151844
In the given circuit, if the ammeter reads $5.0 \mathrm{~A}$ and voltmeter reads $20 \mathrm{~V}$, the value of resistance $R$ is ......... .
1 $4 \Omega$
2 $100 \Omega$
3 $0.25 \Omega$
4 $0.5 \Omega$
Explanation:
A Given, $\mathrm{V}=20 \mathrm{~V}, \mathrm{I}=5 \mathrm{~A}$, According to figure, we can apply Ohm's law, $20=5 \mathrm{R}$ $\mathrm{R}=4 \Omega$
AP EAMCET-25.09.2020
Current Electricity
151845
In the given transistor circuit, the base current is $35 \mu \mathrm{A}$. The value of $R_{b}=$
1 $100 \mathrm{k} \Omega$
2 $200 \mathrm{k} \Omega$
3 $300 \mathrm{k} \Omega$
4 $400 \mathrm{k} \Omega$
Explanation:
B Given, $I_{B}=35 \mu \mathrm{A}, \mathrm{V}_{\mathrm{EB}}=7 \mathrm{~V}, \mathrm{R}_{\mathrm{b}}=$ ? By applying Ohm's law $\mathrm{V}=\mathrm{IR}$ $\mathrm{R}_{\mathrm{b}}=\frac{\mathrm{V}_{\mathrm{EB}}}{\mathrm{I}_{\mathrm{B}}}=\frac{7}{35 \times 10^{-6}}$ $\mathrm{R}_{\mathrm{b}}=200 \mathrm{k} \Omega$
AP EAMCET (Medical)-07.10.2020
Current Electricity
151846
A potential difference of $220 \mathrm{~V}$ is maintained across 12,000 $\Omega$ rheostat as shown in the diagram below. The voltmeter $F$ has a resistance of $6,000 \Omega$ and point $C$ is at one fourth of the distance from a to $b$. The reading of the voltmeter is
151847
Thirteen resistors each of resistance ' $R$ ' are connected in the circuit as shown in the following figure. Net resistance between $A$ and $B$ is
1 $2 \mathrm{R}$
2 $\frac{4 R}{3}$
3 $\frac{2 \mathrm{R}}{3}$
4 $\mathrm{R}$
Explanation:
C In given configuration for the part $A C, R_{1}$ and $R_{2}$ are in series and their resultant is in parallel with the $R_{3}$ $\frac{1}{\left(\mathrm{R}_{\mathrm{AC}}\right)_{\mathrm{eq}}}=\frac{1}{\mathrm{R}}+\frac{1}{2 \mathrm{R}}$ $\left(\mathrm{R}_{\mathrm{AC}}\right)_{\mathrm{eq}}=\frac{2 \mathrm{R}}{3}$ Similarly for part AD, CB, DB $\left(\mathrm{R}_{\mathrm{AD}}\right)_{\mathrm{eq}}=\frac{2 \mathrm{R}}{3},\left(\mathrm{R}_{\mathrm{DB}}\right)_{\mathrm{eq}}=2 \mathrm{R} / 3,\left(\mathrm{R}_{\mathrm{CB}}\right)_{\mathrm{eq}}=\frac{2 \mathrm{R}}{3}$ \therefore$ Due to equal resistance in part $\mathrm{AC}$ and $\mathrm{AD}$ there will be no potential difference across arm $C D$ hence that arm will not contribute in resultant resistor $\therefore$ Also that arm $\mathrm{AC}$ in series with $\mathrm{CB}$ and their resultant is in parallel with the series combination of $\mathrm{AD}$ and DB.s $\text { Also } \mathrm{R}_{1} =\mathrm{R}_{\mathrm{AC}}+\mathrm{R}_{\mathrm{AB}}=\frac{2 \mathrm{R}}{3}+\frac{2 \mathrm{R}}{3}=\frac{4 \mathrm{R}}{3}$ $\frac{1}{\mathrm{R}_{\mathrm{eq}}} =\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}+\mathrm{R}_{\mathrm{DB}}=\frac{2 \mathrm{R}}{3}+\frac{2 \mathrm{R}}{3}=\frac{4 \mathrm{R}}{3}$ $\frac{1}{\mathrm{R}_{\mathrm{eq}}} =\frac{3}{4 \mathrm{R}}+\frac{3}{4 \mathrm{R}}=\frac{6}{4 \mathrm{R}}=\frac{3}{2 \mathrm{R}}$ $\mathrm{R}_{\mathrm{eq}} =\frac{2 \mathrm{R}}{3} \Omega$
NEET Test Series from KOTA - 10 Papers In MS WORD
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Current Electricity
151844
In the given circuit, if the ammeter reads $5.0 \mathrm{~A}$ and voltmeter reads $20 \mathrm{~V}$, the value of resistance $R$ is ......... .
