152483
In a potentiometer arrangement, a cell of emf 1.5 $V$ gives a Balance point an $150 \mathrm{~cm}$ length of the wire. If the cell is replaced by another cell and the balance point shift to $210 \mathrm{~cm}$, what is the emf of the second cell?
1 $3.2 \mathrm{~V}$
2 $1.2 \mathrm{~V}$
3 $4.4 \mathrm{~V}$
4 $2.1 \mathrm{~V}$
Explanation:
D Given $\mathrm{E}_{1}=1.5 \mathrm{volt}, \mathrm{E}_{2}=?$ $l_{1}=150 \mathrm{~cm}, l_{2}=210 \mathrm{~cm}$ We know that $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{l_{1}}{l_{2}}$ $\mathrm{E}_{2}=\frac{l_{2}}{l_{1}} \mathrm{E}_{1}=\frac{210}{150} \times 1.5$ $=2.1 \text { volt }$
GUJCET-PCE- 2021
Current Electricity
152484
If one cell is connected wrongly in a series combination of four cells each of e.m.f $1.5 \mathrm{~V}$ and internal resistance of $0.5 \Omega$, then the equivalent internal resistance of the combination is
1 $0.5 \Omega$
2 $1 \Omega$
3 $1.5 \Omega$
4 $2 \Omega$
5 $2.5 \Omega$
Explanation:
D Given that, $\mathrm{r}=0.5 \Omega=1 / 2$ $\mathrm{E}=1.5 \mathrm{~V}$ And $\mathrm{n}=4$ $r_{1}, r_{2}$ and $r_{3}, r_{4}$ or equivalent $\frac{1}{\mathrm{r}_{\mathrm{eq}}} =\frac{1}{\mathrm{r}_{1}}+\frac{1}{\mathrm{r}_{2}}+\frac{1}{\mathrm{r}_{3}}+\frac{1}{\mathrm{r}_{4}}$ $=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}$ $\frac{1}{\mathrm{r}_{\mathrm{eq}}}=\frac{4}{2}$ $\mathrm{r}_{\mathrm{eq}}=2 \Omega$
Kerala CEE 2021
Current Electricity
152485
In the given circuit, current flowing through it is
1 $5 \mathrm{~A}$
2 $3 \mathrm{~A}$
3 $2 \mathrm{~A}$
4 $4 \mathrm{~A}$
Explanation:
A The two cells are opposing to each other. The net emf in the circuit $=(200-10)=190 \mathrm{~V}$ $\mathrm{R}=38 \Omega$ $\therefore$ Current $(\mathrm{I})=\frac{\mathrm{V}}{\mathrm{R}}=\frac{190}{38}=5 \mathrm{~A}$
MHT-CET 2020
Current Electricity
152486
A cell of e.m.f. ' $E$ ' is connected across a resistance ' $R$ '. The potential difference across the terminals of the cell is $90 \%$ of $E$. The internal resistance of the cell is
1 $9 \mathrm{R}$
2 $\frac{\mathrm{R}}{9}$
3 $\frac{9}{10} \mathrm{R}$
4 $\frac{10}{9} \mathrm{R}$
Explanation:
B Let $r$ be the internal resistance of the cell. $\mathrm{R}$ Terminal potential difference $(\mathrm{V})\lt$ EMF of a cell $(\mathrm{E})$. $\mathrm{V} =\mathrm{E}-\mathrm{Ir}$ $\mathrm{r} =\left(\frac{\mathrm{E}-\mathrm{V}}{\mathrm{V}}\right) \mathrm{R}$ $=\left(\frac{\mathrm{E}}{\mathrm{V}}-1\right) \mathrm{R}=\left(\frac{10 \mathrm{E}}{9 \mathrm{E}}-1\right) \mathrm{R}$ $\mathrm{r} =\frac{\mathrm{R}}{9}$
MHT-CET 2020
Current Electricity
152487
Consider the circuit shown. If all the cells have negligible internal resistance, what will be the current through the $2 \Omega$ resistor when steady state is reached?
1 $0.66 \mathrm{~A}$
2 $0.29 \mathrm{~A}$
3 $0 \mathrm{~A}$
4 $0.14 \mathrm{~A}$
Explanation:
C The emfs due to two of the cancel out. The capacitor in steady state. Hence, circuit is open. So, there is no current passes through it.
