152810
An ammeter of resistance $20 \Omega$ gives full scale deflection when $1 \mathrm{~mA}$ current flows through it. What is the maximum current that can be measured by connecting 4 resistors each of $16 \Omega$ in parallel with the meter?
152811
The range of the voltmeter is ' $V$ ' when $50 \Omega$ resistance is connected in series. Its range gets doubled when $500 \Omega$ resistance is connected in series. The resistance of voltmeter is
1 $600 \Omega$
2 $400 \Omega$
3 $200 \Omega$
4 $800 \Omega$
Explanation:
B Let galvanometer resistance be R \& current through it is $I_{G}$ Then, $\quad \mathrm{V}=\mathrm{I}_{\mathrm{G}}(\mathrm{G}+\mathrm{R})$ When $\mathrm{R}=50 \Omega$, Voltmeter range $=\mathrm{V}$ $\therefore \quad \mathrm{V}=\mathrm{I}_{\mathrm{G}}(\mathrm{G}+50)$ When $\mathrm{R}=500$, Voltmeter range is $=2 \mathrm{~V}$ $\therefore \quad 2 \mathrm{~V}=\mathrm{I}_{\mathrm{G}}(\mathrm{G}+500)$ substituting the values of $\mathrm{V}$ form equation (i) to (ii) $2(\mathrm{G}+50)=(\mathrm{G}+500)$ $2 \mathrm{G}+100=\mathrm{G}+500$ $\mathrm{G}=400 \Omega$
MHT-CET 2020
Current Electricity
152812
An ammeter is obtained by shunting ' $n$ ' $\Omega$ galvanometer with ' $n$ ' $\Omega$ resistance. The additional shunt required to be connected across it to double the range is
1 $\mathrm{n}$
2 $\frac{n}{4}$
3 $\frac{\mathrm{n}}{3}$
4 $\frac{\mathrm{n}}{2}$
Explanation:
D $: \because \mathrm{S}=\frac{\mathrm{i}_{\mathrm{g}}}{\mathrm{I}-\mathrm{i}_{\mathrm{g}}} \times \mathrm{G}$ $n=\frac{i_{g}}{I-i_{g}} \times n$ $i_{g}=I-i_{g}$ $I=2 i_{g}$ When range is doubled $\mathrm{I}^{\prime}=2 \mathrm{I}==4 \mathrm{i}_{\mathrm{g}}$ $\quad \mathrm{S}^{\prime}=\frac{\mathrm{i}_{\mathrm{g}}}{\mathrm{I}^{\prime}-\mathrm{i}_{\mathrm{g}}} \mathrm{G}$ $=\frac{\mathrm{i}_{\mathrm{g}}}{4 \mathrm{i}_{\mathrm{g}}-\mathrm{i}_{\mathrm{g}}} \times \mathrm{n}=\frac{\mathrm{n}}{3}$ Since, shunt is connected in parallel so, additional shunt is required $\therefore \quad \frac{1}{\mathrm{~S}^{\prime}} =\frac{1}{\mathrm{~S}}+\frac{1}{\mathrm{X}}$ $\frac{1}{\mathrm{X}} =\frac{1}{\mathrm{~S}^{\prime}}-\frac{1}{\mathrm{~S}}=\frac{3}{\mathrm{n}}-\frac{1}{\mathrm{n}}=\frac{2}{\mathrm{n}}$ $\mathrm{X} =\frac{\mathrm{n}}{2}$
MHT-CET 2020
Current Electricity
152813
Three voltmeters all having different resistances are joined as shown. When some potential difference is applied across $P$ and $Q$, their readings are $V_{1}, V_{2}$ and $V_{3}$ respectively. Then:
1 $\mathrm{V}_{1}=\mathrm{V}_{2}$
2 $V_{1} \neq V_{2}$
3 $\mathrm{V}_{1}+\mathrm{V}_{2}=\mathrm{V}_{3}$
4 $\mathrm{V}_{1}+\mathrm{V}_{2}>\mathrm{V}_{3}$
Explanation:
C They have the same potential difference across them, So, \(\mathrm{V}_1+\mathrm{V}_2=\mathrm{V}_3\) \(\text { Ohm's law }\) \(\mathrm{V}_1=\mathrm{I}_1 \mathrm{R}_1\) \(\mathrm{~V}_2=\mathrm{IR}_2\) \(\mathrm{~V}_{\mathrm{AB}}=\mathrm{V}_1+\mathrm{V}_2\) \(\mathrm{~V}_{\mathrm{AB}}=\mathrm{V}_3\) \(\mathrm{~V}_1+\mathrm{V}_2=\mathrm{V}_3\)
152810
An ammeter of resistance $20 \Omega$ gives full scale deflection when $1 \mathrm{~mA}$ current flows through it. What is the maximum current that can be measured by connecting 4 resistors each of $16 \Omega$ in parallel with the meter?