1 $4 \Omega$
2 $100 \Omega$
3 $0.25 \Omega$
4 $0.5 \Omega$
Explanation:
A Given, $\mathrm{V}=20 \mathrm{~V}, \mathrm{I}=5 \mathrm{~A}$, According to figure, we can apply Ohm's law, $20=5 \mathrm{R}$ $\mathrm{R}=4 \Omega$
AP EAMCET-25.09.2020
Current Electricity
151845
In the given transistor circuit, the base current is $35 \mu \mathrm{A}$. The value of $R_{b}=$
1 $100 \mathrm{k} \Omega$
2 $200 \mathrm{k} \Omega$
3 $300 \mathrm{k} \Omega$
4 $400 \mathrm{k} \Omega$
Explanation:
B Given, $I_{B}=35 \mu \mathrm{A}, \mathrm{V}_{\mathrm{EB}}=7 \mathrm{~V}, \mathrm{R}_{\mathrm{b}}=$ ? By applying Ohm's law $\mathrm{V}=\mathrm{IR}$ $\mathrm{R}_{\mathrm{b}}=\frac{\mathrm{V}_{\mathrm{EB}}}{\mathrm{I}_{\mathrm{B}}}=\frac{7}{35 \times 10^{-6}}$ $\mathrm{R}_{\mathrm{b}}=200 \mathrm{k} \Omega$
AP EAMCET (Medical)-07.10.2020
Current Electricity
151846
A potential difference of $220 \mathrm{~V}$ is maintained across 12,000 $\Omega$ rheostat as shown in the diagram below. The voltmeter $F$ has a resistance of $6,000 \Omega$ and point $C$ is at one fourth of the distance from a to $b$. The reading of the voltmeter is
151847
Thirteen resistors each of resistance ' $R$ ' are connected in the circuit as shown in the following figure. Net resistance between $A$ and $B$ is
1 $2 \mathrm{R}$
2 $\frac{4 R}{3}$
3 $\frac{2 \mathrm{R}}{3}$
4 $\mathrm{R}$
Explanation:
C In given configuration for the part $A C, R_{1}$ and $R_{2}$ are in series and their resultant is in parallel with the $R_{3}$ $\frac{1}{\left(\mathrm{R}_{\mathrm{AC}}\right)_{\mathrm{eq}}}=\frac{1}{\mathrm{R}}+\frac{1}{2 \mathrm{R}}$ $\left(\mathrm{R}_{\mathrm{AC}}\right)_{\mathrm{eq}}=\frac{2 \mathrm{R}}{3}$ Similarly for part AD, CB, DB $\left(\mathrm{R}_{\mathrm{AD}}\right)_{\mathrm{eq}}=\frac{2 \mathrm{R}}{3},\left(\mathrm{R}_{\mathrm{DB}}\right)_{\mathrm{eq}}=2 \mathrm{R} / 3,\left(\mathrm{R}_{\mathrm{CB}}\right)_{\mathrm{eq}}=\frac{2 \mathrm{R}}{3}$ \therefore$ Due to equal resistance in part $\mathrm{AC}$ and $\mathrm{AD}$ there will be no potential difference across arm $C D$ hence that arm will not contribute in resultant resistor $\therefore$ Also that arm $\mathrm{AC}$ in series with $\mathrm{CB}$ and their resultant is in parallel with the series combination of $\mathrm{AD}$ and DB.s $\text { Also } \mathrm{R}_{1} =\mathrm{R}_{\mathrm{AC}}+\mathrm{R}_{\mathrm{AB}}=\frac{2 \mathrm{R}}{3}+\frac{2 \mathrm{R}}{3}=\frac{4 \mathrm{R}}{3}$ $\frac{1}{\mathrm{R}_{\mathrm{eq}}} =\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}+\mathrm{R}_{\mathrm{DB}}=\frac{2 \mathrm{R}}{3}+\frac{2 \mathrm{R}}{3}=\frac{4 \mathrm{R}}{3}$ $\frac{1}{\mathrm{R}_{\mathrm{eq}}} =\frac{3}{4 \mathrm{R}}+\frac{3}{4 \mathrm{R}}=\frac{6}{4 \mathrm{R}}=\frac{3}{2 \mathrm{R}}$ $\mathrm{R}_{\mathrm{eq}} =\frac{2 \mathrm{R}}{3} \Omega$
151844
In the given circuit, if the ammeter reads $5.0 \mathrm{~A}$ and voltmeter reads $20 \mathrm{~V}$, the value of resistance $R$ is ......... .