152483
In a potentiometer arrangement, a cell of emf 1.5 $V$ gives a Balance point an $150 \mathrm{~cm}$ length of the wire. If the cell is replaced by another cell and the balance point shift to $210 \mathrm{~cm}$, what is the emf of the second cell?
1 $3.2 \mathrm{~V}$
2 $1.2 \mathrm{~V}$
3 $4.4 \mathrm{~V}$
4 $2.1 \mathrm{~V}$
Explanation:
D Given $\mathrm{E}_{1}=1.5 \mathrm{volt}, \mathrm{E}_{2}=?$ $l_{1}=150 \mathrm{~cm}, l_{2}=210 \mathrm{~cm}$ We know that $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{l_{1}}{l_{2}}$ $\mathrm{E}_{2}=\frac{l_{2}}{l_{1}} \mathrm{E}_{1}=\frac{210}{150} \times 1.5$ $=2.1 \text { volt }$
GUJCET-PCE- 2021
Current Electricity
152484
If one cell is connected wrongly in a series combination of four cells each of e.m.f $1.5 \mathrm{~V}$ and internal resistance of $0.5 \Omega$, then the equivalent internal resistance of the combination is
1 $0.5 \Omega$
2 $1 \Omega$
3 $1.5 \Omega$
4 $2 \Omega$
5 $2.5 \Omega$
Explanation:
D Given that, $\mathrm{r}=0.5 \Omega=1 / 2$ $\mathrm{E}=1.5 \mathrm{~V}$ And $\mathrm{n}=4$ $r_{1}, r_{2}$ and $r_{3}, r_{4}$ or equivalent $\frac{1}{\mathrm{r}_{\mathrm{eq}}} =\frac{1}{\mathrm{r}_{1}}+\frac{1}{\mathrm{r}_{2}}+\frac{1}{\mathrm{r}_{3}}+\frac{1}{\mathrm{r}_{4}}$ $=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}$ $\frac{1}{\mathrm{r}_{\mathrm{eq}}}=\frac{4}{2}$ $\mathrm{r}_{\mathrm{eq}}=2 \Omega$
Kerala CEE 2021
Current Electricity
152485
In the given circuit, current flowing through it is
1 $5 \mathrm{~A}$
2 $3 \mathrm{~A}$
3 $2 \mathrm{~A}$
4 $4 \mathrm{~A}$
Explanation:
A The two cells are opposing to each other. The net emf in the circuit $=(200-10)=190 \mathrm{~V}$ $\mathrm{R}=38 \Omega$ $\therefore$ Current $(\mathrm{I})=\frac{\mathrm{V}}{\mathrm{R}}=\frac{190}{38}=5 \mathrm{~A}$
MHT-CET 2020
Current Electricity
152486
A cell of e.m.f. ' $E$ ' is connected across a resistance ' $R$ '. The potential difference across the terminals of the cell is $90 \%$ of $E$. The internal resistance of the cell is
1 $9 \mathrm{R}$
2 $\frac{\mathrm{R}}{9}$
3 $\frac{9}{10} \mathrm{R}$
4 $\frac{10}{9} \mathrm{R}$
Explanation:
B Let $r$ be the internal resistance of the cell. $\mathrm{R}$ Terminal potential difference $(\mathrm{V})\lt$ EMF of a cell $(\mathrm{E})$. $\mathrm{V} =\mathrm{E}-\mathrm{Ir}$ $\mathrm{r} =\left(\frac{\mathrm{E}-\mathrm{V}}{\mathrm{V}}\right) \mathrm{R}$ $=\left(\frac{\mathrm{E}}{\mathrm{V}}-1\right) \mathrm{R}=\left(\frac{10 \mathrm{E}}{9 \mathrm{E}}-1\right) \mathrm{R}$ $\mathrm{r} =\frac{\mathrm{R}}{9}$
MHT-CET 2020
Current Electricity
152487
Consider the circuit shown. If all the cells have negligible internal resistance, what will be the current through the $2 \Omega$ resistor when steady state is reached?
1 $0.66 \mathrm{~A}$
2 $0.29 \mathrm{~A}$
3 $0 \mathrm{~A}$
4 $0.14 \mathrm{~A}$
Explanation:
C The emfs due to two of the cancel out. The capacitor in steady state. Hence, circuit is open. So, there is no current passes through it.
