152572
An electrical meter of internal resistance $20 \Omega$ gives a full scales deflection when one miliampere current, that can be measured by using three resistors of resistance $12 \Omega$ each, in milliampere is:
1 10
2 8
3 6
4 4
Explanation:
C Given that, Meter has a resistance $(r)=20 \Omega$ Current in full deflection $\left(\mathrm{I}_{\mathrm{f}}\right)=1 \times 10^{-3} \mathrm{~A}$ Now, when resistance connected in shunt then three resistors are in parallel then, $\frac{1}{\mathrm{R}_{\text {eq }}}=\frac{1}{12}+\frac{1}{12}+\frac{1}{12}$ $\mathrm{R}_{\mathrm{eq}}=4 \Omega$ Then, $\quad \frac{r \cdot I_{f}}{I-I_{f}}=R_{e q}$ $4 =\frac{1 \times 10^{-3} \times 20}{\mathrm{I}-1 \times 10^{-3}}$ $4 \mathrm{I} =20 \times 10^{-3}+4 \times 10^{-3}=24 \times 10^{-3}$ So, $\quad \mathrm{I}=\frac{24 \times 10^{-3}}{4}=6 \times 10^{-3} \mathrm{~A}$ $\mathrm{I}=6 \mathrm{~mA}$
AP EAMCET(Medical)-2004
Current Electricity
152573
In the circuit shown below, a voltmeter of internal resistance $R$, when connected across $B$ and $C$ reads $\frac{100}{3} \mathrm{~V}$. Neglecting the internal resistance of the cell, the value of $R$ is
152574
The emf of a deniel cell is $1.08 \mathrm{~V}$. When the terminals of the cell are connected to resistance of $3 \Omega$, the potential difference across the terminals is found to be $0.6 \mathrm{~V}$. Then, the internal resistance of the cell is
1 $1.8 \Omega$
2 $2.4 \Omega$
3 $3.24 \Omega$
4 $0.2 \Omega$
Explanation:
B Given that, $\mathrm{E}=1.08 \mathrm{~V}$ $\mathrm{~V}=0.6 \mathrm{~V}$ $\mathrm{R}=3 \Omega$ For internal resistance $\mathrm{V}=\mathrm{E}-\mathrm{Ir}$ $0.6=1.08-\mathrm{Ir}$ $\mathrm{Ir}=0.48$ But when external resistance is added then $\mathrm{E}=\mathrm{I}(\mathrm{r}+\mathrm{R})$ $\frac{1.08}{(3+\mathrm{r})}=\mathrm{I}$ From equation (i) - $\frac{1.08 \times \mathrm{r}}{(3+\mathrm{r})}=0.48$ Then, $\quad 0.6 \mathrm{r}=0.48 \times 3$ $\mathrm{r}=\frac{0.48 \times 3}{0.6}=2.4 \Omega$
AP EAMCET(Medical)-1997
Current Electricity
152575
The balancing length for a cell is $560 \mathrm{~cm}$ in a potentiometer experiment. When an external resistance of $10 \Omega$ is connected in parallel to a cell, the balancing length changes by $60 \mathrm{~cm}$. The internal resistance of the cell in $\mathrm{ohm}$, is
1 1.6
2 1.4
3 1.2
4 0.12
Explanation:
C Given that, $\mathrm{R}=10 \Omega$ $l_{1}=560 \mathrm{~cm}$ $l_{2}=560-60=500 \mathrm{~cm}$ Then internal Resistance $r=R\left(\frac{l_{1}}{l_{2}}-1\right)$ $r=R\left(\frac{560}{500}-1\right)$ $r=10 \times \frac{60}{500}=\frac{6}{5}$ $r=1.2 \Omega$
152572
An electrical meter of internal resistance $20 \Omega$ gives a full scales deflection when one miliampere current, that can be measured by using three resistors of resistance $12 \Omega$ each, in milliampere is:
1 10
2 8
3 6
4 4
Explanation:
C Given that, Meter has a resistance $(r)=20 \Omega$ Current in full deflection $\left(\mathrm{I}_{\mathrm{f}}\right)=1 \times 10^{-3} \mathrm{~A}$ Now, when resistance connected in shunt then three resistors are in parallel then, $\frac{1}{\mathrm{R}_{\text {eq }}}=\frac{1}{12}+\frac{1}{12}+\frac{1}{12}$ $\mathrm{R}_{\mathrm{eq}}=4 \Omega$ Then, $\quad \frac{r \cdot I_{f}}{I-I_{f}}=R_{e q}$ $4 =\frac{1 \times 10^{-3} \times 20}{\mathrm{I}-1 \times 10^{-3}}$ $4 \mathrm{I} =20 \times 10^{-3}+4 \times 10^{-3}=24 \times 10^{-3}$ So, $\quad \mathrm{I}=\frac{24 \times 10^{-3}}{4}=6 \times 10^{-3} \mathrm{~A}$ $\mathrm{I}=6 \mathrm{~mA}$
