154242
If the ammeter $A$ shows a zero reading in the circuit shown below, the value of resistance $R$ is
1 $500 \Omega$
2 $125 \Omega$
3 $100 \Omega$
4 $41.5 \Omega$
5 $4 \Omega$
Explanation:
B Apply KVL in loop 1, $10-500 \mathrm{I}-\mathrm{RI}=0$ $10=\mathrm{I}(500+\mathrm{R})$ Apply KVL in loop 2, $10-500 \mathrm{I}-2=0$ $500 \mathrm{I}=8$ $\mathrm{I}=\frac{8}{500}$ Putting the value of I in equation (i), $10=\frac{8}{500}(500+\mathrm{R})$ $10=8+\frac{8}{500} \mathrm{R}$ $2=\frac{8}{500} \mathrm{R}$ $\mathrm{R}=\frac{2 \times 500}{8}$ $\mathrm{R}=125$
Kerala CEE- 2014
Magnetism and Matter
154243
The shunt resistance required to allow $4 \%$ of the main current through the galvanometer of resistance $48 \Omega$ is
1 $1 \Omega$
2 $2 \Omega$
3 $3 \Omega$
4 $4 \Omega$
5 $5 \Omega$
Explanation:
B Current through the galvanometer is $4 \%$ of main current $\mathrm{I}_{\mathrm{G}}=\frac{4 \mathrm{I}}{100}=\frac{\mathrm{I}}{25}$ Resistance of galvanometer $(\mathrm{G})=48 \Omega$ $\mathrm{R}_{\mathrm{S}}=\frac{\mathrm{GI}_{\mathrm{G}}}{\mathrm{I}-\mathrm{I}_{\mathrm{G}}}$ $\mathrm{R}_{\mathrm{S}}=\frac{48 \times \frac{\mathrm{I}}{25}}{\mathrm{I}-\frac{\mathrm{I}}{25}}$ $\mathrm{R}_{\mathrm{S}}=\frac{48}{24}=2$ $\mathrm{R}_{\mathrm{S}}=2 \Omega$
Kerala CEE 2012
Magnetism and Matter
154244
A galvanometer of resistance $100 \Omega$ is converted to a voltmeter of range $10 \mathrm{~V}$ by connecting a resistance of $10 \mathrm{k} \Omega$. The resistance required to convert the same galvanometer to an ammeter of range $1 \mathrm{~A}$ is
1 $0.4 \Omega$
2 $0.3 \Omega$
3 $1.2 \Omega$
4 $0.2 \Omega$
5 $0.1 \Omega$
Explanation:
E Given that, $\mathrm{G}=100 \Omega$ $\mathrm{V}=10 \mathrm{~V}$ $\mathrm{R}=10 \mathrm{k} \Omega$ $I_{G}=\frac{V}{G+R}$ $\mathrm{I}_{\mathrm{G}}=\frac{10}{100+10 \times 10^{3}}$ $\mathrm{I}_{\mathrm{G}}=\frac{1}{1010}$ When shunt $\mathrm{S}$ is connected in parallel $S =\frac{\mathrm{GI}_{G}}{\mathrm{I}-\mathrm{I}_{\mathrm{G}}}$ $\mathrm{S} =\frac{100 \times \frac{1}{1010}}{1-\frac{1}{1010}}$ $\mathrm{~S} =\frac{100}{1009}$ $\mathrm{~S} =0.099 \approx 0.1 \Omega$
Kerala CEE - 2010
Magnetism and Matter
154245
A galvanometer of resistance $20 \Omega$ shows a deflection of 10 division when a current of 1 $\mathrm{mA}$ is passed through it. If a shunt of $4 \Omega$ is connected and there are 50 division on the scale, the range of the galvanometer is
1 $1 \mathrm{~A}$
2 $3 \mathrm{~A}$
3 $10 \mathrm{~mA}$
4 $30 \mathrm{~A}$
5 $30 \mathrm{~mA}$
Explanation:
E Current for 10 division $=1 \mathrm{~mA}$ Current for 50 division $\mathrm{I}_{\mathrm{g}}=\frac{1 \times 50}{10}=5 \mathrm{~mA}$ $\mathrm{I}_{\mathrm{g}} \mathrm{G}=\left(\mathrm{I}-\mathrm{I}_{\mathrm{g}}\right) \mathrm{S}$ $5 \times 10^{-3} \times 20=\left(\mathrm{I}-5 \times 10^{-3}\right) 4$ $25 \times 10^{-3}=\mathrm{I}-5 \times 10^{-3}$ $\mathrm{I}=30 \times 10^{-3}$ $\mathrm{I}=30 \mathrm{~mA}$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Magnetism and Matter
