154218
In a circuit, 5 percent of total current passes through a galvanometer. If resistance of the galvanometer is $\mathbf{G}$. Then the value of the shunt is-
1 $19 \mathrm{G}$
2 $20 \mathrm{G}$
3 $\frac{G}{20}$
4 $\frac{\mathrm{G}}{19}$
Explanation:
D Given, $\mathrm{I}_{\mathrm{G}}=5 \% \text { of } \mathrm{I}$ $\frac{\mathrm{I}}{\mathrm{I}_{\mathrm{G}}}=\frac{100}{5}$ Shunt resistance, $\mathrm{R}_{\mathrm{S}}=\frac{\mathrm{G}}{\left(\frac{\mathrm{I}}{\mathrm{I}_{\mathrm{G}}}-1\right)}$ $\mathrm{R}_{\mathrm{S}}=\frac{\mathrm{G}}{\left(\frac{100}{5}-1\right)}=\frac{\mathrm{G}}{19}$
MHT-CET 2020
Magnetism and Matter
154219
A moving coil galvanometer has 28 turns and area of coil is $4 \times 10^{-2} \mathrm{~m}^{2}$. If the magnetic field is $0.2 \mathrm{~T}$, then to increase the sensitivity by $25 \%$ without changing area and field, the number of turns should be changed to
1 24
2 35
3 60
4 54
Explanation:
B Number of turns (n) $=28$ turns Area $(A)=4 \times 10^{-2} \mathrm{~m}^{2}$ Magnetic field $(\mathrm{B})=0.2 \mathrm{~T}$ Increasing sensitivity $=25 \%$ $\mathrm{I}_{\mathrm{S}_{2}}=1.25 \mathrm{I}_{\mathrm{S}_{1}}$ $\frac{\mathrm{I}_{\mathrm{S}_{1}}}{\mathrm{I}_{\mathrm{S}_{2}}}=\frac{1}{1.25}=\frac{100}{125}$ Current Sensitivity, $I_{S}=\frac{\theta}{i}=\frac{N A B}{k}$ So, sensitivity is directly proportional to number of turns $\frac{\mathrm{I}_{\mathrm{S}_{1}}}{\mathrm{I}_{\mathrm{S}_{2}}}=\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}$ $\frac{100}{125}=\frac{28}{\mathrm{~N}_{2}}$ $\mathrm{~N}_{2}=28 \times \frac{125}{100}=35$
COMEDK 2020
Magnetism and Matter
154220
A moving coil galvanometer of resistance $100 \Omega$ is used as an ammeter using a resistance $0.1 \Omega$. The maximum deflection current in the galvanometer is $100 \mu \mathrm{A}$. Find the minimum current in the circuit, so that ammeter shows maximum deflection?
154222
A galvanometer of resistance $G \Omega$, is shunted by a resistance $S \Omega$. To keep the main current in the circuit unchanged, the resistance to be connected in series with the galvanometer is
1 $\frac{\mathrm{G}^{2}}{\mathrm{~S}+\mathrm{G}}$
2 $\frac{\mathrm{S}}{\mathrm{S}+\mathrm{G}}$
3 $\frac{\mathrm{S}^{2}}{\mathrm{~S}+\mathrm{G}}$
4 $\frac{\mathrm{SG}}{\mathrm{S}+\mathrm{G}}$
Explanation:
A Galvanometer resistance $G$ and shunt resistance $S$ are in parallel. So, equivalent resistance $R_{P}=\frac{G S}{G+S}$ Now, the resistance $\mathrm{R}$ to be connected in series with Galvanometer to unchanged circuit resistance $G=R_{P}+R$ $G=\left(\frac{G S}{G+S}\right)+R$ $R=G-\frac{G S}{G+S}$ $R=G\left(1-\frac{S}{G+S}\right)$ $R=G\left(\frac{G+S-S}{G+S}\right)$ $R=\frac{G^{2}}{G+S}$
