362955
A galvanometer of resistance, \(G\) is shunted by a resistance \(S\) ohm. To keep the main current in the circuit unchanged, the resistance to be put in series with the galvanometer is
1 \(\dfrac{S^{2}}{(S+G)}\)
2 \(\dfrac{S G}{(S+G)}\)
3 \(\dfrac{G^{2}}{(S+G)}\)
4 \(\dfrac{G}{(S+G)}\)
Explanation:
Refer to figures \((a)\) and \((b).\) To keep the main current \(I\) in the circuit unchanged, a resistance \(R\) is connected in series with a shunted galvanometer (figure (b)). It will be so if \(G = \frac{{GS}}{{G + S}} + R\) or \(R = G - \frac{{GS}}{{G + S}} = \frac{{{G^2} + GS - GS}}{{G + S}} = \frac{{{G^2}}}{{G + S}}\)
PHXII04:MOVING CHARGES AND MAGNETISM
362956
An ammeter reads upto \(1\;A\). Its internal resistance is \(0.81 \Omega\). To increase the range to \(10\;A\), the value of the required shunt is
1 \(0.09\, \Omega\)
2 \(0.03\, \Omega\)
3 \(0.3\, \Omega\)
4 \(0.9\, \Omega\)
Explanation:
Here, \(G=0.81\, \Omega, I_{g}=1 \mathrm{~A}, I=10 \mathrm{~A}\) Shunt resistance is given by, \(S=\dfrac{G I_{g}}{I-I_{g}}\). \(S=\dfrac{0.81 \times 1}{10-1}=\dfrac{0.81}{9}=0.09\, \Omega\)
PHXII04:MOVING CHARGES AND MAGNETISM
362957
In an ammeter \(0.2 \%\) of main current passes through the galvanometer. If resistance of galvanometer is \(G\), the resistance of ammeter will be
1 \(\dfrac{499}{500} G\)
2 \(\dfrac{1}{499} G\)
3 \(\dfrac{500}{499} G\)
4 \(\dfrac{1}{500} G\)
Explanation:
For ammeter \(0.002 I \times G=0.998 I \times r_{s}\) \(r_{s}=\dfrac{0.002}{0.998} G \Rightarrow r_{s}=\dfrac{1}{499} \times G\) Equivalent resistance of ammeter, \(\begin{aligned}& \dfrac{1}{R}=\dfrac{1}{G}+\dfrac{1}{r_{s}} \\& \therefore \quad \dfrac{1}{R}=\dfrac{1}{G}+\dfrac{1}{G / 499} \Rightarrow R=\dfrac{G}{500}\end{aligned}\)
PHXII04:MOVING CHARGES AND MAGNETISM
362958
A galvanometer has a resistance of \(100\,\Omega \). A potential difference of \(100\,mV\) between its terminals gives a full scale deflection. The shunt resistance required to convert it into an ammeter reading upto \(5\,A\) is
1 \(0.01\,\Omega \)
2 \(0.02\,\Omega \)
3 \(0.03\,\Omega \)
4 \(0.04\,\Omega \)
Explanation:
The current required for full scale deflection is \(i_{g}=\dfrac{\text { PD across galvanometer }}{\text { Resistance of galvanometer }}\) \( \Rightarrow {i_g} = \frac{{100 \times {{10}^{ - 3}}}}{{100}} = {10^{ - 3}}\;A\) The shunt resistance required is \(S=\left(\dfrac{i_{g}}{i-i_{g}}\right) G=\left(\dfrac{10^{-3}}{5-10^{-3}}\right) \times 100\) \(=0.02 \Omega\) So, correct option is (2)
362955
A galvanometer of resistance, \(G\) is shunted by a resistance \(S\) ohm. To keep the main current in the circuit unchanged, the resistance to be put in series with the galvanometer is
1 \(\dfrac{S^{2}}{(S+G)}\)
2 \(\dfrac{S G}{(S+G)}\)
3 \(\dfrac{G^{2}}{(S+G)}\)
4 \(\dfrac{G}{(S+G)}\)
Explanation:
Refer to figures \((a)\) and \((b).\) To keep the main current \(I\) in the circuit unchanged, a resistance \(R\) is connected in series with a shunted galvanometer (figure (b)). It will be so if \(G = \frac{{GS}}{{G + S}} + R\) or \(R = G - \frac{{GS}}{{G + S}} = \frac{{{G^2} + GS - GS}}{{G + S}} = \frac{{{G^2}}}{{G + S}}\)
PHXII04:MOVING CHARGES AND MAGNETISM
362956
An ammeter reads upto \(1\;A\). Its internal resistance is \(0.81 \Omega\). To increase the range to \(10\;A\), the value of the required shunt is
1 \(0.09\, \Omega\)
2 \(0.03\, \Omega\)