1 $4 \Omega$
2 $100 \Omega$
3 $0.25 \Omega$
4 $0.5 \Omega$
Explanation:
A Given, $\mathrm{V}=20 \mathrm{~V}, \mathrm{I}=5 \mathrm{~A}$, According to figure, we can apply Ohm's law, $20=5 \mathrm{R}$ $\mathrm{R}=4 \Omega$
AP EAMCET-25.09.2020
Current Electricity
151845
In the given transistor circuit, the base current is $35 \mu \mathrm{A}$. The value of $R_{b}=$
1 $100 \mathrm{k} \Omega$
2 $200 \mathrm{k} \Omega$
3 $300 \mathrm{k} \Omega$
4 $400 \mathrm{k} \Omega$
Explanation:
B Given, $I_{B}=35 \mu \mathrm{A}, \mathrm{V}_{\mathrm{EB}}=7 \mathrm{~V}, \mathrm{R}_{\mathrm{b}}=$ ? By applying Ohm's law $\mathrm{V}=\mathrm{IR}$ $\mathrm{R}_{\mathrm{b}}=\frac{\mathrm{V}_{\mathrm{EB}}}{\mathrm{I}_{\mathrm{B}}}=\frac{7}{35 \times 10^{-6}}$ $\mathrm{R}_{\mathrm{b}}=200 \mathrm{k} \Omega$
AP EAMCET (Medical)-07.10.2020
Current Electricity
151846
A potential difference of $220 \mathrm{~V}$ is maintained across 12,000 $\Omega$ rheostat as shown in the diagram below. The voltmeter $F$ has a resistance of $6,000 \Omega$ and point $C$ is at one fourth of the distance from a to $b$. The reading of the voltmeter is
151847
Thirteen resistors each of resistance ' $R$ ' are connected in the circuit as shown in the following figure. Net resistance between $A$ and $B$ is
1 $2 \mathrm{R}$
2 $\frac{4 R}{3}$
3 $\frac{2 \mathrm{R}}{3}$
4 $\mathrm{R}$
Explanation:
C In given configuration for the part $A C, R_{1}$ and $R_{2}$ are in series and their resultant is in parallel with the $R_{3}$ $\frac{1}{\left(\mathrm{R}_{\mathrm{AC}}\right)_{\mathrm{eq}}}=\frac{1}{\mathrm{R}}+\frac{1}{2 \mathrm{R}}$ $\left(\mathrm{R}_{\mathrm{AC}}\right)_{\mathrm{eq}}=\frac{2 \mathrm{R}}{3}$ Similarly for part AD, CB, DB $\left(\mathrm{R}_{\mathrm{AD}}\right)_{\mathrm{eq}}=\frac{2 \mathrm{R}}{3},\left(\mathrm{R}_{\mathrm{DB}}\right)_{\mathrm{eq}}=2 \mathrm{R} / 3,\left(\mathrm{R}_{\mathrm{CB}}\right)_{\mathrm{eq}}=\frac{2 \mathrm{R}}{3}$ \therefore$ Due to equal resistance in part $\mathrm{AC}$ and $\mathrm{AD}$ there will be no potential difference across arm $C D$ hence that arm will not contribute in resultant resistor $\therefore$ Also that arm $\mathrm{AC}$ in series with $\mathrm{CB}$ and their resultant is in parallel with the series combination of $\mathrm{AD}$ and DB.s $\text { Also } \mathrm{R}_{1} =\mathrm{R}_{\mathrm{AC}}+\mathrm{R}_{\mathrm{AB}}=\frac{2 \mathrm{R}}{3}+\frac{2 \mathrm{R}}{3}=\frac{4 \mathrm{R}}{3}$ $\frac{1}{\mathrm{R}_{\mathrm{eq}}} =\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}+\mathrm{R}_{\mathrm{DB}}=\frac{2 \mathrm{R}}{3}+\frac{2 \mathrm{R}}{3}=\frac{4 \mathrm{R}}{3}$ $\frac{1}{\mathrm{R}_{\mathrm{eq}}} =\frac{3}{4 \mathrm{R}}+\frac{3}{4 \mathrm{R}}=\frac{6}{4 \mathrm{R}}=\frac{3}{2 \mathrm{R}}$ $\mathrm{R}_{\mathrm{eq}} =\frac{2 \mathrm{R}}{3} \Omega$
151844
In the given circuit, if the ammeter reads $5.0 \mathrm{~A}$ and voltmeter reads $20 \mathrm{~V}$, the value of resistance $R$ is ......... .