154218
In a circuit, 5 percent of total current passes through a galvanometer. If resistance of the galvanometer is $\mathbf{G}$. Then the value of the shunt is-
1 $19 \mathrm{G}$
2 $20 \mathrm{G}$
3 $\frac{G}{20}$
4 $\frac{\mathrm{G}}{19}$
Explanation:
D Given, $\mathrm{I}_{\mathrm{G}}=5 \% \text { of } \mathrm{I}$ $\frac{\mathrm{I}}{\mathrm{I}_{\mathrm{G}}}=\frac{100}{5}$ Shunt resistance, $\mathrm{R}_{\mathrm{S}}=\frac{\mathrm{G}}{\left(\frac{\mathrm{I}}{\mathrm{I}_{\mathrm{G}}}-1\right)}$ $\mathrm{R}_{\mathrm{S}}=\frac{\mathrm{G}}{\left(\frac{100}{5}-1\right)}=\frac{\mathrm{G}}{19}$
MHT-CET 2020
Magnetism and Matter
154219
A moving coil galvanometer has 28 turns and area of coil is $4 \times 10^{-2} \mathrm{~m}^{2}$. If the magnetic field is $0.2 \mathrm{~T}$, then to increase the sensitivity by $25 \%$ without changing area and field, the number of turns should be changed to
1 24
2 35
3 60
4 54
Explanation:
B Number of turns (n) $=28$ turns Area $(A)=4 \times 10^{-2} \mathrm{~m}^{2}$ Magnetic field $(\mathrm{B})=0.2 \mathrm{~T}$ Increasing sensitivity $=25 \%$ $\mathrm{I}_{\mathrm{S}_{2}}=1.25 \mathrm{I}_{\mathrm{S}_{1}}$ $\frac{\mathrm{I}_{\mathrm{S}_{1}}}{\mathrm{I}_{\mathrm{S}_{2}}}=\frac{1}{1.25}=\frac{100}{125}$ Current Sensitivity, $I_{S}=\frac{\theta}{i}=\frac{N A B}{k}$ So, sensitivity is directly proportional to number of turns $\frac{\mathrm{I}_{\mathrm{S}_{1}}}{\mathrm{I}_{\mathrm{S}_{2}}}=\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}$ $\frac{100}{125}=\frac{28}{\mathrm{~N}_{2}}$ $\mathrm{~N}_{2}=28 \times \frac{125}{100}=35$
COMEDK 2020
Magnetism and Matter
154220
A moving coil galvanometer of resistance $100 \Omega$ is used as an ammeter using a resistance $0.1 \Omega$. The maximum deflection current in the galvanometer is $100 \mu \mathrm{A}$. Find the minimum current in the circuit, so that ammeter shows maximum deflection?
154222
A galvanometer of resistance $G \Omega$, is shunted by a resistance $S \Omega$. To keep the main current in the circuit unchanged, the resistance to be connected in series with the galvanometer is
1 $\frac{\mathrm{G}^{2}}{\mathrm{~S}+\mathrm{G}}$
2 $\frac{\mathrm{S}}{\mathrm{S}+\mathrm{G}}$
3 $\frac{\mathrm{S}^{2}}{\mathrm{~S}+\mathrm{G}}$
4 $\frac{\mathrm{SG}}{\mathrm{S}+\mathrm{G}}$
Explanation:
A Galvanometer resistance $G$ and shunt resistance $S$ are in parallel. So, equivalent resistance $R_{P}=\frac{G S}{G+S}$ Now, the resistance $\mathrm{R}$ to be connected in series with Galvanometer to unchanged circuit resistance $G=R_{P}+R$ $G=\left(\frac{G S}{G+S}\right)+R$ $R=G-\frac{G S}{G+S}$ $R=G\left(1-\frac{S}{G+S}\right)$ $R=G\left(\frac{G+S-S}{G+S}\right)$ $R=\frac{G^{2}}{G+S}$
