NEET Test Series from KOTA - 10 Papers In MS WORD
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Current Electricity
152479 A battery of emf $E$ and internal resistance $r$ is connected with an external resistance $R$ as shown in the figure. The battery will act as a constant voltage source if
1 $r\lt\ltR$
2 $r>>R$
3 $r=R$
4 It will never act as a constant voltage source.
Explanation:
A The current in the circuit $I=\frac{E}{R+r}$ $v=I . R=\frac{E R}{R+r}=\frac{E}{1+\frac{r}{R}}$ If $\mathrm{r}\lt\lt\mathrm{R}$ $\mathrm{V} = \mathrm{E}$
WB JEE 2022
Current Electricity
152480
A current of $2 \mathrm{~A}$ flows through a $2 \Omega$ resistor when connected across a battery. The same battery supplies a current of $0.5 \mathrm{~A}$. When connected across a $9 \Omega$ resistor. The internal resistance of the battery is-
1 $\frac{1}{3} \Omega$
2 $\frac{1}{4} \Omega$
3 $5 \Omega$
4 $0.5 \Omega$
Explanation:
A Given, $\mathrm{I}_{1}=2 \mathrm{~A}$ $\mathrm{R}_{1}=2 \Omega$ $\mathrm{I}_{2}=0.5 \mathrm{~A}$ $\mathrm{R}_{2}=9 \Omega$ Let, $\mathrm{E}=$ emf of the cell $r=\text { internal resistance of the cell }$ In first condition, $I_{1}=\frac{E}{r+R_{1}}$ $2=\frac{E}{r+2}$ In second condition, $I_{2} =\frac{E}{r+R_{2}}$ $0.5 =\frac{E}{r+9}$ From equation (i) divided by equation (ii), we have - $\frac{2}{0.5}=\frac{r+9}{r+2}$ $4 =\frac{\mathrm{r}+9}{\mathrm{r}+2}$ $4 \mathrm{r}+8 =\mathrm{r}+9$ $3 \mathrm{r} =1$ $\mathrm{r} =\frac{1}{3} \Omega$
TS EAMCET 30.07.2022
Current Electricity
152481
A battery of 54 cells each of emf $1.5 \mathrm{~V}$ and internal resistance $2 \Omega$ is to be connected in order to send a maximum current through a 12 $\Omega$ resistor. The correct arrangement has to be -
1 2 rows of 27 cells connected in parallel.
2 18 rows of 3 cells connected in parallel
3 9 rows of 6 cells connected in parallel.
4 3 rows of 18 cells connected in parallel.
Explanation:
D Given, No. of cells $(\mathrm{n} \times \mathrm{m})=54$ Each voltages $(\mathrm{E})=1.5 \mathrm{~V}$ Internal resistance $(\mathrm{r})=2 \Omega$ External resistance $(\mathrm{R})=12 \Omega$ Let the number of row $=m$ Suppose that rows are connected in parallel. Maximum current in the external resistance, $\mathrm{R}=\frac{\mathrm{nr}}{\mathrm{m}}$ $12=\frac{\mathrm{n} 2}{\mathrm{~m}}$ $6 \mathrm{~m}=\mathrm{n}$ Total cells $=\mathrm{n} \times \mathrm{m}$ $54=\mathrm{n} . \mathrm{m}$ $54=6 \mathrm{~m} \cdot \mathrm{m}$ $6 \mathrm{~m}^{2}=54$ $\mathrm{m}=3$ $\mathrm{n}=6 \mathrm{~m}$ $=6 \times 3=18$ 3 rows of 18 cells connected in parallel.
AP EAMCET-07.09.2021
Current Electricity
152482
A cell of emf 1.8 volts gives a current of $17 \mathrm{~A}$ when directly connected to an ammeter of resistance $0.06 \Omega$. Internal resistance of the cell is
1 $0.046 \Omega$
2 $0.066 \Omega$
3 $0.10 \Omega$
4 $10 \Omega$
Explanation:
A Given, Voltage $(\mathrm{E})=1.8 \mathrm{~V}$ Current $(\mathrm{I})=17 \mathrm{~A}$ External Resistance $(\mathrm{R})=0.06 \Omega$ $\mathrm{E}=\mathrm{V}+\mathrm{Ir}$ $\mathrm{E}=\mathrm{IR}+\mathrm{Ir}$ $I=\frac{E}{(R+r)}$ $17=\frac{1.8}{(0.06+\mathrm{r})}$ $(0.06+r)=\frac{1.8}{17}$ $0.06+r=0.1058$ $r=0.0458$ $r=0.046 \Omega$ Internal resistance of the battery is $0.046 \Omega$.
