_2} = {V_3}\)
JEE - 2024
PHXII04:MOVING CHARGES AND MAGNETISM
362970
A voltmeter having resistance of \(50 \times {10^3}ohm\) is used to measure the voltage in a circuit. To increase the range of measurement 3 times the additional series resistance required is
1 \({10^5}ohm\)
2 \(9 \times {10^6}ohm\)
3 \(150k.ohm\)
4 \(900k.ohm\)
Explanation:
Given that \({R_v} = 50 \times {10^3}ohm\) The range of a voltmeter is \(\mathrm{V}=\mathrm{i}_{\mathrm{g}} \mathrm{R}_{\mathrm{v}}\) As the range is to be made three times \(3 \mathrm{~V}=\mathrm{i}_{\mathrm{g}}\left(\mathrm{R}_{\mathrm{v}}+\mathrm{S}\right)\) From both the equations we get \(\mathrm{S}=10^{5} \Omega\)
PHXII04:MOVING CHARGES AND MAGNETISM
362971
A galvanometer has a resistance of \(50\,\Omega \) and it allows maximum current of \(5\,mA.\) It can be converted into voltmeter to measure upto \(100\,V\) by connecting it in series to a resistor of resistance
1 \(20050\,\Omega \)
2 \(19500\,\Omega \)
3 \(5975\,\Omega \)
4 \(19950\,\Omega \)
Explanation:
Given, \({R_G} = 50\,\Omega ,I = 5\;mA,\;V = 100\;V\) Let the resistance is \(R\). \(\left(R+R_{G}\right) I=V\) \((R+50) \times 5 \times 10^{-3}=100\) \(R = 20000 - 50 = 19950\,\Omega \)
