152589
A torch battery consisting of two cells of $1.45 \mathrm{~V}$ and an internal resistance $0.15 \Omega$, each cell sending currents through the filament of the lamps having resistance $1.5 \Omega$. The value of current will be
1 $16.11 \mathrm{~A}$
2 $1.611 \mathrm{~A}$
3 $0.1611 \mathrm{~A}$
4 $2.6 \mathrm{~A}$
Explanation:
B Applying KVL $-I(0.15)+1.45-\mathrm{I}(0.15)+1.45-1.5 \mathrm{I}=0$ $\mathrm{I}=\frac{2.9}{1.8}$ $\mathrm{I}=1.611 \mathrm{~A}$
CG PET- 2007
Current Electricity
152591
A voltmeter having a resistance of $998 \Omega$ is connected to a cell of emf $2 \mathrm{~V}$ and internal resistance $2 \Omega$. The error in the measurement of emf will be
1 $4 \times 10^{-1} \mathrm{~V}$
2 $2 \times 10^{-3} \mathrm{~V}$
3 $4 \times 10^{-3} \mathrm{~V}$
4 $2 \times 10^{-1} \mathrm{~V}$
Explanation:
C Given, \(\operatorname{emf}(\mathrm{E})=2 \mathrm{~V}\) Resistance of voltmeter \((\mathrm{R})=998 \Omega\) Internal resistance \((r)=2 \Omega\) Actual value of emf of cell \(=2 \mathrm{~V}\) \(\mathrm{I} =\frac{\mathrm{E}}{\mathrm{r}+\mathrm{R}} \Rightarrow \frac{1}{500} \mathrm{~A}\) \(\mathrm{~V}=\mathrm{E}-\mathrm{Ir}\) \(\mathrm{V} =2 \times 2 \mathrm{I}\) \(=2-2 \times \frac{1}{500}=\frac{998}{500} \mathrm{~V}\) \(\because\) Error \(=\) Actual value - measured value \(\text { Error }=2-\frac{998}{500}=\frac{2}{500}=4 \times 10^{-3} \text { Volt }\)
CG PET- 2004
Current Electricity
152592
A current of two amperes is flowing through a cell of emf $5 \mathrm{~V}$ and internal resistance $0.5 \Omega$ from negative to positive electrode. If the potential of negative electrode is $10 \mathrm{~V}$, the potential of positive electrode will be
1 $5 \mathrm{~V}$
2 $14 \mathrm{~V}$
3 $15 \mathrm{~V}$
4 $16 \mathrm{~V}$
Explanation:
B Given, $\operatorname{Emf}(\mathrm{E})=5 \mathrm{~V}$ Current $(\mathrm{I})=2 \mathrm{~A}$ Internal resistance $(\mathrm{r})=0.5 \Omega$ Voltage across internal resistance $\left(\mathrm{v}_{\mathrm{r}}\right)=\mathrm{Ir}=1 \mathrm{~V}$ From circuit diagram, current inside cell is from negative terminal to positive terminal. So, current outside at the cell is from positive terminal to negative terminal. $\mathrm{V}_{\text {positive }}>\mathrm{V}_{\text {negative }}$ then, Potential of postive electrode $\mathrm{V}_{+\mathrm{ve}}=\mathrm{V}_{-\mathrm{ve}}+\mathrm{Emf}-\mathrm{V}_{\mathrm{r}}$ $\mathrm{V}_{+\mathrm{ve}}=10+5-\mathrm{Ir}$ $\mathrm{V}_{+\mathrm{ve}}=10+5-2 \times 0.5$ $\mathrm{V}_{+\mathrm{ve}}=14 \mathrm{~V}$
CG PET- 2004
Current Electricity
152593
When connected across the terminals of a cell, a voltmeter measure $5 \mathrm{~V}$ and a connected ammeter measures 10 A of current. A resistance of $2 \Omega$ is connected across the terminals of the cell. The current flowing through this resistance will be
1 $2.5 \mathrm{~A}$
2 $2.0 \mathrm{~A}$
3 $5.0 \mathrm{~A}$
4 $7.5 \mathrm{~A}$
Explanation:
B Given, $\operatorname{Emf}(\mathrm{E})=5 \mathrm{~V}$ When ammeter connected, then Current $(\mathrm{I})=10 \mathrm{~A}$ Then, internal resistance $\mathrm{r}=\frac{\mathrm{V}}{\mathrm{I}}=\frac{5}{10}=0.5 \Omega$ We know, $I=\frac{E}{r+R}$ $I=\frac{5}{0.5+2}=\frac{5}{2.5}=2 \mathrm{~A}$