154218
In a circuit, 5 percent of total current passes through a galvanometer. If resistance of the galvanometer is $\mathbf{G}$. Then the value of the shunt is-
1 $19 \mathrm{G}$
2 $20 \mathrm{G}$
3 $\frac{G}{20}$
4 $\frac{\mathrm{G}}{19}$
Explanation:
D Given, $\mathrm{I}_{\mathrm{G}}=5 \% \text { of } \mathrm{I}$ $\frac{\mathrm{I}}{\mathrm{I}_{\mathrm{G}}}=\frac{100}{5}$ Shunt resistance, $\mathrm{R}_{\mathrm{S}}=\frac{\mathrm{G}}{\left(\frac{\mathrm{I}}{\mathrm{I}_{\mathrm{G}}}-1\right)}$ $\mathrm{R}_{\mathrm{S}}=\frac{\mathrm{G}}{\left(\frac{100}{5}-1\right)}=\frac{\mathrm{G}}{19}$
MHT-CET 2020
Magnetism and Matter
154219
A moving coil galvanometer has 28 turns and area of coil is $4 \times 10^{-2} \mathrm{~m}^{2}$. If the magnetic field is $0.2 \mathrm{~T}$, then to increase the sensitivity by $25 \%$ without changing area and field, the number of turns should be changed to
1 24
2 35
3 60
4 54
Explanation:
B Number of turns (n) $=28$ turns Area $(A)=4 \times 10^{-2} \mathrm{~m}^{2}$ Magnetic field $(\mathrm{B})=0.2 \mathrm{~T}$ Increasing sensitivity $=25 \%$ $\mathrm{I}_{\mathrm{S}_{2}}=1.25 \mathrm{I}_{\mathrm{S}_{1}}$ $\frac{\mathrm{I}_{\mathrm{S}_{1}}}{\mathrm{I}_{\mathrm{S}_{2}}}=\frac{1}{1.25}=\frac{100}{125}$ Current Sensitivity, $I_{S}=\frac{\theta}{i}=\frac{N A B}{k}$ So, sensitivity is directly proportional to number of turns $\frac{\mathrm{I}_{\mathrm{S}_{1}}}{\mathrm{I}_{\mathrm{S}_{2}}}=\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}$ $\frac{100}{125}=\frac{28}{\mathrm{~N}_{2}}$ $\mathrm{~N}_{2}=28 \times \frac{125}{100}=35$
COMEDK 2020
Magnetism and Matter
154220
A moving coil galvanometer of resistance $100 \Omega$ is used as an ammeter using a resistance $0.1 \Omega$. The maximum deflection current in the galvanometer is $100 \mu \mathrm{A}$. Find the minimum current in the circuit, so that ammeter shows maximum deflection?
154222
A galvanometer of resistance $G \Omega$, is shunted by a resistance $S \Omega$. To keep the main current in the circuit unchanged, the resistance to be connected in series with the galvanometer is
1 $\frac{\mathrm{G}^{2}}{\mathrm{~S}+\mathrm{G}}$
2 $\frac{\mathrm{S}}{\mathrm{S}+\mathrm{G}}$
3 $\frac{\mathrm{S}^{2}}{\mathrm{~S}+\mathrm{G}}$
4 $\frac{\mathrm{SG}}{\mathrm{S}+\mathrm{G}}$
Explanation:
A Galvanometer resistance $G$ and shunt resistance $S$ are in parallel. So, equivalent resistance $R_{P}=\frac{G S}{G+S}$ Now, the resistance $\mathrm{R}$ to be connected in series with Galvanometer to unchanged circuit resistance $G=R_{P}+R$ $G=\left(\frac{G S}{G+S}\right)+R$ $R=G-\frac{G S}{G+S}$ $R=G\left(1-\frac{S}{G+S}\right)$ $R=G\left(\frac{G+S-S}{G+S}\right)$ $R=\frac{G^{2}}{G+S}$