152479 A battery of emf $E$ and internal resistance $r$ is connected with an external resistance $R$ as shown in the figure. The battery will act as a constant voltage source if
1 $r\lt\ltR$
2 $r>>R$
3 $r=R$
4 It will never act as a constant voltage source.
Explanation:
A The current in the circuit $I=\frac{E}{R+r}$ $v=I . R=\frac{E R}{R+r}=\frac{E}{1+\frac{r}{R}}$ If $\mathrm{r}\lt\lt\mathrm{R}$ $\mathrm{V} = \mathrm{E}$
WB JEE 2022
Current Electricity
152480
A current of $2 \mathrm{~A}$ flows through a $2 \Omega$ resistor when connected across a battery. The same battery supplies a current of $0.5 \mathrm{~A}$. When connected across a $9 \Omega$ resistor. The internal resistance of the battery is-
1 $\frac{1}{3} \Omega$
2 $\frac{1}{4} \Omega$
3 $5 \Omega$
4 $0.5 \Omega$
Explanation:
A Given, $\mathrm{I}_{1}=2 \mathrm{~A}$ $\mathrm{R}_{1}=2 \Omega$ $\mathrm{I}_{2}=0.5 \mathrm{~A}$ $\mathrm{R}_{2}=9 \Omega$ Let, $\mathrm{E}=$ emf of the cell $r=\text { internal resistance of the cell }$ In first condition, $I_{1}=\frac{E}{r+R_{1}}$ $2=\frac{E}{r+2}$ In second condition, $I_{2} =\frac{E}{r+R_{2}}$ $0.5 =\frac{E}{r+9}$ From equation (i) divided by equation (ii), we have - $\frac{2}{0.5}=\frac{r+9}{r+2}$ $4 =\frac{\mathrm{r}+9}{\mathrm{r}+2}$ $4 \mathrm{r}+8 =\mathrm{r}+9$ $3 \mathrm{r} =1$ $\mathrm{r} =\frac{1}{3} \Omega$
TS EAMCET 30.07.2022
Current Electricity
152481
A battery of 54 cells each of emf $1.5 \mathrm{~V}$ and internal resistance $2 \Omega$ is to be connected in order to send a maximum current through a 12 $\Omega$ resistor. The correct arrangement has to be -
1 2 rows of 27 cells connected in parallel.
2 18 rows of 3 cells connected in parallel
3 9 rows of 6 cells connected in parallel.
4 3 rows of 18 cells connected in parallel.
Explanation:
D Given, No. of cells $(\mathrm{n} \times \mathrm{m})=54$ Each voltages $(\mathrm{E})=1.5 \mathrm{~V}$ Internal resistance $(\mathrm{r})=2 \Omega$ External resistance $(\mathrm{R})=12 \Omega$ Let the number of row $=m$ Suppose that rows are connected in parallel. Maximum current in the external resistance, $\mathrm{R}=\frac{\mathrm{nr}}{\mathrm{m}}$ $12=\frac{\mathrm{n} 2}{\mathrm{~m}}$ $6 \mathrm{~m}=\mathrm{n}$ Total cells $=\mathrm{n} \times \mathrm{m}$ $54=\mathrm{n} . \mathrm{m}$ $54=6 \mathrm{~m} \cdot \mathrm{m}$ $6 \mathrm{~m}^{2}=54$ $\mathrm{m}=3$ $\mathrm{n}=6 \mathrm{~m}$ $=6 \times 3=18$ 3 rows of 18 cells connected in parallel.
AP EAMCET-07.09.2021
Current Electricity
152482
A cell of emf 1.8 volts gives a current of $17 \mathrm{~A}$ when directly connected to an ammeter of resistance $0.06 \Omega$. Internal resistance of the cell is
1 $0.046 \Omega$
2 $0.066 \Omega$
3 $0.10 \Omega$
4 $10 \Omega$
Explanation:
A Given, Voltage $(\mathrm{E})=1.8 \mathrm{~V}$ Current $(\mathrm{I})=17 \mathrm{~A}$ External Resistance $(\mathrm{R})=0.06 \Omega$ $\mathrm{E}=\mathrm{V}+\mathrm{Ir}$ $\mathrm{E}=\mathrm{IR}+\mathrm{Ir}$ $I=\frac{E}{(R+r)}$ $17=\frac{1.8}{(0.06+\mathrm{r})}$ $(0.06+r)=\frac{1.8}{17}$ $0.06+r=0.1058$ $r=0.0458$ $r=0.046 \Omega$ Internal resistance of the battery is $0.046 \Omega$.