AP EAMCET(Medical)-2004
Current Electricity
152573
In the circuit shown below, a voltmeter of internal resistance $R$, when connected across $B$ and $C$ reads $\frac{100}{3} \mathrm{~V}$. Neglecting the internal resistance of the cell, the value of $R$ is
152574
The emf of a deniel cell is $1.08 \mathrm{~V}$. When the terminals of the cell are connected to resistance of $3 \Omega$, the potential difference across the terminals is found to be $0.6 \mathrm{~V}$. Then, the internal resistance of the cell is
1 $1.8 \Omega$
2 $2.4 \Omega$
3 $3.24 \Omega$
4 $0.2 \Omega$
Explanation:
B Given that, $\mathrm{E}=1.08 \mathrm{~V}$ $\mathrm{~V}=0.6 \mathrm{~V}$ $\mathrm{R}=3 \Omega$ For internal resistance $\mathrm{V}=\mathrm{E}-\mathrm{Ir}$ $0.6=1.08-\mathrm{Ir}$ $\mathrm{Ir}=0.48$ But when external resistance is added then $\mathrm{E}=\mathrm{I}(\mathrm{r}+\mathrm{R})$ $\frac{1.08}{(3+\mathrm{r})}=\mathrm{I}$ From equation (i) - $\frac{1.08 \times \mathrm{r}}{(3+\mathrm{r})}=0.48$ Then, $\quad 0.6 \mathrm{r}=0.48 \times 3$ $\mathrm{r}=\frac{0.48 \times 3}{0.6}=2.4 \Omega$
AP EAMCET(Medical)-1997
Current Electricity
152575
The balancing length for a cell is $560 \mathrm{~cm}$ in a potentiometer experiment. When an external resistance of $10 \Omega$ is connected in parallel to a cell, the balancing length changes by $60 \mathrm{~cm}$. The internal resistance of the cell in $\mathrm{ohm}$, is
1 1.6
2 1.4
3 1.2
4 0.12
Explanation:
C Given that, $\mathrm{R}=10 \Omega$ $l_{1}=560 \mathrm{~cm}$ $l_{2}=560-60=500 \mathrm{~cm}$ Then internal Resistance $r=R\left(\frac{l_{1}}{l_{2}}-1\right)$ $r=R\left(\frac{560}{500}-1\right)$ $r=10 \times \frac{60}{500}=\frac{6}{5}$ $r=1.2 \Omega$
152572
An electrical meter of internal resistance $20 \Omega$ gives a full scales deflection when one miliampere current, that can be measured by using three resistors of resistance $12 \Omega$ each, in milliampere is:
1 10
2 8
3 6
4 4
Explanation:
C Given that, Meter has a resistance $(r)=20 \Omega$ Current in full deflection $\left(\mathrm{I}_{\mathrm{f}}\right)=1 \times 10^{-3} \mathrm{~A}$ Now, when resistance connected in shunt then three resistors are in parallel then, $\frac{1}{\mathrm{R}_{\text {eq }}}=\frac{1}{12}+\frac{1}{12}+\frac{1}{12}$ $\mathrm{R}_{\mathrm{eq}}=4 \Omega$ Then, $\quad \frac{r \cdot I_{f}}{I-I_{f}}=R_{e q}$ $4 =\frac{1 \times 10^{-3} \times 20}{\mathrm{I}-1 \times 10^{-3}}$ $4 \mathrm{I} =20 \times 10^{-3}+4 \times 10^{-3}=24 \times 10^{-3}$ So, $\quad \mathrm{I}=\frac{24 \times 10^{-3}}{4}=6 \times 10^{-3} \mathrm{~A}$ $\mathrm{I}=6 \mathrm{~mA}$
AP EAMCET(Medical)-2004
Current Electricity
152573
In the circuit shown below, a voltmeter of internal resistance $R$, when connected across $B$ and $C$ reads $\frac{100}{3} \mathrm{~V}$. Neglecting the internal resistance of the cell, the value of $R$ is
152574
The emf of a deniel cell is $1.08 \mathrm{~V}$. When the terminals of the cell are connected to resistance of $3 \Omega$, the potential difference across the terminals is found to be $0.6 \mathrm{~V}$. Then, the internal resistance of the cell is
1 $1.8 \Omega$
2 $2.4 \Omega$
3 $3.24 \Omega$
4 $0.2 \Omega$
Explanation:
B Given that, $\mathrm{E}=1.08 \mathrm{~V}$ $\mathrm{~V}=0.6 \mathrm{~V}$ $\mathrm{R}=3 \Omega$ For internal resistance $\mathrm{V}=\mathrm{E}-\mathrm{Ir}$ $0.6=1.08-\mathrm{Ir}$ $\mathrm{Ir}=0.48$ But when external resistance is added then $\mathrm{E}=\mathrm{I}(\mathrm{r}+\mathrm{R})$ $\frac{1.08}{(3+\mathrm{r})}=\mathrm{I}$ From equation (i) - $\frac{1.08 \times \mathrm{r}}{(3+\mathrm{r})}=0.48$ Then, $\quad 0.6 \mathrm{r}=0.48 \times 3$ $\mathrm{r}=\frac{0.48 \times 3}{0.6}=2.4 \Omega$
