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.
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.