152811
The range of the voltmeter is ' $V$ ' when $50 \Omega$ resistance is connected in series. Its range gets doubled when $500 \Omega$ resistance is connected in series. The resistance of voltmeter is
1 $600 \Omega$
2 $400 \Omega$
3 $200 \Omega$
4 $800 \Omega$
Explanation:
B Let galvanometer resistance be R \& current through it is $I_{G}$ Then, $\quad \mathrm{V}=\mathrm{I}_{\mathrm{G}}(\mathrm{G}+\mathrm{R})$ When $\mathrm{R}=50 \Omega$, Voltmeter range $=\mathrm{V}$ $\therefore \quad \mathrm{V}=\mathrm{I}_{\mathrm{G}}(\mathrm{G}+50)$ When $\mathrm{R}=500$, Voltmeter range is $=2 \mathrm{~V}$ $\therefore \quad 2 \mathrm{~V}=\mathrm{I}_{\mathrm{G}}(\mathrm{G}+500)$ substituting the values of $\mathrm{V}$ form equation (i) to (ii) $2(\mathrm{G}+50)=(\mathrm{G}+500)$ $2 \mathrm{G}+100=\mathrm{G}+500$ $\mathrm{G}=400 \Omega$
MHT-CET 2020
Current Electricity
152812
An ammeter is obtained by shunting ' $n$ ' $\Omega$ galvanometer with ' $n$ ' $\Omega$ resistance. The additional shunt required to be connected across it to double the range is
1 $\mathrm{n}$
2 $\frac{n}{4}$
3 $\frac{\mathrm{n}}{3}$
4 $\frac{\mathrm{n}}{2}$
Explanation:
D $: \because \mathrm{S}=\frac{\mathrm{i}_{\mathrm{g}}}{\mathrm{I}-\mathrm{i}_{\mathrm{g}}} \times \mathrm{G}$ $n=\frac{i_{g}}{I-i_{g}} \times n$ $i_{g}=I-i_{g}$ $I=2 i_{g}$ When range is doubled $\mathrm{I}^{\prime}=2 \mathrm{I}==4 \mathrm{i}_{\mathrm{g}}$ $\quad \mathrm{S}^{\prime}=\frac{\mathrm{i}_{\mathrm{g}}}{\mathrm{I}^{\prime}-\mathrm{i}_{\mathrm{g}}} \mathrm{G}$ $=\frac{\mathrm{i}_{\mathrm{g}}}{4 \mathrm{i}_{\mathrm{g}}-\mathrm{i}_{\mathrm{g}}} \times \mathrm{n}=\frac{\mathrm{n}}{3}$ Since, shunt is connected in parallel so, additional shunt is required $\therefore \quad \frac{1}{\mathrm{~S}^{\prime}} =\frac{1}{\mathrm{~S}}+\frac{1}{\mathrm{X}}$ $\frac{1}{\mathrm{X}} =\frac{1}{\mathrm{~S}^{\prime}}-\frac{1}{\mathrm{~S}}=\frac{3}{\mathrm{n}}-\frac{1}{\mathrm{n}}=\frac{2}{\mathrm{n}}$ $\mathrm{X} =\frac{\mathrm{n}}{2}$
MHT-CET 2020
Current Electricity
152813
Three voltmeters all having different resistances are joined as shown. When some potential difference is applied across $P$ and $Q$, their readings are $V_{1}, V_{2}$ and $V_{3}$ respectively. Then:
1 $\mathrm{V}_{1}=\mathrm{V}_{2}$
2 $V_{1} \neq V_{2}$
3 $\mathrm{V}_{1}+\mathrm{V}_{2}=\mathrm{V}_{3}$
4 $\mathrm{V}_{1}+\mathrm{V}_{2}>\mathrm{V}_{3}$
Explanation:
C They have the same potential difference across them, So, \(\mathrm{V}_1+\mathrm{V}_2=\mathrm{V}_3\) \(\text { Ohm's law }\) \(\mathrm{V}_1=\mathrm{I}_1 \mathrm{R}_1\) \(\mathrm{~V}_2=\mathrm{IR}_2\) \(\mathrm{~V}_{\mathrm{AB}}=\mathrm{V}_1+\mathrm{V}_2\) \(\mathrm{~V}_{\mathrm{AB}}=\mathrm{V}_3\) \(\mathrm{~V}_1+\mathrm{V}_2=\mathrm{V}_3\)
152810
An ammeter of resistance $20 \Omega$ gives full scale deflection when $1 \mathrm{~mA}$ current flows through it. What is the maximum current that can be measured by connecting 4 resistors each of $16 \Omega$ in parallel with the meter?