154242
If the ammeter $A$ shows a zero reading in the circuit shown below, the value of resistance $R$ is
1 $500 \Omega$
2 $125 \Omega$
3 $100 \Omega$
4 $41.5 \Omega$
5 $4 \Omega$
Explanation:
B Apply KVL in loop 1, $10-500 \mathrm{I}-\mathrm{RI}=0$ $10=\mathrm{I}(500+\mathrm{R})$ Apply KVL in loop 2, $10-500 \mathrm{I}-2=0$ $500 \mathrm{I}=8$ $\mathrm{I}=\frac{8}{500}$ Putting the value of I in equation (i), $10=\frac{8}{500}(500+\mathrm{R})$ $10=8+\frac{8}{500} \mathrm{R}$ $2=\frac{8}{500} \mathrm{R}$ $\mathrm{R}=\frac{2 \times 500}{8}$ $\mathrm{R}=125$
Kerala CEE- 2014
Magnetism and Matter
154243
The shunt resistance required to allow $4 \%$ of the main current through the galvanometer of resistance $48 \Omega$ is
1 $1 \Omega$
2 $2 \Omega$
3 $3 \Omega$
4 $4 \Omega$
5 $5 \Omega$
Explanation:
B Current through the galvanometer is $4 \%$ of main current $\mathrm{I}_{\mathrm{G}}=\frac{4 \mathrm{I}}{100}=\frac{\mathrm{I}}{25}$ Resistance of galvanometer $(\mathrm{G})=48 \Omega$ $\mathrm{R}_{\mathrm{S}}=\frac{\mathrm{GI}_{\mathrm{G}}}{\mathrm{I}-\mathrm{I}_{\mathrm{G}}}$ $\mathrm{R}_{\mathrm{S}}=\frac{48 \times \frac{\mathrm{I}}{25}}{\mathrm{I}-\frac{\mathrm{I}}{25}}$ $\mathrm{R}_{\mathrm{S}}=\frac{48}{24}=2$ $\mathrm{R}_{\mathrm{S}}=2 \Omega$
Kerala CEE 2012
Magnetism and Matter
154244
A galvanometer of resistance $100 \Omega$ is converted to a voltmeter of range $10 \mathrm{~V}$ by connecting a resistance of $10 \mathrm{k} \Omega$. The resistance required to convert the same galvanometer to an ammeter of range $1 \mathrm{~A}$ is
1 $0.4 \Omega$
2 $0.3 \Omega$
3 $1.2 \Omega$
4 $0.2 \Omega$
5 $0.1 \Omega$
Explanation:
E Given that, $\mathrm{G}=100 \Omega$ $\mathrm{V}=10 \mathrm{~V}$ $\mathrm{R}=10 \mathrm{k} \Omega$ $I_{G}=\frac{V}{G+R}$ $\mathrm{I}_{\mathrm{G}}=\frac{10}{100+10 \times 10^{3}}$ $\mathrm{I}_{\mathrm{G}}=\frac{1}{1010}$ When shunt $\mathrm{S}$ is connected in parallel $S =\frac{\mathrm{GI}_{G}}{\mathrm{I}-\mathrm{I}_{\mathrm{G}}}$ $\mathrm{S} =\frac{100 \times \frac{1}{1010}}{1-\frac{1}{1010}}$ $\mathrm{~S} =\frac{100}{1009}$ $\mathrm{~S} =0.099 \approx 0.1 \Omega$
Kerala CEE - 2010
Magnetism and Matter
154245
A galvanometer of resistance $20 \Omega$ shows a deflection of 10 division when a current of 1 $\mathrm{mA}$ is passed through it. If a shunt of $4 \Omega$ is connected and there are 50 division on the scale, the range of the galvanometer is
1 $1 \mathrm{~A}$
2 $3 \mathrm{~A}$
3 $10 \mathrm{~mA}$
4 $30 \mathrm{~A}$
5 $30 \mathrm{~mA}$
Explanation:
E Current for 10 division $=1 \mathrm{~mA}$ Current for 50 division $\mathrm{I}_{\mathrm{g}}=\frac{1 \times 50}{10}=5 \mathrm{~mA}$ $\mathrm{I}_{\mathrm{g}} \mathrm{G}=\left(\mathrm{I}-\mathrm{I}_{\mathrm{g}}\right) \mathrm{S}$ $5 \times 10^{-3} \times 20=\left(\mathrm{I}-5 \times 10^{-3}\right) 4$ $25 \times 10^{-3}=\mathrm{I}-5 \times 10^{-3}$ $\mathrm{I}=30 \times 10^{-3}$ $\mathrm{I}=30 \mathrm{~mA}$