152479 A battery of emf $E$ and internal resistance $r$ is connected with an external resistance $R$ as shown in the figure. The battery will act as a constant voltage source if
1 $r\lt\ltR$
2 $r>>R$
3 $r=R$
4 It will never act as a constant voltage source.
Explanation:
A The current in the circuit $I=\frac{E}{R+r}$ $v=I . R=\frac{E R}{R+r}=\frac{E}{1+\frac{r}{R}}$ If $\mathrm{r}\lt\lt\mathrm{R}$ $\mathrm{V} = \mathrm{E}$
WB JEE 2022
Current Electricity
152480
A current of $2 \mathrm{~A}$ flows through a $2 \Omega$ resistor when connected across a battery. The same battery supplies a current of $0.5 \mathrm{~A}$. When connected across a $9 \Omega$ resistor. The internal resistance of the battery is-
1 $\frac{1}{3} \Omega$
2 $\frac{1}{4} \Omega$
3 $5 \Omega$
4 $0.5 \Omega$
Explanation:
A Given, $\mathrm{I}_{1}=2 \mathrm{~A}$ $\mathrm{R}_{1}=2 \Omega$ $\mathrm{I}_{2}=0.5 \mathrm{~A}$ $\mathrm{R}_{2}=9 \Omega$ Let, $\mathrm{E}=$ emf of the cell $r=\text { internal resistance of the cell }$ In first condition, $I_{1}=\frac{E}{r+R_{1}}$ $2=\frac{E}{r+2}$ In second condition, $I_{2} =\frac{E}{r+R_{2}}$ $0.5 =\frac{E}{r+9}$ From equation (i) divided by equation (ii), we have - $\frac{2}{0.5}=\frac{r+9}{r+2}$ $4 =\frac{\mathrm{r}+9}{\mathrm{r}+2}$ $4 \mathrm{r}+8 =\mathrm{r}+9$ $3 \mathrm{r} =1$ $\mathrm{r} =\frac{1}{3} \Omega$
TS EAMCET 30.07.2022
Current Electricity
152481
A battery of 54 cells each of emf $1.5 \mathrm{~V}$ and internal resistance $2 \Omega$ is to be connected in order to send a maximum current through a 12 $\Omega$ resistor. The correct arrangement has to be -
1 2 rows of 27 cells connected in parallel.
2 18 rows of 3 cells connected in parallel
3 9 rows of 6 cells connected in parallel.
4 3 rows of 18 cells connected in parallel.
Explanation:
D Given, No. of cells $(\mathrm{n} \times \mathrm{m})=54$ Each voltages $(\mathrm{E})=1.5 \mathrm{~V}$ Internal resistance $(\mathrm{r})=2 \Omega$ External resistance $(\mathrm{R})=12 \Omega$ Let the number of row $=m$ Suppose that rows are connected in parallel. Maximum current in the external resistance, $\mathrm{R}=\frac{\mathrm{nr}}{\mathrm{m}}$ $12=\frac{\mathrm{n} 2}{\mathrm{~m}}$ $6 \mathrm{~m}=\mathrm{n}$ Total cells $=\mathrm{n} \times \mathrm{m}$ $54=\mathrm{n} . \mathrm{m}$ $54=6 \mathrm{~m} \cdot \mathrm{m}$ $6 \mathrm{~m}^{2}=54$ $\mathrm{m}=3$ $\mathrm{n}=6 \mathrm{~m}$ $=6 \times 3=18$ 3 rows of 18 cells connected in parallel.
AP EAMCET-07.09.2021
Current Electricity
152482
A cell of emf 1.8 volts gives a current of $17 \mathrm{~A}$ when directly connected to an ammeter of resistance $0.06 \Omega$. Internal resistance of the cell is
1 $0.046 \Omega$
2 $0.066 \Omega$
3 $0.10 \Omega$
4 $10 \Omega$
Explanation:
A Given, Voltage $(\mathrm{E})=1.8 \mathrm{~V}$ Current $(\mathrm{I})=17 \mathrm{~A}$ External Resistance $(\mathrm{R})=0.06 \Omega$ $\mathrm{E}=\mathrm{V}+\mathrm{Ir}$ $\mathrm{E}=\mathrm{IR}+\mathrm{Ir}$ $I=\frac{E}{(R+r)}$ $17=\frac{1.8}{(0.06+\mathrm{r})}$ $(0.06+r)=\frac{1.8}{17}$ $0.06+r=0.1058$ $r=0.0458$ $r=0.046 \Omega$ Internal resistance of the battery is $0.046 \Omega$.