NEET Test Series from KOTA - 10 Papers In MS WORD
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Current Electricity
152483
In a potentiometer arrangement, a cell of emf 1.5 $V$ gives a Balance point an $150 \mathrm{~cm}$ length of the wire. If the cell is replaced by another cell and the balance point shift to $210 \mathrm{~cm}$, what is the emf of the second cell?
1 $3.2 \mathrm{~V}$
2 $1.2 \mathrm{~V}$
3 $4.4 \mathrm{~V}$
4 $2.1 \mathrm{~V}$
Explanation:
D Given $\mathrm{E}_{1}=1.5 \mathrm{volt}, \mathrm{E}_{2}=?$ $l_{1}=150 \mathrm{~cm}, l_{2}=210 \mathrm{~cm}$ We know that $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{l_{1}}{l_{2}}$ $\mathrm{E}_{2}=\frac{l_{2}}{l_{1}} \mathrm{E}_{1}=\frac{210}{150} \times 1.5$ $=2.1 \text { volt }$
GUJCET-PCE- 2021
Current Electricity
152484
If one cell is connected wrongly in a series combination of four cells each of e.m.f $1.5 \mathrm{~V}$ and internal resistance of $0.5 \Omega$, then the equivalent internal resistance of the combination is
1 $0.5 \Omega$
2 $1 \Omega$
3 $1.5 \Omega$
4 $2 \Omega$
5 $2.5 \Omega$
Explanation:
D Given that, $\mathrm{r}=0.5 \Omega=1 / 2$ $\mathrm{E}=1.5 \mathrm{~V}$ And $\mathrm{n}=4$ $r_{1}, r_{2}$ and $r_{3}, r_{4}$ or equivalent $\frac{1}{\mathrm{r}_{\mathrm{eq}}} =\frac{1}{\mathrm{r}_{1}}+\frac{1}{\mathrm{r}_{2}}+\frac{1}{\mathrm{r}_{3}}+\frac{1}{\mathrm{r}_{4}}$ $=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}$ $\frac{1}{\mathrm{r}_{\mathrm{eq}}}=\frac{4}{2}$ $\mathrm{r}_{\mathrm{eq}}=2 \Omega$
Kerala CEE 2021
Current Electricity
152485
In the given circuit, current flowing through it is
1 $5 \mathrm{~A}$
2 $3 \mathrm{~A}$
3 $2 \mathrm{~A}$
4 $4 \mathrm{~A}$
Explanation:
A The two cells are opposing to each other. The net emf in the circuit $=(200-10)=190 \mathrm{~V}$ $\mathrm{R}=38 \Omega$ $\therefore$ Current $(\mathrm{I})=\frac{\mathrm{V}}{\mathrm{R}}=\frac{190}{38}=5 \mathrm{~A}$
MHT-CET 2020
Current Electricity
152486
A cell of e.m.f. ' $E$ ' is connected across a resistance ' $R$ '. The potential difference across the terminals of the cell is $90 \%$ of $E$. The internal resistance of the cell is
1 $9 \mathrm{R}$
2 $\frac{\mathrm{R}}{9}$
3 $\frac{9}{10} \mathrm{R}$
4 $\frac{10}{9} \mathrm{R}$
Explanation:
B Let $r$ be the internal resistance of the cell. $\mathrm{R}$ Terminal potential difference $(\mathrm{V})\lt$ EMF of a cell $(\mathrm{E})$. $\mathrm{V} =\mathrm{E}-\mathrm{Ir}$ $\mathrm{r} =\left(\frac{\mathrm{E}-\mathrm{V}}{\mathrm{V}}\right) \mathrm{R}$ $=\left(\frac{\mathrm{E}}{\mathrm{V}}-1\right) \mathrm{R}=\left(\frac{10 \mathrm{E}}{9 \mathrm{E}}-1\right) \mathrm{R}$ $\mathrm{r} =\frac{\mathrm{R}}{9}$
MHT-CET 2020
Current Electricity
152487
Consider the circuit shown. If all the cells have negligible internal resistance, what will be the current through the $2 \Omega$ resistor when steady state is reached?
1 $0.66 \mathrm{~A}$
2 $0.29 \mathrm{~A}$
3 $0 \mathrm{~A}$
4 $0.14 \mathrm{~A}$
Explanation:
C The emfs due to two of the cancel out. The capacitor in steady state. Hence, circuit is open. So, there is no current passes through it.