154242
If the ammeter $A$ shows a zero reading in the circuit shown below, the value of resistance $R$ is
1 $500 \Omega$
2 $125 \Omega$
3 $100 \Omega$
4 $41.5 \Omega$
5 $4 \Omega$
Explanation:
B Apply KVL in loop 1, $10-500 \mathrm{I}-\mathrm{RI}=0$ $10=\mathrm{I}(500+\mathrm{R})$ Apply KVL in loop 2, $10-500 \mathrm{I}-2=0$ $500 \mathrm{I}=8$ $\mathrm{I}=\frac{8}{500}$ Putting the value of I in equation (i), $10=\frac{8}{500}(500+\mathrm{R})$ $10=8+\frac{8}{500} \mathrm{R}$ $2=\frac{8}{500} \mathrm{R}$ $\mathrm{R}=\frac{2 \times 500}{8}$ $\mathrm{R}=125$
Kerala CEE- 2014
Magnetism and Matter
154243
The shunt resistance required to allow $4 \%$ of the main current through the galvanometer of resistance $48 \Omega$ is
1 $1 \Omega$
2 $2 \Omega$
3 $3 \Omega$
4 $4 \Omega$
5 $5 \Omega$
Explanation:
B Current through the galvanometer is $4 \%$ of main current $\mathrm{I}_{\mathrm{G}}=\frac{4 \mathrm{I}}{100}=\frac{\mathrm{I}}{25}$ Resistance of galvanometer $(\mathrm{G})=48 \Omega$ $\mathrm{R}_{\mathrm{S}}=\frac{\mathrm{GI}_{\mathrm{G}}}{\mathrm{I}-\mathrm{I}_{\mathrm{G}}}$ $\mathrm{R}_{\mathrm{S}}=\frac{48 \times \frac{\mathrm{I}}{25}}{\mathrm{I}-\frac{\mathrm{I}}{25}}$ $\mathrm{R}_{\mathrm{S}}=\frac{48}{24}=2$ $\mathrm{R}_{\mathrm{S}}=2 \Omega$
Kerala CEE 2012
Magnetism and Matter
154244
A galvanometer of resistance $100 \Omega$ is converted to a voltmeter of range $10 \mathrm{~V}$ by connecting a resistance of $10 \mathrm{k} \Omega$. The resistance required to convert the same galvanometer to an ammeter of range $1 \mathrm{~A}$ is
1 $0.4 \Omega$
2 $0.3 \Omega$
3 $1.2 \Omega$
4 $0.2 \Omega$
5 $0.1 \Omega$
Explanation:
E Given that, $\mathrm{G}=100 \Omega$ $\mathrm{V}=10 \mathrm{~V}$ $\mathrm{R}=10 \mathrm{k} \Omega$ $I_{G}=\frac{V}{G+R}$ $\mathrm{I}_{\mathrm{G}}=\frac{10}{100+10 \times 10^{3}}$ $\mathrm{I}_{\mathrm{G}}=\frac{1}{1010}$ When shunt $\mathrm{S}$ is connected in parallel $S =\frac{\mathrm{GI}_{G}}{\mathrm{I}-\mathrm{I}_{\mathrm{G}}}$ $\mathrm{S} =\frac{100 \times \frac{1}{1010}}{1-\frac{1}{1010}}$ $\mathrm{~S} =\frac{100}{1009}$ $\mathrm{~S} =0.099 \approx 0.1 \Omega$
Kerala CEE - 2010
Magnetism and Matter
154245
A galvanometer of resistance $20 \Omega$ shows a deflection of 10 division when a current of 1 $\mathrm{mA}$ is passed through it. If a shunt of $4 \Omega$ is connected and there are 50 division on the scale, the range of the galvanometer is
1 $1 \mathrm{~A}$
2 $3 \mathrm{~A}$
3 $10 \mathrm{~mA}$
4 $30 \mathrm{~A}$
5 $30 \mathrm{~mA}$
Explanation:
E Current for 10 division $=1 \mathrm{~mA}$ Current for 50 division $\mathrm{I}_{\mathrm{g}}=\frac{1 \times 50}{10}=5 \mathrm{~mA}$ $\mathrm{I}_{\mathrm{g}} \mathrm{G}=\left(\mathrm{I}-\mathrm{I}_{\mathrm{g}}\right) \mathrm{S}$ $5 \times 10^{-3} \times 20=\left(\mathrm{I}-5 \times 10^{-3}\right) 4$ $25 \times 10^{-3}=\mathrm{I}-5 \times 10^{-3}$ $\mathrm{I}=30 \times 10^{-3}$ $\mathrm{I}=30 \mathrm{~mA}$