152589
A torch battery consisting of two cells of $1.45 \mathrm{~V}$ and an internal resistance $0.15 \Omega$, each cell sending currents through the filament of the lamps having resistance $1.5 \Omega$. The value of current will be
1 $16.11 \mathrm{~A}$
2 $1.611 \mathrm{~A}$
3 $0.1611 \mathrm{~A}$
4 $2.6 \mathrm{~A}$
Explanation:
B Applying KVL $-I(0.15)+1.45-\mathrm{I}(0.15)+1.45-1.5 \mathrm{I}=0$ $\mathrm{I}=\frac{2.9}{1.8}$ $\mathrm{I}=1.611 \mathrm{~A}$
CG PET- 2007
Current Electricity
152591
A voltmeter having a resistance of $998 \Omega$ is connected to a cell of emf $2 \mathrm{~V}$ and internal resistance $2 \Omega$. The error in the measurement of emf will be
1 $4 \times 10^{-1} \mathrm{~V}$
2 $2 \times 10^{-3} \mathrm{~V}$
3 $4 \times 10^{-3} \mathrm{~V}$
4 $2 \times 10^{-1} \mathrm{~V}$
Explanation:
C Given, \(\operatorname{emf}(\mathrm{E})=2 \mathrm{~V}\) Resistance of voltmeter \((\mathrm{R})=998 \Omega\) Internal resistance \((r)=2 \Omega\) Actual value of emf of cell \(=2 \mathrm{~V}\) \(\mathrm{I} =\frac{\mathrm{E}}{\mathrm{r}+\mathrm{R}} \Rightarrow \frac{1}{500} \mathrm{~A}\) \(\mathrm{~V}=\mathrm{E}-\mathrm{Ir}\) \(\mathrm{V} =2 \times 2 \mathrm{I}\) \(=2-2 \times \frac{1}{500}=\frac{998}{500} \mathrm{~V}\) \(\because\) Error \(=\) Actual value - measured value \(\text { Error }=2-\frac{998}{500}=\frac{2}{500}=4 \times 10^{-3} \text { Volt }\)
CG PET- 2004
Current Electricity
152592
A current of two amperes is flowing through a cell of emf $5 \mathrm{~V}$ and internal resistance $0.5 \Omega$ from negative to positive electrode. If the potential of negative electrode is $10 \mathrm{~V}$, the potential of positive electrode will be
1 $5 \mathrm{~V}$
2 $14 \mathrm{~V}$
3 $15 \mathrm{~V}$
4 $16 \mathrm{~V}$
Explanation:
B Given, $\operatorname{Emf}(\mathrm{E})=5 \mathrm{~V}$ Current $(\mathrm{I})=2 \mathrm{~A}$ Internal resistance $(\mathrm{r})=0.5 \Omega$ Voltage across internal resistance $\left(\mathrm{v}_{\mathrm{r}}\right)=\mathrm{Ir}=1 \mathrm{~V}$ From circuit diagram, current inside cell is from negative terminal to positive terminal. So, current outside at the cell is from positive terminal to negative terminal. $\mathrm{V}_{\text {positive }}>\mathrm{V}_{\text {negative }}$ then, Potential of postive electrode $\mathrm{V}_{+\mathrm{ve}}=\mathrm{V}_{-\mathrm{ve}}+\mathrm{Emf}-\mathrm{V}_{\mathrm{r}}$ $\mathrm{V}_{+\mathrm{ve}}=10+5-\mathrm{Ir}$ $\mathrm{V}_{+\mathrm{ve}}=10+5-2 \times 0.5$ $\mathrm{V}_{+\mathrm{ve}}=14 \mathrm{~V}$
CG PET- 2004
Current Electricity
152593
When connected across the terminals of a cell, a voltmeter measure $5 \mathrm{~V}$ and a connected ammeter measures 10 A of current. A resistance of $2 \Omega$ is connected across the terminals of the cell. The current flowing through this resistance will be
1 $2.5 \mathrm{~A}$
2 $2.0 \mathrm{~A}$
3 $5.0 \mathrm{~A}$
4 $7.5 \mathrm{~A}$
Explanation:
B Given, $\operatorname{Emf}(\mathrm{E})=5 \mathrm{~V}$ When ammeter connected, then Current $(\mathrm{I})=10 \mathrm{~A}$ Then, internal resistance $\mathrm{r}=\frac{\mathrm{V}}{\mathrm{I}}=\frac{5}{10}=0.5 \Omega$ We know, $I=\frac{E}{r+R}$ $I=\frac{5}{0.5+2}=\frac{5}{2.5}=2 \mathrm{~A}$
