152492
Internal resistance of a cell is independent of
1 the circuit elements connected to it
2 surface area of the electrode
3 distance between the electrode
4 concentration of the electrolytes
5 temperature of the electrolytes
Explanation:
A The internal resistance of a cell depends on the *Surface area of its electrodes. *Separation between its electrodes. *Nature concentration and temperature of electrolyte.
Kerala CEE 2020
Current Electricity
152501
A cell of emf $E$ and internal resistance $r$ is connected across a variable load resistance $R$. the graph drawn between its terminal voltage and resistance $R$ is
1
2
3
4
Explanation:
A (i) Graph between the terminal voltage V and Variable load resistor $\mathrm{R}$ (ii) Graph between the terminal voltage $V$ and current $i$
AP EAMCET (23.04.2019) Shift-I
Current Electricity
152522
A cell of emf $E$ is connected across a resistance R. The potential difference between the terminals of the cell is found to be $V$. The internal resistance of the cell must be:
C Given, emf of cell $=\mathrm{E}$ External resistance $=r$ Potential difference between the terminal $=\mathrm{V}$ Let, internal resistance $=\mathrm{r}$ We know that, $E=I R+I r$ $E=V+I r$ $\left(\because V=I R \text { or } I=\frac{V}{R}\right)$ $E-V=\frac{V}{R} r$ $\frac{(E-V)}{V} R=r$ $r=\frac{(E-V)}{V} R$
Assam CEE-2017
Current Electricity
152528
What is the maximum power output that can be obtained from a cell of emf $E$ and internal resistance $r$ ?
1 $2 \mathrm{E}^{2} / \mathrm{r}$
2 $E^{2} / 2 r$
3 $E^{2} / 4 r$
4 None of these
Explanation:
C Suppose, cell have emf E and internal resistance $r$ be connected with the series resistance $R$ Then, $\quad I=\frac{E}{r+R}$ $\mathrm{P}_{\max }=\frac{\mathrm{E}^{2} \mathrm{r}}{(\mathrm{r}+\mathrm{r})^{2}} \quad(\because \mathrm{R}=\mathrm{r} \text { for maximum }$ power) $P_{\max }=\frac{E^{2}}{4 r}$
152492
Internal resistance of a cell is independent of
1 the circuit elements connected to it
2 surface area of the electrode
3 distance between the electrode
4 concentration of the electrolytes
5 temperature of the electrolytes
Explanation:
A The internal resistance of a cell depends on the *Surface area of its electrodes. *Separation between its electrodes. *Nature concentration and temperature of electrolyte.
Kerala CEE 2020
Current Electricity
152501
A cell of emf $E$ and internal resistance $r$ is connected across a variable load resistance $R$. the graph drawn between its terminal voltage and resistance $R$ is
1
2
3
4
Explanation:
A (i) Graph between the terminal voltage V and Variable load resistor $\mathrm{R}$ (ii) Graph between the terminal voltage $V$ and current $i$
AP EAMCET (23.04.2019) Shift-I
Current Electricity
152522
A cell of emf $E$ is connected across a resistance R. The potential difference between the terminals of the cell is found to be $V$. The internal resistance of the cell must be:
C Given, emf of cell $=\mathrm{E}$ External resistance $=r$ Potential difference between the terminal $=\mathrm{V}$ Let, internal resistance $=\mathrm{r}$ We know that, $E=I R+I r$ $E=V+I r$ $\left(\because V=I R \text { or } I=\frac{V}{R}\right)$ $E-V=\frac{V}{R} r$ $\frac{(E-V)}{V} R=r$ $r=\frac{(E-V)}{V} R$
Assam CEE-2017
Current Electricity
152528
What is the maximum power output that can be obtained from a cell of emf $E$ and internal resistance $r$ ?
1 $2 \mathrm{E}^{2} / \mathrm{r}$
2 $E^{2} / 2 r$
3 $E^{2} / 4 r$
4 None of these
Explanation:
C Suppose, cell have emf E and internal resistance $r$ be connected with the series resistance $R$ Then, $\quad I=\frac{E}{r+R}$ $\mathrm{P}_{\max }=\frac{\mathrm{E}^{2} \mathrm{r}}{(\mathrm{r}+\mathrm{r})^{2}} \quad(\because \mathrm{R}=\mathrm{r} \text { for maximum }$ power) $P_{\max }=\frac{E^{2}}{4 r}$
152492
Internal resistance of a cell is independent of
1 the circuit elements connected to it
2 surface area of the electrode
3 distance between the electrode
4 concentration of the electrolytes
5 temperature of the electrolytes
Explanation:
A The internal resistance of a cell depends on the *Surface area of its electrodes. *Separation between its electrodes. *Nature concentration and temperature of electrolyte.