3 \(0.3\, \Omega\)
4 \(0.9\, \Omega\)
Explanation:
Here, \(G=0.81\, \Omega, I_{g}=1 \mathrm{~A}, I=10 \mathrm{~A}\) Shunt resistance is given by, \(S=\dfrac{G I_{g}}{I-I_{g}}\). \(S=\dfrac{0.81 \times 1}{10-1}=\dfrac{0.81}{9}=0.09\, \Omega\)
PHXII04:MOVING CHARGES AND MAGNETISM
362957
In an ammeter \(0.2 \%\) of main current passes through the galvanometer. If resistance of galvanometer is \(G\), the resistance of ammeter will be
1 \(\dfrac{499}{500} G\)
2 \(\dfrac{1}{499} G\)
3 \(\dfrac{500}{499} G\)
4 \(\dfrac{1}{500} G\)
Explanation:
For ammeter \(0.002 I \times G=0.998 I \times r_{s}\) \(r_{s}=\dfrac{0.002}{0.998} G \Rightarrow r_{s}=\dfrac{1}{499} \times G\) Equivalent resistance of ammeter, \(\begin{aligned}& \dfrac{1}{R}=\dfrac{1}{G}+\dfrac{1}{r_{s}} \\& \therefore \quad \dfrac{1}{R}=\dfrac{1}{G}+\dfrac{1}{G / 499} \Rightarrow R=\dfrac{G}{500}\end{aligned}\)
PHXII04:MOVING CHARGES AND MAGNETISM
362958
A galvanometer has a resistance of \(100\,\Omega \). A potential difference of \(100\,mV\) between its terminals gives a full scale deflection. The shunt resistance required to convert it into an ammeter reading upto \(5\,A\) is
1 \(0.01\,\Omega \)
2 \(0.02\,\Omega \)
3 \(0.03\,\Omega \)
4 \(0.04\,\Omega \)
Explanation:
The current required for full scale deflection is \(i_{g}=\dfrac{\text { PD across galvanometer }}{\text { Resistance of galvanometer }}\) \( \Rightarrow {i_g} = \frac{{100 \times {{10}^{ - 3}}}}{{100}} = {10^{ - 3}}\;A\) The shunt resistance required is \(S=\left(\dfrac{i_{g}}{i-i_{g}}\right) G=\left(\dfrac{10^{-3}}{5-10^{-3}}\right) \times 100\) \(=0.02 \Omega\) So, correct option is (2)
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PHXII04:MOVING CHARGES AND MAGNETISM
362955
A galvanometer of resistance, \(G\) is shunted by a resistance \(S\) ohm. To keep the main current in the circuit unchanged, the resistance to be put in series with the galvanometer is
1 \(\dfrac{S^{2}}{(S+G)}\)
2 \(\dfrac{S G}{(S+G)}\)
3 \(\dfrac{G^{2}}{(S+G)}\)
4 \(\dfrac{G}{(S+G)}\)
Explanation:
Refer to figures \((a)\) and \((b).\) To keep the main current \(I\) in the circuit unchanged, a resistance \(R\) is connected in series with a shunted galvanometer (figure (b)). It will be so if \(G = \frac{{GS}}{{G + S}} + R\) or \(R = G - \frac{{GS}}{{G + S}} = \frac{{{G^2} + GS - GS}}{{G + S}} = \frac{{{G^2}}}{{G + S}}\)
PHXII04:MOVING CHARGES AND MAGNETISM
362956
An ammeter reads upto \(1\;A\). Its internal resistance is \(0.81 \Omega\). To increase the range to \(10\;A\), the value of the required shunt is
1 \(0.09\, \Omega\)
2 \(0.03\, \Omega\)
3 \(0.3\, \Omega\)
4 \(0.9\, \Omega\)
Explanation:
Here, \(G=0.81\, \Omega, I_{g}=1 \mathrm{~A}, I=10 \mathrm{~A}\) Shunt resistance is given by, \(S=\dfrac{G I_{g}}{I-I_{g}}\). \(S=\dfrac{0.81 \times 1}{10-1}=\dfrac{0.81}{9}=0.09\, \Omega\)
PHXII04:MOVING CHARGES AND MAGNETISM
362957
In an ammeter \(0.2 \%\) of main current passes through the galvanometer. If resistance of galvanometer is \(G\), the resistance of ammeter will be
1 \(\dfrac{499}{500} G\)
2 \(\dfrac{1}{499} G\)
3 \(\dfrac{500}{499} G\)
4 \(\dfrac{1}{500} G\)
Explanation:
For ammeter \(0.002 I \times G=0.998 I \times r_{s}\) \(r_{s}=\dfrac{0.002}{0.998} G \Rightarrow r_{s}=\dfrac{1}{499} \times G\) Equivalent resistance of ammeter, \(\begin{aligned}& \dfrac{1}{R}=\dfrac{1}{G}+\dfrac{1}{r_{s}} \\& \therefore \quad \dfrac{1}{R}=\dfrac{1}{G}+\dfrac{1}{G / 499} \Rightarrow R=\dfrac{G}{500}\end{aligned}\)