1 $4 \Omega$
2 $100 \Omega$
3 $0.25 \Omega$
4 $0.5 \Omega$
Explanation:
A Given, $\mathrm{V}=20 \mathrm{~V}, \mathrm{I}=5 \mathrm{~A}$, According to figure, we can apply Ohm's law, $20=5 \mathrm{R}$ $\mathrm{R}=4 \Omega$
AP EAMCET-25.09.2020
Current Electricity
151845
In the given transistor circuit, the base current is $35 \mu \mathrm{A}$. The value of $R_{b}=$
1 $100 \mathrm{k} \Omega$
2 $200 \mathrm{k} \Omega$
3 $300 \mathrm{k} \Omega$
4 $400 \mathrm{k} \Omega$
Explanation:
B Given, $I_{B}=35 \mu \mathrm{A}, \mathrm{V}_{\mathrm{EB}}=7 \mathrm{~V}, \mathrm{R}_{\mathrm{b}}=$ ? By applying Ohm's law $\mathrm{V}=\mathrm{IR}$ $\mathrm{R}_{\mathrm{b}}=\frac{\mathrm{V}_{\mathrm{EB}}}{\mathrm{I}_{\mathrm{B}}}=\frac{7}{35 \times 10^{-6}}$ $\mathrm{R}_{\mathrm{b}}=200 \mathrm{k} \Omega$
AP EAMCET (Medical)-07.10.2020
Current Electricity
151846
A potential difference of $220 \mathrm{~V}$ is maintained across 12,000 $\Omega$ rheostat as shown in the diagram below. The voltmeter $F$ has a resistance of $6,000 \Omega$ and point $C$ is at one fourth of the distance from a to $b$. The reading of the voltmeter is
151847
Thirteen resistors each of resistance ' $R$ ' are connected in the circuit as shown in the following figure. Net resistance between $A$ and $B$ is
1 $2 \mathrm{R}$
2 $\frac{4 R}{3}$
3 $\frac{2 \mathrm{R}}{3}$
4 $\mathrm{R}$
Explanation:
C In given configuration for the part $A C, R_{1}$ and $R_{2}$ are in series and their resultant is in parallel with the $R_{3}$ $\frac{1}{\left(\mathrm{R}_{\mathrm{AC}}\right)_{\mathrm{eq}}}=\frac{1}{\mathrm{R}}+\frac{1}{2 \mathrm{R}}$ $\left(\mathrm{R}_{\mathrm{AC}}\right)_{\mathrm{eq}}=\frac{2 \mathrm{R}}{3}$ Similarly for part AD, CB, DB $\left(\mathrm{R}_{\mathrm{AD}}\right)_{\mathrm{eq}}=\frac{2 \mathrm{R}}{3},\left(\mathrm{R}_{\mathrm{DB}}\right)_{\mathrm{eq}}=2 \mathrm{R} / 3,\left(\mathrm{R}_{\mathrm{CB}}\right)_{\mathrm{eq}}=\frac{2 \mathrm{R}}{3}$ \therefore$ Due to equal resistance in part $\mathrm{AC}$ and $\mathrm{AD}$ there will be no potential difference across arm $C D$ hence that arm will not contribute in resultant resistor $\therefore$ Also that arm $\mathrm{AC}$ in series with $\mathrm{CB}$ and their resultant is in parallel with the series combination of $\mathrm{AD}$ and DB.s $\text { Also } \mathrm{R}_{1} =\mathrm{R}_{\mathrm{AC}}+\mathrm{R}_{\mathrm{AB}}=\frac{2 \mathrm{R}}{3}+\frac{2 \mathrm{R}}{3}=\frac{4 \mathrm{R}}{3}$ $\frac{1}{\mathrm{R}_{\mathrm{eq}}} =\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}+\mathrm{R}_{\mathrm{DB}}=\frac{2 \mathrm{R}}{3}+\frac{2 \mathrm{R}}{3}=\frac{4 \mathrm{R}}{3}$ $\frac{1}{\mathrm{R}_{\mathrm{eq}}} =\frac{3}{4 \mathrm{R}}+\frac{3}{4 \mathrm{R}}=\frac{6}{4 \mathrm{R}}=\frac{3}{2 \mathrm{R}}$ $\mathrm{R}_{\mathrm{eq}} =\frac{2 \mathrm{R}}{3} \Omega$
