152496
In the circuit shown, find the internal resistance value ' $r$ ' of the ac generator, if the average rate at which energy is dissipated in $R$ is maximum
1 $\mathrm{R}$
2 $\mathrm{R}^{2}$
3 $\sqrt{\mathrm{R}}$
4 $\frac{R}{2}$
Explanation:
D : We know $I=\frac{E}{r+\frac{R}{2}}$ $I=E\left(r+\frac{R}{2}\right)$ And $\text { Power }(P)=I^{2} R$ $P=\frac{E^{2}}{\left(r+\frac{R}{2}\right)^{2}}$ For maximum power, $\frac{\mathrm{dP}}{\mathrm{dR}}=0$ $\mathrm{E}^{2}\left[\frac{\left(\mathrm{r}+\frac{\mathrm{R}}{2}\right)^{2}-2 \mathrm{R}\left(\mathrm{r}+\frac{\mathrm{R}}{2}\right) \frac{1}{2}}{\left(\mathrm{r}+\frac{\mathrm{R}}{2}\right)^{4}}\right]=0$ $\left(r+\frac{R}{2}\right)^{2}=\frac{2 R}{2}\left(r+\frac{R}{2}\right)$ $r+\frac{R}{2}=R$ $r=R-\frac{R}{2}$ $r=\frac{R}{2}$
TS EAMCET 09.05.2019
Current Electricity
152497
A cell of internal resistance $r$ is connected across an external resistance $n r$. Then the ratio of the terminal voltage to the emf of the cell is
1 $\frac{1}{n}$
2 $\frac{1}{n+1}$
3 $\frac{\mathrm{n}}{\mathrm{n}+1}$
4 $\frac{\mathrm{n}-1}{\mathrm{n}}$
Explanation:
C Let, $\mathrm{E}=\mathrm{emf}$ of cell Then current, $I=\frac{E}{r+n r}$ Terminal voltage across battery, $\mathrm{V}=\mathrm{E}-\mathrm{Ir}$ $\mathrm{V}=\mathrm{E}-\frac{\mathrm{E}}{\mathrm{r}+\mathrm{nr}} \mathrm{r}$ $\mathrm{V}=\mathrm{E}-\frac{\mathrm{E}}{\mathrm{n}+1}$ $\mathrm{~V}=\mathrm{E}\left[1-\frac{1}{\mathrm{n}+1}\right]$ $\frac{\mathrm{V}}{\mathrm{E}}=\frac{\mathrm{n}+1-1}{\mathrm{n}+1}$ $\frac{\mathrm{V}}{\mathrm{E}}=\frac{\mathrm{n}}{\mathrm{n}+1}$
AIIMS-26.05.2019(E) Shift-2
Current Electricity
152498
4 cells each of emf $2 \mathrm{~V}$ and internal resistance of $1 \Omega$ are connected in parallel to a load resistor of $2 \Omega$. Then the current through the load resistor is
1 $2 \mathrm{~A}$
2 $1.5 \mathrm{~A}$
3 $1 \mathrm{~A}$
4 $0.888 \mathrm{~A}$
Explanation:
D Let $r$ be the internal resistance of each cell. Then, $r=1 \Omega$ $\frac{1}{\mathrm{r}_{\text {eq }}}=\frac{1}{1}+\frac{1}{1}+\frac{1}{1}+\frac{1}{1}$ $\frac{1}{\mathrm{r}_{\text {eq }}}=4$ $\mathrm{r}_{\text {eq }}=\frac{1}{4} \Omega$ $\mathrm{R}^{\prime}=\mathrm{R}+\frac{1}{4}=2+\frac{1}{4}=\frac{9}{4}$ $\therefore$ Current through $2 \Omega$ resistor, $i=\frac{V}{R^{\prime}}=\frac{2}{\frac{9}{4}}=2 \times \frac{4}{9}=\frac{8}{9}$ $i=0.888 \mathrm{~A}$
COMEDK 2019
Current Electricity
152499
Which one of the following equation is the correct. equation for the electrical circuit shown in the figure?
B Applying Kirchhoff's voltage law in loop (1) We, have $E_{2}-i_{2} r_{2}-E_{1}-\left(-i_{1}\right) r_{1}=0$ $\Rightarrow \quad \mathrm{E}_{2}-\mathrm{i}_{2} \mathrm{r}_{2}-\mathrm{E}_{1}+\mathrm{i}_{1} \mathrm{r}_{1}=0$ For loop (2), $E_{1}-\left(i_{1}+i_{2}\right) R-i_{1} r_{1}=0$ Thus, from equation (i) \& (ii). We have the correct match with equation (ii). So option (b) is correct.