PHXII04:MOVING CHARGES AND MAGNETISM
362958
A galvanometer has a resistance of \(100\,\Omega \). A potential difference of \(100\,mV\) between its terminals gives a full scale deflection. The shunt resistance required to convert it into an ammeter reading upto \(5\,A\) is
1 \(0.01\,\Omega \)
2 \(0.02\,\Omega \)
3 \(0.03\,\Omega \)
4 \(0.04\,\Omega \)
Explanation:
The current required for full scale deflection is \(i_{g}=\dfrac{\text { PD across galvanometer }}{\text { Resistance of galvanometer }}\) \( \Rightarrow {i_g} = \frac{{100 \times {{10}^{ - 3}}}}{{100}} = {10^{ - 3}}\;A\) The shunt resistance required is \(S=\left(\dfrac{i_{g}}{i-i_{g}}\right) G=\left(\dfrac{10^{-3}}{5-10^{-3}}\right) \times 100\) \(=0.02 \Omega\) So, correct option is (2)
362955
A galvanometer of resistance, \(G\) is shunted by a resistance \(S\) ohm. To keep the main current in the circuit unchanged, the resistance to be put in series with the galvanometer is
1 \(\dfrac{S^{2}}{(S+G)}\)
2 \(\dfrac{S G}{(S+G)}\)
3 \(\dfrac{G^{2}}{(S+G)}\)
4 \(\dfrac{G}{(S+G)}\)
Explanation:
Refer to figures \((a)\) and \((b).\) To keep the main current \(I\) in the circuit unchanged, a resistance \(R\) is connected in series with a shunted galvanometer (figure (b)). It will be so if \(G = \frac{{GS}}{{G + S}} + R\) or \(R = G - \frac{{GS}}{{G + S}} = \frac{{{G^2} + GS - GS}}{{G + S}} = \frac{{{G^2}}}{{G + S}}\)
PHXII04:MOVING CHARGES AND MAGNETISM
362956
An ammeter reads upto \(1\;A\). Its internal resistance is \(0.81 \Omega\). To increase the range to \(10\;A\), the value of the required shunt is
1 \(0.09\, \Omega\)
2 \(0.03\, \Omega\)
3 \(0.3\, \Omega\)
4 \(0.9\, \Omega\)
Explanation:
Here, \(G=0.81\, \Omega, I_{g}=1 \mathrm{~A}, I=10 \mathrm{~A}\) Shunt resistance is given by, \(S=\dfrac{G I_{g}}{I-I_{g}}\). \(S=\dfrac{0.81 \times 1}{10-1}=\dfrac{0.81}{9}=0.09\, \Omega\)
PHXII04:MOVING CHARGES AND MAGNETISM
362957
In an ammeter \(0.2 \%\) of main current passes through the galvanometer. If resistance of galvanometer is \(G\), the resistance of ammeter will be
1 \(\dfrac{499}{500} G\)
2 \(\dfrac{1}{499} G\)
3 \(\dfrac{500}{499} G\)
4 \(\dfrac{1}{500} G\)
Explanation:
For ammeter \(0.002 I \times G=0.998 I \times r_{s}\) \(r_{s}=\dfrac{0.002}{0.998} G \Rightarrow r_{s}=\dfrac{1}{499} \times G\) Equivalent resistance of ammeter, \(\begin{aligned}& \dfrac{1}{R}=\dfrac{1}{G}+\dfrac{1}{r_{s}} \\& \therefore \quad \dfrac{1}{R}=\dfrac{1}{G}+\dfrac{1}{G / 499} \Rightarrow R=\dfrac{G}{500}\end{aligned}\)
PHXII04:MOVING CHARGES AND MAGNETISM
362958
A galvanometer has a resistance of \(100\,\Omega \). A potential difference of \(100\,mV\) between its terminals gives a full scale deflection. The shunt resistance required to convert it into an ammeter reading upto \(5\,A\) is
1 \(0.01\,\Omega \)
2 \(0.02\,\Omega \)
3 \(0.03\,\Omega \)
4 \(0.04\,\Omega \)
Explanation:
The current required for full scale deflection is \(i_{g}=\dfrac{\text { PD across galvanometer }}{\text { Resistance of galvanometer }}\) \( \Rightarrow {i_g} = \frac{{100 \times {{10}^{ - 3}}}}{{100}} = {10^{ - 3}}\;A\) The shunt resistance required is \(S=\left(\dfrac{i_{g}}{i-i_{g}}\right) G=\left(\dfrac{10^{-3}}{5-10^{-3}}\right) \times 100\) \(=0.02 \Omega\) So, correct option is (2)