154218
In a circuit, 5 percent of total current passes through a galvanometer. If resistance of the galvanometer is $\mathbf{G}$. Then the value of the shunt is-
1 $19 \mathrm{G}$
2 $20 \mathrm{G}$
3 $\frac{G}{20}$
4 $\frac{\mathrm{G}}{19}$
Explanation:
D Given, $\mathrm{I}_{\mathrm{G}}=5 \% \text { of } \mathrm{I}$ $\frac{\mathrm{I}}{\mathrm{I}_{\mathrm{G}}}=\frac{100}{5}$ Shunt resistance, $\mathrm{R}_{\mathrm{S}}=\frac{\mathrm{G}}{\left(\frac{\mathrm{I}}{\mathrm{I}_{\mathrm{G}}}-1\right)}$ $\mathrm{R}_{\mathrm{S}}=\frac{\mathrm{G}}{\left(\frac{100}{5}-1\right)}=\frac{\mathrm{G}}{19}$
MHT-CET 2020
Magnetism and Matter
154219
A moving coil galvanometer has 28 turns and area of coil is $4 \times 10^{-2} \mathrm{~m}^{2}$. If the magnetic field is $0.2 \mathrm{~T}$, then to increase the sensitivity by $25 \%$ without changing area and field, the number of turns should be changed to
1 24
2 35
3 60
4 54
Explanation:
B Number of turns (n) $=28$ turns Area $(A)=4 \times 10^{-2} \mathrm{~m}^{2}$ Magnetic field $(\mathrm{B})=0.2 \mathrm{~T}$ Increasing sensitivity $=25 \%$ $\mathrm{I}_{\mathrm{S}_{2}}=1.25 \mathrm{I}_{\mathrm{S}_{1}}$ $\frac{\mathrm{I}_{\mathrm{S}_{1}}}{\mathrm{I}_{\mathrm{S}_{2}}}=\frac{1}{1.25}=\frac{100}{125}$ Current Sensitivity, $I_{S}=\frac{\theta}{i}=\frac{N A B}{k}$ So, sensitivity is directly proportional to number of turns $\frac{\mathrm{I}_{\mathrm{S}_{1}}}{\mathrm{I}_{\mathrm{S}_{2}}}=\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}$ $\frac{100}{125}=\frac{28}{\mathrm{~N}_{2}}$ $\mathrm{~N}_{2}=28 \times \frac{125}{100}=35$
COMEDK 2020
Magnetism and Matter
154220
A moving coil galvanometer of resistance $100 \Omega$ is used as an ammeter using a resistance $0.1 \Omega$. The maximum deflection current in the galvanometer is $100 \mu \mathrm{A}$. Find the minimum current in the circuit, so that ammeter shows maximum deflection?
154222
A galvanometer of resistance $G \Omega$, is shunted by a resistance $S \Omega$. To keep the main current in the circuit unchanged, the resistance to be connected in series with the galvanometer is
1 $\frac{\mathrm{G}^{2}}{\mathrm{~S}+\mathrm{G}}$
2 $\frac{\mathrm{S}}{\mathrm{S}+\mathrm{G}}$
3 $\frac{\mathrm{S}^{2}}{\mathrm{~S}+\mathrm{G}}$
4 $\frac{\mathrm{SG}}{\mathrm{S}+\mathrm{G}}$
Explanation:
A Galvanometer resistance $G$ and shunt resistance $S$ are in parallel. So, equivalent resistance $R_{P}=\frac{G S}{G+S}$ Now, the resistance $\mathrm{R}$ to be connected in series with Galvanometer to unchanged circuit resistance $G=R_{P}+R$ $G=\left(\frac{G S}{G+S}\right)+R$ $R=G-\frac{G S}{G+S}$ $R=G\left(1-\frac{S}{G+S}\right)$ $R=G\left(\frac{G+S-S}{G+S}\right)$ $R=\frac{G^{2}}{G+S}$