152496
In the circuit shown, find the internal resistance value ' $r$ ' of the ac generator, if the average rate at which energy is dissipated in $R$ is maximum
1 $\mathrm{R}$
2 $\mathrm{R}^{2}$
3 $\sqrt{\mathrm{R}}$
4 $\frac{R}{2}$
Explanation:
D : We know $I=\frac{E}{r+\frac{R}{2}}$ $I=E\left(r+\frac{R}{2}\right)$ And $\text { Power }(P)=I^{2} R$ $P=\frac{E^{2}}{\left(r+\frac{R}{2}\right)^{2}}$ For maximum power, $\frac{\mathrm{dP}}{\mathrm{dR}}=0$ $\mathrm{E}^{2}\left[\frac{\left(\mathrm{r}+\frac{\mathrm{R}}{2}\right)^{2}-2 \mathrm{R}\left(\mathrm{r}+\frac{\mathrm{R}}{2}\right) \frac{1}{2}}{\left(\mathrm{r}+\frac{\mathrm{R}}{2}\right)^{4}}\right]=0$ $\left(r+\frac{R}{2}\right)^{2}=\frac{2 R}{2}\left(r+\frac{R}{2}\right)$ $r+\frac{R}{2}=R$ $r=R-\frac{R}{2}$ $r=\frac{R}{2}$
TS EAMCET 09.05.2019
Current Electricity
152497
A cell of internal resistance $r$ is connected across an external resistance $n r$. Then the ratio of the terminal voltage to the emf of the cell is
1 $\frac{1}{n}$
2 $\frac{1}{n+1}$
3 $\frac{\mathrm{n}}{\mathrm{n}+1}$
4 $\frac{\mathrm{n}-1}{\mathrm{n}}$
Explanation:
C Let, $\mathrm{E}=\mathrm{emf}$ of cell Then current, $I=\frac{E}{r+n r}$ Terminal voltage across battery, $\mathrm{V}=\mathrm{E}-\mathrm{Ir}$ $\mathrm{V}=\mathrm{E}-\frac{\mathrm{E}}{\mathrm{r}+\mathrm{nr}} \mathrm{r}$ $\mathrm{V}=\mathrm{E}-\frac{\mathrm{E}}{\mathrm{n}+1}$ $\mathrm{~V}=\mathrm{E}\left[1-\frac{1}{\mathrm{n}+1}\right]$ $\frac{\mathrm{V}}{\mathrm{E}}=\frac{\mathrm{n}+1-1}{\mathrm{n}+1}$ $\frac{\mathrm{V}}{\mathrm{E}}=\frac{\mathrm{n}}{\mathrm{n}+1}$
AIIMS-26.05.2019(E) Shift-2
Current Electricity
152498
4 cells each of emf $2 \mathrm{~V}$ and internal resistance of $1 \Omega$ are connected in parallel to a load resistor of $2 \Omega$. Then the current through the load resistor is
1 $2 \mathrm{~A}$
2 $1.5 \mathrm{~A}$
3 $1 \mathrm{~A}$
4 $0.888 \mathrm{~A}$
Explanation:
D Let $r$ be the internal resistance of each cell. Then, $r=1 \Omega$ $\frac{1}{\mathrm{r}_{\text {eq }}}=\frac{1}{1}+\frac{1}{1}+\frac{1}{1}+\frac{1}{1}$ $\frac{1}{\mathrm{r}_{\text {eq }}}=4$ $\mathrm{r}_{\text {eq }}=\frac{1}{4} \Omega$ $\mathrm{R}^{\prime}=\mathrm{R}+\frac{1}{4}=2+\frac{1}{4}=\frac{9}{4}$ $\therefore$ Current through $2 \Omega$ resistor, $i=\frac{V}{R^{\prime}}=\frac{2}{\frac{9}{4}}=2 \times \frac{4}{9}=\frac{8}{9}$ $i=0.888 \mathrm{~A}$
COMEDK 2019
Current Electricity
152499
Which one of the following equation is the correct. equation for the electrical circuit shown in the figure?
B Applying Kirchhoff's voltage law in loop (1) We, have $E_{2}-i_{2} r_{2}-E_{1}-\left(-i_{1}\right) r_{1}=0$ $\Rightarrow \quad \mathrm{E}_{2}-\mathrm{i}_{2} \mathrm{r}_{2}-\mathrm{E}_{1}+\mathrm{i}_{1} \mathrm{r}_{1}=0$ For loop (2), $E_{1}-\left(i_{1}+i_{2}\right) R-i_{1} r_{1}=0$ Thus, from equation (i) \& (ii). We have the correct match with equation (ii). So option (b) is correct.