Kerala CEE 2020
Current Electricity
152501
A cell of emf $E$ and internal resistance $r$ is connected across a variable load resistance $R$. the graph drawn between its terminal voltage and resistance $R$ is
1
2
3
4
Explanation:
A (i) Graph between the terminal voltage V and Variable load resistor $\mathrm{R}$ (ii) Graph between the terminal voltage $V$ and current $i$
AP EAMCET (23.04.2019) Shift-I
Current Electricity
152522
A cell of emf $E$ is connected across a resistance R. The potential difference between the terminals of the cell is found to be $V$. The internal resistance of the cell must be:
C Given, emf of cell $=\mathrm{E}$ External resistance $=r$ Potential difference between the terminal $=\mathrm{V}$ Let, internal resistance $=\mathrm{r}$ We know that, $E=I R+I r$ $E=V+I r$ $\left(\because V=I R \text { or } I=\frac{V}{R}\right)$ $E-V=\frac{V}{R} r$ $\frac{(E-V)}{V} R=r$ $r=\frac{(E-V)}{V} R$
Assam CEE-2017
Current Electricity
152528
What is the maximum power output that can be obtained from a cell of emf $E$ and internal resistance $r$ ?
1 $2 \mathrm{E}^{2} / \mathrm{r}$
2 $E^{2} / 2 r$
3 $E^{2} / 4 r$
4 None of these
Explanation:
C Suppose, cell have emf E and internal resistance $r$ be connected with the series resistance $R$ Then, $\quad I=\frac{E}{r+R}$ $\mathrm{P}_{\max }=\frac{\mathrm{E}^{2} \mathrm{r}}{(\mathrm{r}+\mathrm{r})^{2}} \quad(\because \mathrm{R}=\mathrm{r} \text { for maximum }$ power) $P_{\max }=\frac{E^{2}}{4 r}$
152492
Internal resistance of a cell is independent of
1 the circuit elements connected to it
2 surface area of the electrode
3 distance between the electrode
4 concentration of the electrolytes
5 temperature of the electrolytes
Explanation:
A The internal resistance of a cell depends on the *Surface area of its electrodes. *Separation between its electrodes. *Nature concentration and temperature of electrolyte.
Kerala CEE 2020
Current Electricity
152501
A cell of emf $E$ and internal resistance $r$ is connected across a variable load resistance $R$. the graph drawn between its terminal voltage and resistance $R$ is
1
2
3
4
Explanation:
A (i) Graph between the terminal voltage V and Variable load resistor $\mathrm{R}$ (ii) Graph between the terminal voltage $V$ and current $i$
AP EAMCET (23.04.2019) Shift-I
Current Electricity
152522
A cell of emf $E$ is connected across a resistance R. The potential difference between the terminals of the cell is found to be $V$. The internal resistance of the cell must be:
C Given, emf of cell $=\mathrm{E}$ External resistance $=r$ Potential difference between the terminal $=\mathrm{V}$ Let, internal resistance $=\mathrm{r}$ We know that, $E=I R+I r$ $E=V+I r$ $\left(\because V=I R \text { or } I=\frac{V}{R}\right)$ $E-V=\frac{V}{R} r$ $\frac{(E-V)}{V} R=r$ $r=\frac{(E-V)}{V} R$
Assam CEE-2017
Current Electricity
152528
What is the maximum power output that can be obtained from a cell of emf $E$ and internal resistance $r$ ?
1 $2 \mathrm{E}^{2} / \mathrm{r}$
2 $E^{2} / 2 r$
3 $E^{2} / 4 r$
4 None of these
Explanation:
C Suppose, cell have emf E and internal resistance $r$ be connected with the series resistance $R$ Then, $\quad I=\frac{E}{r+R}$ $\mathrm{P}_{\max }=\frac{\mathrm{E}^{2} \mathrm{r}}{(\mathrm{r}+\mathrm{r})^{2}} \quad(\because \mathrm{R}=\mathrm{r} \text { for maximum }$ power) $P_{\max }=\frac{E^{2}}{4 r}$