154242
If the ammeter $A$ shows a zero reading in the circuit shown below, the value of resistance $R$ is
1 $500 \Omega$
2 $125 \Omega$
3 $100 \Omega$
4 $41.5 \Omega$
5 $4 \Omega$
Explanation:
B Apply KVL in loop 1, $10-500 \mathrm{I}-\mathrm{RI}=0$ $10=\mathrm{I}(500+\mathrm{R})$ Apply KVL in loop 2, $10-500 \mathrm{I}-2=0$ $500 \mathrm{I}=8$ $\mathrm{I}=\frac{8}{500}$ Putting the value of I in equation (i), $10=\frac{8}{500}(500+\mathrm{R})$ $10=8+\frac{8}{500} \mathrm{R}$ $2=\frac{8}{500} \mathrm{R}$ $\mathrm{R}=\frac{2 \times 500}{8}$ $\mathrm{R}=125$
Kerala CEE- 2014
Magnetism and Matter
154243
The shunt resistance required to allow $4 \%$ of the main current through the galvanometer of resistance $48 \Omega$ is
1 $1 \Omega$
2 $2 \Omega$
3 $3 \Omega$
4 $4 \Omega$
5 $5 \Omega$
Explanation:
B Current through the galvanometer is $4 \%$ of main current $\mathrm{I}_{\mathrm{G}}=\frac{4 \mathrm{I}}{100}=\frac{\mathrm{I}}{25}$ Resistance of galvanometer $(\mathrm{G})=48 \Omega$ $\mathrm{R}_{\mathrm{S}}=\frac{\mathrm{GI}_{\mathrm{G}}}{\mathrm{I}-\mathrm{I}_{\mathrm{G}}}$ $\mathrm{R}_{\mathrm{S}}=\frac{48 \times \frac{\mathrm{I}}{25}}{\mathrm{I}-\frac{\mathrm{I}}{25}}$ $\mathrm{R}_{\mathrm{S}}=\frac{48}{24}=2$ $\mathrm{R}_{\mathrm{S}}=2 \Omega$
Kerala CEE 2012
Magnetism and Matter
154244
A galvanometer of resistance $100 \Omega$ is converted to a voltmeter of range $10 \mathrm{~V}$ by connecting a resistance of $10 \mathrm{k} \Omega$. The resistance required to convert the same galvanometer to an ammeter of range $1 \mathrm{~A}$ is
1 $0.4 \Omega$
2 $0.3 \Omega$
3 $1.2 \Omega$
4 $0.2 \Omega$
5 $0.1 \Omega$
Explanation:
E Given that, $\mathrm{G}=100 \Omega$ $\mathrm{V}=10 \mathrm{~V}$ $\mathrm{R}=10 \mathrm{k} \Omega$ $I_{G}=\frac{V}{G+R}$ $\mathrm{I}_{\mathrm{G}}=\frac{10}{100+10 \times 10^{3}}$ $\mathrm{I}_{\mathrm{G}}=\frac{1}{1010}$ When shunt $\mathrm{S}$ is connected in parallel $S =\frac{\mathrm{GI}_{G}}{\mathrm{I}-\mathrm{I}_{\mathrm{G}}}$ $\mathrm{S} =\frac{100 \times \frac{1}{1010}}{1-\frac{1}{1010}}$ $\mathrm{~S} =\frac{100}{1009}$ $\mathrm{~S} =0.099 \approx 0.1 \Omega$
Kerala CEE - 2010
Magnetism and Matter
154245
A galvanometer of resistance $20 \Omega$ shows a deflection of 10 division when a current of 1 $\mathrm{mA}$ is passed through it. If a shunt of $4 \Omega$ is connected and there are 50 division on the scale, the range of the galvanometer is
1 $1 \mathrm{~A}$
2 $3 \mathrm{~A}$
3 $10 \mathrm{~mA}$
4 $30 \mathrm{~A}$
5 $30 \mathrm{~mA}$
Explanation:
E Current for 10 division $=1 \mathrm{~mA}$ Current for 50 division $\mathrm{I}_{\mathrm{g}}=\frac{1 \times 50}{10}=5 \mathrm{~mA}$ $\mathrm{I}_{\mathrm{g}} \mathrm{G}=\left(\mathrm{I}-\mathrm{I}_{\mathrm{g}}\right) \mathrm{S}$ $5 \times 10^{-3} \times 20=\left(\mathrm{I}-5 \times 10^{-3}\right) 4$ $25 \times 10^{-3}=\mathrm{I}-5 \times 10^{-3}$ $\mathrm{I}=30 \times 10^{-3}$ $\mathrm{I}=30 \mathrm{~mA}$