152479 A battery of emf $E$ and internal resistance $r$ is connected with an external resistance $R$ as shown in the figure. The battery will act as a constant voltage source if
1 $r\lt\ltR$
2 $r>>R$
3 $r=R$
4 It will never act as a constant voltage source.
Explanation:
A The current in the circuit $I=\frac{E}{R+r}$ $v=I . R=\frac{E R}{R+r}=\frac{E}{1+\frac{r}{R}}$ If $\mathrm{r}\lt\lt\mathrm{R}$ $\mathrm{V} = \mathrm{E}$
WB JEE 2022
Current Electricity
152480
A current of $2 \mathrm{~A}$ flows through a $2 \Omega$ resistor when connected across a battery. The same battery supplies a current of $0.5 \mathrm{~A}$. When connected across a $9 \Omega$ resistor. The internal resistance of the battery is-
1 $\frac{1}{3} \Omega$
2 $\frac{1}{4} \Omega$
3 $5 \Omega$
4 $0.5 \Omega$
Explanation:
A Given, $\mathrm{I}_{1}=2 \mathrm{~A}$ $\mathrm{R}_{1}=2 \Omega$ $\mathrm{I}_{2}=0.5 \mathrm{~A}$ $\mathrm{R}_{2}=9 \Omega$ Let, $\mathrm{E}=$ emf of the cell $r=\text { internal resistance of the cell }$ In first condition, $I_{1}=\frac{E}{r+R_{1}}$ $2=\frac{E}{r+2}$ In second condition, $I_{2} =\frac{E}{r+R_{2}}$ $0.5 =\frac{E}{r+9}$ From equation (i) divided by equation (ii), we have - $\frac{2}{0.5}=\frac{r+9}{r+2}$ $4 =\frac{\mathrm{r}+9}{\mathrm{r}+2}$ $4 \mathrm{r}+8 =\mathrm{r}+9$ $3 \mathrm{r} =1$ $\mathrm{r} =\frac{1}{3} \Omega$
TS EAMCET 30.07.2022
Current Electricity
152481
A battery of 54 cells each of emf $1.5 \mathrm{~V}$ and internal resistance $2 \Omega$ is to be connected in order to send a maximum current through a 12 $\Omega$ resistor. The correct arrangement has to be -
1 2 rows of 27 cells connected in parallel.
2 18 rows of 3 cells connected in parallel
3 9 rows of 6 cells connected in parallel.
4 3 rows of 18 cells connected in parallel.
Explanation:
D Given, No. of cells $(\mathrm{n} \times \mathrm{m})=54$ Each voltages $(\mathrm{E})=1.5 \mathrm{~V}$ Internal resistance $(\mathrm{r})=2 \Omega$ External resistance $(\mathrm{R})=12 \Omega$ Let the number of row $=m$ Suppose that rows are connected in parallel. Maximum current in the external resistance, $\mathrm{R}=\frac{\mathrm{nr}}{\mathrm{m}}$ $12=\frac{\mathrm{n} 2}{\mathrm{~m}}$ $6 \mathrm{~m}=\mathrm{n}$ Total cells $=\mathrm{n} \times \mathrm{m}$ $54=\mathrm{n} . \mathrm{m}$ $54=6 \mathrm{~m} \cdot \mathrm{m}$ $6 \mathrm{~m}^{2}=54$ $\mathrm{m}=3$ $\mathrm{n}=6 \mathrm{~m}$ $=6 \times 3=18$ 3 rows of 18 cells connected in parallel.
AP EAMCET-07.09.2021
Current Electricity
152482
A cell of emf 1.8 volts gives a current of $17 \mathrm{~A}$ when directly connected to an ammeter of resistance $0.06 \Omega$. Internal resistance of the cell is
1 $0.046 \Omega$
2 $0.066 \Omega$
3 $0.10 \Omega$
4 $10 \Omega$
Explanation:
A Given, Voltage $(\mathrm{E})=1.8 \mathrm{~V}$ Current $(\mathrm{I})=17 \mathrm{~A}$ External Resistance $(\mathrm{R})=0.06 \Omega$ $\mathrm{E}=\mathrm{V}+\mathrm{Ir}$ $\mathrm{E}=\mathrm{IR}+\mathrm{Ir}$ $I=\frac{E}{(R+r)}$ $17=\frac{1.8}{(0.06+\mathrm{r})}$ $(0.06+r)=\frac{1.8}{17}$ $0.06+r=0.1058$ $r=0.0458$ $r=0.046 \Omega$ Internal resistance of the battery is $0.046 \Omega$.