AP EAMCET(Medical)-1997
Current Electricity
152575
The balancing length for a cell is $560 \mathrm{~cm}$ in a potentiometer experiment. When an external resistance of $10 \Omega$ is connected in parallel to a cell, the balancing length changes by $60 \mathrm{~cm}$. The internal resistance of the cell in $\mathrm{ohm}$, is
1 1.6
2 1.4
3 1.2
4 0.12
Explanation:
C Given that, $\mathrm{R}=10 \Omega$ $l_{1}=560 \mathrm{~cm}$ $l_{2}=560-60=500 \mathrm{~cm}$ Then internal Resistance $r=R\left(\frac{l_{1}}{l_{2}}-1\right)$ $r=R\left(\frac{560}{500}-1\right)$ $r=10 \times \frac{60}{500}=\frac{6}{5}$ $r=1.2 \Omega$
152572
An electrical meter of internal resistance $20 \Omega$ gives a full scales deflection when one miliampere current, that can be measured by using three resistors of resistance $12 \Omega$ each, in milliampere is:
1 10
2 8
3 6
4 4
Explanation:
C Given that, Meter has a resistance $(r)=20 \Omega$ Current in full deflection $\left(\mathrm{I}_{\mathrm{f}}\right)=1 \times 10^{-3} \mathrm{~A}$ Now, when resistance connected in shunt then three resistors are in parallel then, $\frac{1}{\mathrm{R}_{\text {eq }}}=\frac{1}{12}+\frac{1}{12}+\frac{1}{12}$ $\mathrm{R}_{\mathrm{eq}}=4 \Omega$ Then, $\quad \frac{r \cdot I_{f}}{I-I_{f}}=R_{e q}$ $4 =\frac{1 \times 10^{-3} \times 20}{\mathrm{I}-1 \times 10^{-3}}$ $4 \mathrm{I} =20 \times 10^{-3}+4 \times 10^{-3}=24 \times 10^{-3}$ So, $\quad \mathrm{I}=\frac{24 \times 10^{-3}}{4}=6 \times 10^{-3} \mathrm{~A}$ $\mathrm{I}=6 \mathrm{~mA}$
AP EAMCET(Medical)-2004
Current Electricity
152573
In the circuit shown below, a voltmeter of internal resistance $R$, when connected across $B$ and $C$ reads $\frac{100}{3} \mathrm{~V}$. Neglecting the internal resistance of the cell, the value of $R$ is
152574
The emf of a deniel cell is $1.08 \mathrm{~V}$. When the terminals of the cell are connected to resistance of $3 \Omega$, the potential difference across the terminals is found to be $0.6 \mathrm{~V}$. Then, the internal resistance of the cell is
1 $1.8 \Omega$
2 $2.4 \Omega$
3 $3.24 \Omega$
4 $0.2 \Omega$
Explanation:
B Given that, $\mathrm{E}=1.08 \mathrm{~V}$ $\mathrm{~V}=0.6 \mathrm{~V}$ $\mathrm{R}=3 \Omega$ For internal resistance $\mathrm{V}=\mathrm{E}-\mathrm{Ir}$ $0.6=1.08-\mathrm{Ir}$ $\mathrm{Ir}=0.48$ But when external resistance is added then $\mathrm{E}=\mathrm{I}(\mathrm{r}+\mathrm{R})$ $\frac{1.08}{(3+\mathrm{r})}=\mathrm{I}$ From equation (i) - $\frac{1.08 \times \mathrm{r}}{(3+\mathrm{r})}=0.48$ Then, $\quad 0.6 \mathrm{r}=0.48 \times 3$ $\mathrm{r}=\frac{0.48 \times 3}{0.6}=2.4 \Omega$
AP EAMCET(Medical)-1997
Current Electricity
152575
The balancing length for a cell is $560 \mathrm{~cm}$ in a potentiometer experiment. When an external resistance of $10 \Omega$ is connected in parallel to a cell, the balancing length changes by $60 \mathrm{~cm}$. The internal resistance of the cell in $\mathrm{ohm}$, is
1 1.6
2 1.4
3 1.2
4 0.12
Explanation:
C Given that, $\mathrm{R}=10 \Omega$ $l_{1}=560 \mathrm{~cm}$ $l_{2}=560-60=500 \mathrm{~cm}$ Then internal Resistance $r=R\left(\frac{l_{1}}{l_{2}}-1\right)$ $r=R\left(\frac{560}{500}-1\right)$ $r=10 \times \frac{60}{500}=\frac{6}{5}$ $r=1.2 \Omega$