152811
The range of the voltmeter is ' $V$ ' when $50 \Omega$ resistance is connected in series. Its range gets doubled when $500 \Omega$ resistance is connected in series. The resistance of voltmeter is
1 $600 \Omega$
2 $400 \Omega$
3 $200 \Omega$
4 $800 \Omega$
Explanation:
B Let galvanometer resistance be R \& current through it is $I_{G}$ Then, $\quad \mathrm{V}=\mathrm{I}_{\mathrm{G}}(\mathrm{G}+\mathrm{R})$ When $\mathrm{R}=50 \Omega$, Voltmeter range $=\mathrm{V}$ $\therefore \quad \mathrm{V}=\mathrm{I}_{\mathrm{G}}(\mathrm{G}+50)$ When $\mathrm{R}=500$, Voltmeter range is $=2 \mathrm{~V}$ $\therefore \quad 2 \mathrm{~V}=\mathrm{I}_{\mathrm{G}}(\mathrm{G}+500)$ substituting the values of $\mathrm{V}$ form equation (i) to (ii) $2(\mathrm{G}+50)=(\mathrm{G}+500)$ $2 \mathrm{G}+100=\mathrm{G}+500$ $\mathrm{G}=400 \Omega$
MHT-CET 2020
Current Electricity
152812
An ammeter is obtained by shunting ' $n$ ' $\Omega$ galvanometer with ' $n$ ' $\Omega$ resistance. The additional shunt required to be connected across it to double the range is
1 $\mathrm{n}$
2 $\frac{n}{4}$
3 $\frac{\mathrm{n}}{3}$
4 $\frac{\mathrm{n}}{2}$
Explanation:
D $: \because \mathrm{S}=\frac{\mathrm{i}_{\mathrm{g}}}{\mathrm{I}-\mathrm{i}_{\mathrm{g}}} \times \mathrm{G}$ $n=\frac{i_{g}}{I-i_{g}} \times n$ $i_{g}=I-i_{g}$ $I=2 i_{g}$ When range is doubled $\mathrm{I}^{\prime}=2 \mathrm{I}==4 \mathrm{i}_{\mathrm{g}}$ $\quad \mathrm{S}^{\prime}=\frac{\mathrm{i}_{\mathrm{g}}}{\mathrm{I}^{\prime}-\mathrm{i}_{\mathrm{g}}} \mathrm{G}$ $=\frac{\mathrm{i}_{\mathrm{g}}}{4 \mathrm{i}_{\mathrm{g}}-\mathrm{i}_{\mathrm{g}}} \times \mathrm{n}=\frac{\mathrm{n}}{3}$ Since, shunt is connected in parallel so, additional shunt is required $\therefore \quad \frac{1}{\mathrm{~S}^{\prime}} =\frac{1}{\mathrm{~S}}+\frac{1}{\mathrm{X}}$ $\frac{1}{\mathrm{X}} =\frac{1}{\mathrm{~S}^{\prime}}-\frac{1}{\mathrm{~S}}=\frac{3}{\mathrm{n}}-\frac{1}{\mathrm{n}}=\frac{2}{\mathrm{n}}$ $\mathrm{X} =\frac{\mathrm{n}}{2}$
MHT-CET 2020
Current Electricity
152813
Three voltmeters all having different resistances are joined as shown. When some potential difference is applied across $P$ and $Q$, their readings are $V_{1}, V_{2}$ and $V_{3}$ respectively. Then:
1 $\mathrm{V}_{1}=\mathrm{V}_{2}$
2 $V_{1} \neq V_{2}$
3 $\mathrm{V}_{1}+\mathrm{V}_{2}=\mathrm{V}_{3}$
4 $\mathrm{V}_{1}+\mathrm{V}_{2}>\mathrm{V}_{3}$
Explanation:
C They have the same potential difference across them, So, \(\mathrm{V}_1+\mathrm{V}_2=\mathrm{V}_3\) \(\text { Ohm's law }\) \(\mathrm{V}_1=\mathrm{I}_1 \mathrm{R}_1\) \(\mathrm{~V}_2=\mathrm{IR}_2\) \(\mathrm{~V}_{\mathrm{AB}}=\mathrm{V}_1+\mathrm{V}_2\) \(\mathrm{~V}_{\mathrm{AB}}=\mathrm{V}_3\) \(\mathrm{~V}_1+\mathrm{V}_2=\mathrm{V}_3\)
152810
An ammeter of resistance $20 \Omega$ gives full scale deflection when $1 \mathrm{~mA}$ current flows through it. What is the maximum current that can be measured by connecting 4 resistors each of $16 \Omega$ in parallel with the meter?