152483
In a potentiometer arrangement, a cell of emf 1.5 $V$ gives a Balance point an $150 \mathrm{~cm}$ length of the wire. If the cell is replaced by another cell and the balance point shift to $210 \mathrm{~cm}$, what is the emf of the second cell?
1 $3.2 \mathrm{~V}$
2 $1.2 \mathrm{~V}$
3 $4.4 \mathrm{~V}$
4 $2.1 \mathrm{~V}$
Explanation:
D Given $\mathrm{E}_{1}=1.5 \mathrm{volt}, \mathrm{E}_{2}=?$ $l_{1}=150 \mathrm{~cm}, l_{2}=210 \mathrm{~cm}$ We know that $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{l_{1}}{l_{2}}$ $\mathrm{E}_{2}=\frac{l_{2}}{l_{1}} \mathrm{E}_{1}=\frac{210}{150} \times 1.5$ $=2.1 \text { volt }$
GUJCET-PCE- 2021
Current Electricity
152484
If one cell is connected wrongly in a series combination of four cells each of e.m.f $1.5 \mathrm{~V}$ and internal resistance of $0.5 \Omega$, then the equivalent internal resistance of the combination is
1 $0.5 \Omega$
2 $1 \Omega$
3 $1.5 \Omega$
4 $2 \Omega$
5 $2.5 \Omega$
Explanation:
D Given that, $\mathrm{r}=0.5 \Omega=1 / 2$ $\mathrm{E}=1.5 \mathrm{~V}$ And $\mathrm{n}=4$ $r_{1}, r_{2}$ and $r_{3}, r_{4}$ or equivalent $\frac{1}{\mathrm{r}_{\mathrm{eq}}} =\frac{1}{\mathrm{r}_{1}}+\frac{1}{\mathrm{r}_{2}}+\frac{1}{\mathrm{r}_{3}}+\frac{1}{\mathrm{r}_{4}}$ $=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}$ $\frac{1}{\mathrm{r}_{\mathrm{eq}}}=\frac{4}{2}$ $\mathrm{r}_{\mathrm{eq}}=2 \Omega$
Kerala CEE 2021
Current Electricity
152485
In the given circuit, current flowing through it is
1 $5 \mathrm{~A}$
2 $3 \mathrm{~A}$
3 $2 \mathrm{~A}$
4 $4 \mathrm{~A}$
Explanation:
A The two cells are opposing to each other. The net emf in the circuit $=(200-10)=190 \mathrm{~V}$ $\mathrm{R}=38 \Omega$ $\therefore$ Current $(\mathrm{I})=\frac{\mathrm{V}}{\mathrm{R}}=\frac{190}{38}=5 \mathrm{~A}$
MHT-CET 2020
Current Electricity
152486
A cell of e.m.f. ' $E$ ' is connected across a resistance ' $R$ '. The potential difference across the terminals of the cell is $90 \%$ of $E$. The internal resistance of the cell is
1 $9 \mathrm{R}$
2 $\frac{\mathrm{R}}{9}$
3 $\frac{9}{10} \mathrm{R}$
4 $\frac{10}{9} \mathrm{R}$
Explanation:
B Let $r$ be the internal resistance of the cell. $\mathrm{R}$ Terminal potential difference $(\mathrm{V})\lt$ EMF of a cell $(\mathrm{E})$. $\mathrm{V} =\mathrm{E}-\mathrm{Ir}$ $\mathrm{r} =\left(\frac{\mathrm{E}-\mathrm{V}}{\mathrm{V}}\right) \mathrm{R}$ $=\left(\frac{\mathrm{E}}{\mathrm{V}}-1\right) \mathrm{R}=\left(\frac{10 \mathrm{E}}{9 \mathrm{E}}-1\right) \mathrm{R}$ $\mathrm{r} =\frac{\mathrm{R}}{9}$
MHT-CET 2020
Current Electricity
152487
Consider the circuit shown. If all the cells have negligible internal resistance, what will be the current through the $2 \Omega$ resistor when steady state is reached?
1 $0.66 \mathrm{~A}$
2 $0.29 \mathrm{~A}$
3 $0 \mathrm{~A}$
4 $0.14 \mathrm{~A}$
Explanation:
C The emfs due to two of the cancel out. The capacitor in steady state. Hence, circuit is open. So, there is no current passes through it.