NEET Test Series from KOTA - 10 Papers In MS WORD
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Current Electricity
152496
In the circuit shown, find the internal resistance value ' $r$ ' of the ac generator, if the average rate at which energy is dissipated in $R$ is maximum
1 $\mathrm{R}$
2 $\mathrm{R}^{2}$
3 $\sqrt{\mathrm{R}}$
4 $\frac{R}{2}$
Explanation:
D : We know $I=\frac{E}{r+\frac{R}{2}}$ $I=E\left(r+\frac{R}{2}\right)$ And $\text { Power }(P)=I^{2} R$ $P=\frac{E^{2}}{\left(r+\frac{R}{2}\right)^{2}}$ For maximum power, $\frac{\mathrm{dP}}{\mathrm{dR}}=0$ $\mathrm{E}^{2}\left[\frac{\left(\mathrm{r}+\frac{\mathrm{R}}{2}\right)^{2}-2 \mathrm{R}\left(\mathrm{r}+\frac{\mathrm{R}}{2}\right) \frac{1}{2}}{\left(\mathrm{r}+\frac{\mathrm{R}}{2}\right)^{4}}\right]=0$ $\left(r+\frac{R}{2}\right)^{2}=\frac{2 R}{2}\left(r+\frac{R}{2}\right)$ $r+\frac{R}{2}=R$ $r=R-\frac{R}{2}$ $r=\frac{R}{2}$
TS EAMCET 09.05.2019
Current Electricity
152497
A cell of internal resistance $r$ is connected across an external resistance $n r$. Then the ratio of the terminal voltage to the emf of the cell is
1 $\frac{1}{n}$
2 $\frac{1}{n+1}$
3 $\frac{\mathrm{n}}{\mathrm{n}+1}$
4 $\frac{\mathrm{n}-1}{\mathrm{n}}$
Explanation:
C Let, $\mathrm{E}=\mathrm{emf}$ of cell Then current, $I=\frac{E}{r+n r}$ Terminal voltage across battery, $\mathrm{V}=\mathrm{E}-\mathrm{Ir}$ $\mathrm{V}=\mathrm{E}-\frac{\mathrm{E}}{\mathrm{r}+\mathrm{nr}} \mathrm{r}$ $\mathrm{V}=\mathrm{E}-\frac{\mathrm{E}}{\mathrm{n}+1}$ $\mathrm{~V}=\mathrm{E}\left[1-\frac{1}{\mathrm{n}+1}\right]$ $\frac{\mathrm{V}}{\mathrm{E}}=\frac{\mathrm{n}+1-1}{\mathrm{n}+1}$ $\frac{\mathrm{V}}{\mathrm{E}}=\frac{\mathrm{n}}{\mathrm{n}+1}$
AIIMS-26.05.2019(E) Shift-2
Current Electricity
152498
4 cells each of emf $2 \mathrm{~V}$ and internal resistance of $1 \Omega$ are connected in parallel to a load resistor of $2 \Omega$. Then the current through the load resistor is
1 $2 \mathrm{~A}$
2 $1.5 \mathrm{~A}$
3 $1 \mathrm{~A}$
4 $0.888 \mathrm{~A}$
Explanation:
D Let $r$ be the internal resistance of each cell. Then, $r=1 \Omega$ $\frac{1}{\mathrm{r}_{\text {eq }}}=\frac{1}{1}+\frac{1}{1}+\frac{1}{1}+\frac{1}{1}$ $\frac{1}{\mathrm{r}_{\text {eq }}}=4$ $\mathrm{r}_{\text {eq }}=\frac{1}{4} \Omega$ $\mathrm{R}^{\prime}=\mathrm{R}+\frac{1}{4}=2+\frac{1}{4}=\frac{9}{4}$ $\therefore$ Current through $2 \Omega$ resistor, $i=\frac{V}{R^{\prime}}=\frac{2}{\frac{9}{4}}=2 \times \frac{4}{9}=\frac{8}{9}$ $i=0.888 \mathrm{~A}$
COMEDK 2019
Current Electricity
152499
Which one of the following equation is the correct. equation for the electrical circuit shown in the figure?
B Applying Kirchhoff's voltage law in loop (1) We, have $E_{2}-i_{2} r_{2}-E_{1}-\left(-i_{1}\right) r_{1}=0$ $\Rightarrow \quad \mathrm{E}_{2}-\mathrm{i}_{2} \mathrm{r}_{2}-\mathrm{E}_{1}+\mathrm{i}_{1} \mathrm{r}_{1}=0$ For loop (2), $E_{1}-\left(i_{1}+i_{2}\right) R-i_{1} r_{1}=0$ Thus, from equation (i) \& (ii). We have the correct match with equation (ii). So option (b) is correct.