152811
The range of the voltmeter is ' $V$ ' when $50 \Omega$ resistance is connected in series. Its range gets doubled when $500 \Omega$ resistance is connected in series. The resistance of voltmeter is
1 $600 \Omega$
2 $400 \Omega$
3 $200 \Omega$
4 $800 \Omega$
Explanation:
B Let galvanometer resistance be R \& current through it is $I_{G}$ Then, $\quad \mathrm{V}=\mathrm{I}_{\mathrm{G}}(\mathrm{G}+\mathrm{R})$ When $\mathrm{R}=50 \Omega$, Voltmeter range $=\mathrm{V}$ $\therefore \quad \mathrm{V}=\mathrm{I}_{\mathrm{G}}(\mathrm{G}+50)$ When $\mathrm{R}=500$, Voltmeter range is $=2 \mathrm{~V}$ $\therefore \quad 2 \mathrm{~V}=\mathrm{I}_{\mathrm{G}}(\mathrm{G}+500)$ substituting the values of $\mathrm{V}$ form equation (i) to (ii) $2(\mathrm{G}+50)=(\mathrm{G}+500)$ $2 \mathrm{G}+100=\mathrm{G}+500$ $\mathrm{G}=400 \Omega$
MHT-CET 2020
Current Electricity
152812
An ammeter is obtained by shunting ' $n$ ' $\Omega$ galvanometer with ' $n$ ' $\Omega$ resistance. The additional shunt required to be connected across it to double the range is
1 $\mathrm{n}$
2 $\frac{n}{4}$
3 $\frac{\mathrm{n}}{3}$
4 $\frac{\mathrm{n}}{2}$
Explanation:
D $: \because \mathrm{S}=\frac{\mathrm{i}_{\mathrm{g}}}{\mathrm{I}-\mathrm{i}_{\mathrm{g}}} \times \mathrm{G}$ $n=\frac{i_{g}}{I-i_{g}} \times n$ $i_{g}=I-i_{g}$ $I=2 i_{g}$ When range is doubled $\mathrm{I}^{\prime}=2 \mathrm{I}==4 \mathrm{i}_{\mathrm{g}}$ $\quad \mathrm{S}^{\prime}=\frac{\mathrm{i}_{\mathrm{g}}}{\mathrm{I}^{\prime}-\mathrm{i}_{\mathrm{g}}} \mathrm{G}$ $=\frac{\mathrm{i}_{\mathrm{g}}}{4 \mathrm{i}_{\mathrm{g}}-\mathrm{i}_{\mathrm{g}}} \times \mathrm{n}=\frac{\mathrm{n}}{3}$ Since, shunt is connected in parallel so, additional shunt is required $\therefore \quad \frac{1}{\mathrm{~S}^{\prime}} =\frac{1}{\mathrm{~S}}+\frac{1}{\mathrm{X}}$ $\frac{1}{\mathrm{X}} =\frac{1}{\mathrm{~S}^{\prime}}-\frac{1}{\mathrm{~S}}=\frac{3}{\mathrm{n}}-\frac{1}{\mathrm{n}}=\frac{2}{\mathrm{n}}$ $\mathrm{X} =\frac{\mathrm{n}}{2}$
MHT-CET 2020
Current Electricity
152813
Three voltmeters all having different resistances are joined as shown. When some potential difference is applied across $P$ and $Q$, their readings are $V_{1}, V_{2}$ and $V_{3}$ respectively. Then:
1 $\mathrm{V}_{1}=\mathrm{V}_{2}$
2 $V_{1} \neq V_{2}$
3 $\mathrm{V}_{1}+\mathrm{V}_{2}=\mathrm{V}_{3}$
4 $\mathrm{V}_{1}+\mathrm{V}_{2}>\mathrm{V}_{3}$
Explanation:
C They have the same potential difference across them, So, \(\mathrm{V}_1+\mathrm{V}_2=\mathrm{V}_3\) \(\text { Ohm's law }\) \(\mathrm{V}_1=\mathrm{I}_1 \mathrm{R}_1\) \(\mathrm{~V}_2=\mathrm{IR}_2\) \(\mathrm{~V}_{\mathrm{AB}}=\mathrm{V}_1+\mathrm{V}_2\) \(\mathrm{~V}_{\mathrm{AB}}=\mathrm{V}_3\) \(\mathrm{~V}_1+\mathrm{V}_2=\mathrm{V}_3\)