152483
In a potentiometer arrangement, a cell of emf 1.5 $V$ gives a Balance point an $150 \mathrm{~cm}$ length of the wire. If the cell is replaced by another cell and the balance point shift to $210 \mathrm{~cm}$, what is the emf of the second cell?
1 $3.2 \mathrm{~V}$
2 $1.2 \mathrm{~V}$
3 $4.4 \mathrm{~V}$
4 $2.1 \mathrm{~V}$
Explanation:
D Given $\mathrm{E}_{1}=1.5 \mathrm{volt}, \mathrm{E}_{2}=?$ $l_{1}=150 \mathrm{~cm}, l_{2}=210 \mathrm{~cm}$ We know that $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{l_{1}}{l_{2}}$ $\mathrm{E}_{2}=\frac{l_{2}}{l_{1}} \mathrm{E}_{1}=\frac{210}{150} \times 1.5$ $=2.1 \text { volt }$
GUJCET-PCE- 2021
Current Electricity
152484
If one cell is connected wrongly in a series combination of four cells each of e.m.f $1.5 \mathrm{~V}$ and internal resistance of $0.5 \Omega$, then the equivalent internal resistance of the combination is
1 $0.5 \Omega$
2 $1 \Omega$
3 $1.5 \Omega$
4 $2 \Omega$
5 $2.5 \Omega$
Explanation:
D Given that, $\mathrm{r}=0.5 \Omega=1 / 2$ $\mathrm{E}=1.5 \mathrm{~V}$ And $\mathrm{n}=4$ $r_{1}, r_{2}$ and $r_{3}, r_{4}$ or equivalent $\frac{1}{\mathrm{r}_{\mathrm{eq}}} =\frac{1}{\mathrm{r}_{1}}+\frac{1}{\mathrm{r}_{2}}+\frac{1}{\mathrm{r}_{3}}+\frac{1}{\mathrm{r}_{4}}$ $=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}$ $\frac{1}{\mathrm{r}_{\mathrm{eq}}}=\frac{4}{2}$ $\mathrm{r}_{\mathrm{eq}}=2 \Omega$
Kerala CEE 2021
Current Electricity
152485
In the given circuit, current flowing through it is
1 $5 \mathrm{~A}$
2 $3 \mathrm{~A}$
3 $2 \mathrm{~A}$
4 $4 \mathrm{~A}$
Explanation:
A The two cells are opposing to each other. The net emf in the circuit $=(200-10)=190 \mathrm{~V}$ $\mathrm{R}=38 \Omega$ $\therefore$ Current $(\mathrm{I})=\frac{\mathrm{V}}{\mathrm{R}}=\frac{190}{38}=5 \mathrm{~A}$
MHT-CET 2020
Current Electricity
152486
A cell of e.m.f. ' $E$ ' is connected across a resistance ' $R$ '. The potential difference across the terminals of the cell is $90 \%$ of $E$. The internal resistance of the cell is
1 $9 \mathrm{R}$
2 $\frac{\mathrm{R}}{9}$
3 $\frac{9}{10} \mathrm{R}$
4 $\frac{10}{9} \mathrm{R}$
Explanation:
B Let $r$ be the internal resistance of the cell. $\mathrm{R}$ Terminal potential difference $(\mathrm{V})\lt$ EMF of a cell $(\mathrm{E})$. $\mathrm{V} =\mathrm{E}-\mathrm{Ir}$ $\mathrm{r} =\left(\frac{\mathrm{E}-\mathrm{V}}{\mathrm{V}}\right) \mathrm{R}$ $=\left(\frac{\mathrm{E}}{\mathrm{V}}-1\right) \mathrm{R}=\left(\frac{10 \mathrm{E}}{9 \mathrm{E}}-1\right) \mathrm{R}$ $\mathrm{r} =\frac{\mathrm{R}}{9}$
MHT-CET 2020
Current Electricity
152487
Consider the circuit shown. If all the cells have negligible internal resistance, what will be the current through the $2 \Omega$ resistor when steady state is reached?
1 $0.66 \mathrm{~A}$
2 $0.29 \mathrm{~A}$
3 $0 \mathrm{~A}$
4 $0.14 \mathrm{~A}$
Explanation:
C The emfs due to two of the cancel out. The capacitor in steady state. Hence, circuit is open. So, there is no current passes through it.