152589
A torch battery consisting of two cells of $1.45 \mathrm{~V}$ and an internal resistance $0.15 \Omega$, each cell sending currents through the filament of the lamps having resistance $1.5 \Omega$. The value of current will be
1 $16.11 \mathrm{~A}$
2 $1.611 \mathrm{~A}$
3 $0.1611 \mathrm{~A}$
4 $2.6 \mathrm{~A}$
Explanation:
B Applying KVL $-I(0.15)+1.45-\mathrm{I}(0.15)+1.45-1.5 \mathrm{I}=0$ $\mathrm{I}=\frac{2.9}{1.8}$ $\mathrm{I}=1.611 \mathrm{~A}$
CG PET- 2007
Current Electricity
152591
A voltmeter having a resistance of $998 \Omega$ is connected to a cell of emf $2 \mathrm{~V}$ and internal resistance $2 \Omega$. The error in the measurement of emf will be
1 $4 \times 10^{-1} \mathrm{~V}$
2 $2 \times 10^{-3} \mathrm{~V}$
3 $4 \times 10^{-3} \mathrm{~V}$
4 $2 \times 10^{-1} \mathrm{~V}$
Explanation:
C Given, \(\operatorname{emf}(\mathrm{E})=2 \mathrm{~V}\) Resistance of voltmeter \((\mathrm{R})=998 \Omega\) Internal resistance \((r)=2 \Omega\) Actual value of emf of cell \(=2 \mathrm{~V}\) \(\mathrm{I} =\frac{\mathrm{E}}{\mathrm{r}+\mathrm{R}} \Rightarrow \frac{1}{500} \mathrm{~A}\) \(\mathrm{~V}=\mathrm{E}-\mathrm{Ir}\) \(\mathrm{V} =2 \times 2 \mathrm{I}\) \(=2-2 \times \frac{1}{500}=\frac{998}{500} \mathrm{~V}\) \(\because\) Error \(=\) Actual value - measured value \(\text { Error }=2-\frac{998}{500}=\frac{2}{500}=4 \times 10^{-3} \text { Volt }\)
CG PET- 2004
Current Electricity
152592
A current of two amperes is flowing through a cell of emf $5 \mathrm{~V}$ and internal resistance $0.5 \Omega$ from negative to positive electrode. If the potential of negative electrode is $10 \mathrm{~V}$, the potential of positive electrode will be
1 $5 \mathrm{~V}$
2 $14 \mathrm{~V}$
3 $15 \mathrm{~V}$
4 $16 \mathrm{~V}$
Explanation:
B Given, $\operatorname{Emf}(\mathrm{E})=5 \mathrm{~V}$ Current $(\mathrm{I})=2 \mathrm{~A}$ Internal resistance $(\mathrm{r})=0.5 \Omega$ Voltage across internal resistance $\left(\mathrm{v}_{\mathrm{r}}\right)=\mathrm{Ir}=1 \mathrm{~V}$ From circuit diagram, current inside cell is from negative terminal to positive terminal. So, current outside at the cell is from positive terminal to negative terminal. $\mathrm{V}_{\text {positive }}>\mathrm{V}_{\text {negative }}$ then, Potential of postive electrode $\mathrm{V}_{+\mathrm{ve}}=\mathrm{V}_{-\mathrm{ve}}+\mathrm{Emf}-\mathrm{V}_{\mathrm{r}}$ $\mathrm{V}_{+\mathrm{ve}}=10+5-\mathrm{Ir}$ $\mathrm{V}_{+\mathrm{ve}}=10+5-2 \times 0.5$ $\mathrm{V}_{+\mathrm{ve}}=14 \mathrm{~V}$
CG PET- 2004
Current Electricity
152593
When connected across the terminals of a cell, a voltmeter measure $5 \mathrm{~V}$ and a connected ammeter measures 10 A of current. A resistance of $2 \Omega$ is connected across the terminals of the cell. The current flowing through this resistance will be
1 $2.5 \mathrm{~A}$
2 $2.0 \mathrm{~A}$
3 $5.0 \mathrm{~A}$
4 $7.5 \mathrm{~A}$
Explanation:
B Given, $\operatorname{Emf}(\mathrm{E})=5 \mathrm{~V}$ When ammeter connected, then Current $(\mathrm{I})=10 \mathrm{~A}$ Then, internal resistance $\mathrm{r}=\frac{\mathrm{V}}{\mathrm{I}}=\frac{5}{10}=0.5 \Omega$ We know, $I=\frac{E}{r+R}$ $I=\frac{5}{0.5+2}=\frac{5}{2.5}=2 \mathrm{~A}$