152496
In the circuit shown, find the internal resistance value ' $r$ ' of the ac generator, if the average rate at which energy is dissipated in $R$ is maximum
1 $\mathrm{R}$
2 $\mathrm{R}^{2}$
3 $\sqrt{\mathrm{R}}$
4 $\frac{R}{2}$
Explanation:
D : We know $I=\frac{E}{r+\frac{R}{2}}$ $I=E\left(r+\frac{R}{2}\right)$ And $\text { Power }(P)=I^{2} R$ $P=\frac{E^{2}}{\left(r+\frac{R}{2}\right)^{2}}$ For maximum power, $\frac{\mathrm{dP}}{\mathrm{dR}}=0$ $\mathrm{E}^{2}\left[\frac{\left(\mathrm{r}+\frac{\mathrm{R}}{2}\right)^{2}-2 \mathrm{R}\left(\mathrm{r}+\frac{\mathrm{R}}{2}\right) \frac{1}{2}}{\left(\mathrm{r}+\frac{\mathrm{R}}{2}\right)^{4}}\right]=0$ $\left(r+\frac{R}{2}\right)^{2}=\frac{2 R}{2}\left(r+\frac{R}{2}\right)$ $r+\frac{R}{2}=R$ $r=R-\frac{R}{2}$ $r=\frac{R}{2}$
TS EAMCET 09.05.2019
Current Electricity
152497
A cell of internal resistance $r$ is connected across an external resistance $n r$. Then the ratio of the terminal voltage to the emf of the cell is
1 $\frac{1}{n}$
2 $\frac{1}{n+1}$
3 $\frac{\mathrm{n}}{\mathrm{n}+1}$
4 $\frac{\mathrm{n}-1}{\mathrm{n}}$
Explanation:
C Let, $\mathrm{E}=\mathrm{emf}$ of cell Then current, $I=\frac{E}{r+n r}$ Terminal voltage across battery, $\mathrm{V}=\mathrm{E}-\mathrm{Ir}$ $\mathrm{V}=\mathrm{E}-\frac{\mathrm{E}}{\mathrm{r}+\mathrm{nr}} \mathrm{r}$ $\mathrm{V}=\mathrm{E}-\frac{\mathrm{E}}{\mathrm{n}+1}$ $\mathrm{~V}=\mathrm{E}\left[1-\frac{1}{\mathrm{n}+1}\right]$ $\frac{\mathrm{V}}{\mathrm{E}}=\frac{\mathrm{n}+1-1}{\mathrm{n}+1}$ $\frac{\mathrm{V}}{\mathrm{E}}=\frac{\mathrm{n}}{\mathrm{n}+1}$
AIIMS-26.05.2019(E) Shift-2
Current Electricity
152498
4 cells each of emf $2 \mathrm{~V}$ and internal resistance of $1 \Omega$ are connected in parallel to a load resistor of $2 \Omega$. Then the current through the load resistor is
1 $2 \mathrm{~A}$
2 $1.5 \mathrm{~A}$
3 $1 \mathrm{~A}$
4 $0.888 \mathrm{~A}$
Explanation:
D Let $r$ be the internal resistance of each cell. Then, $r=1 \Omega$ $\frac{1}{\mathrm{r}_{\text {eq }}}=\frac{1}{1}+\frac{1}{1}+\frac{1}{1}+\frac{1}{1}$ $\frac{1}{\mathrm{r}_{\text {eq }}}=4$ $\mathrm{r}_{\text {eq }}=\frac{1}{4} \Omega$ $\mathrm{R}^{\prime}=\mathrm{R}+\frac{1}{4}=2+\frac{1}{4}=\frac{9}{4}$ $\therefore$ Current through $2 \Omega$ resistor, $i=\frac{V}{R^{\prime}}=\frac{2}{\frac{9}{4}}=2 \times \frac{4}{9}=\frac{8}{9}$ $i=0.888 \mathrm{~A}$
COMEDK 2019
Current Electricity
152499
Which one of the following equation is the correct. equation for the electrical circuit shown in the figure?
B Applying Kirchhoff's voltage law in loop (1) We, have $E_{2}-i_{2} r_{2}-E_{1}-\left(-i_{1}\right) r_{1}=0$ $\Rightarrow \quad \mathrm{E}_{2}-\mathrm{i}_{2} \mathrm{r}_{2}-\mathrm{E}_{1}+\mathrm{i}_{1} \mathrm{r}_{1}=0$ For loop (2), $E_{1}-\left(i_{1}+i_{2}\right) R-i_{1} r_{1}=0$ Thus, from equation (i) \& (ii). We have the correct match with equation (ii). So option (b) is correct.