152589
A torch battery consisting of two cells of $1.45 \mathrm{~V}$ and an internal resistance $0.15 \Omega$, each cell sending currents through the filament of the lamps having resistance $1.5 \Omega$. The value of current will be
1 $16.11 \mathrm{~A}$
2 $1.611 \mathrm{~A}$
3 $0.1611 \mathrm{~A}$
4 $2.6 \mathrm{~A}$
Explanation:
B Applying KVL $-I(0.15)+1.45-\mathrm{I}(0.15)+1.45-1.5 \mathrm{I}=0$ $\mathrm{I}=\frac{2.9}{1.8}$ $\mathrm{I}=1.611 \mathrm{~A}$
CG PET- 2007
Current Electricity
152591
A voltmeter having a resistance of $998 \Omega$ is connected to a cell of emf $2 \mathrm{~V}$ and internal resistance $2 \Omega$. The error in the measurement of emf will be
1 $4 \times 10^{-1} \mathrm{~V}$
2 $2 \times 10^{-3} \mathrm{~V}$
3 $4 \times 10^{-3} \mathrm{~V}$
4 $2 \times 10^{-1} \mathrm{~V}$
Explanation:
C Given, \(\operatorname{emf}(\mathrm{E})=2 \mathrm{~V}\) Resistance of voltmeter \((\mathrm{R})=998 \Omega\) Internal resistance \((r)=2 \Omega\) Actual value of emf of cell \(=2 \mathrm{~V}\) \(\mathrm{I} =\frac{\mathrm{E}}{\mathrm{r}+\mathrm{R}} \Rightarrow \frac{1}{500} \mathrm{~A}\) \(\mathrm{~V}=\mathrm{E}-\mathrm{Ir}\) \(\mathrm{V} =2 \times 2 \mathrm{I}\) \(=2-2 \times \frac{1}{500}=\frac{998}{500} \mathrm{~V}\) \(\because\) Error \(=\) Actual value - measured value \(\text { Error }=2-\frac{998}{500}=\frac{2}{500}=4 \times 10^{-3} \text { Volt }\)
CG PET- 2004
Current Electricity
152592
A current of two amperes is flowing through a cell of emf $5 \mathrm{~V}$ and internal resistance $0.5 \Omega$ from negative to positive electrode. If the potential of negative electrode is $10 \mathrm{~V}$, the potential of positive electrode will be
1 $5 \mathrm{~V}$
2 $14 \mathrm{~V}$
3 $15 \mathrm{~V}$
4 $16 \mathrm{~V}$
Explanation:
B Given, $\operatorname{Emf}(\mathrm{E})=5 \mathrm{~V}$ Current $(\mathrm{I})=2 \mathrm{~A}$ Internal resistance $(\mathrm{r})=0.5 \Omega$ Voltage across internal resistance $\left(\mathrm{v}_{\mathrm{r}}\right)=\mathrm{Ir}=1 \mathrm{~V}$ From circuit diagram, current inside cell is from negative terminal to positive terminal. So, current outside at the cell is from positive terminal to negative terminal. $\mathrm{V}_{\text {positive }}>\mathrm{V}_{\text {negative }}$ then, Potential of postive electrode $\mathrm{V}_{+\mathrm{ve}}=\mathrm{V}_{-\mathrm{ve}}+\mathrm{Emf}-\mathrm{V}_{\mathrm{r}}$ $\mathrm{V}_{+\mathrm{ve}}=10+5-\mathrm{Ir}$ $\mathrm{V}_{+\mathrm{ve}}=10+5-2 \times 0.5$ $\mathrm{V}_{+\mathrm{ve}}=14 \mathrm{~V}$
CG PET- 2004
Current Electricity
152593
When connected across the terminals of a cell, a voltmeter measure $5 \mathrm{~V}$ and a connected ammeter measures 10 A of current. A resistance of $2 \Omega$ is connected across the terminals of the cell. The current flowing through this resistance will be
1 $2.5 \mathrm{~A}$
2 $2.0 \mathrm{~A}$
3 $5.0 \mathrm{~A}$
4 $7.5 \mathrm{~A}$
Explanation:
B Given, $\operatorname{Emf}(\mathrm{E})=5 \mathrm{~V}$ When ammeter connected, then Current $(\mathrm{I})=10 \mathrm{~A}$ Then, internal resistance $\mathrm{r}=\frac{\mathrm{V}}{\mathrm{I}}=\frac{5}{10}=0.5 \Omega$ We know, $I=\frac{E}{r+R}$ $I=\frac{5}{0.5+2}=\frac{5}{2.5}=2 \mathrm{~A}$
