152582
Two similar accumulators each of emf $E$ and internal resistance $r$ are connected as shown in the following figure. Then, the potential difference between $x$ and $y$ is
1 $2 \mathrm{E}$
2 $\mathrm{E}$
3 zero
4 None of these
Explanation:
C Let, 'I' be the current in the above given circuit, then $E=E=I r+I r=2 I r$ $\text { Current, } I=\frac{E}{r}$ Potential difference $\left(\mathrm{V}_{\mathrm{x}}-\mathrm{V}_{\mathrm{y}}\right)=\mathrm{E}-\mathrm{Ir}$ $\mathrm{V}_{\mathrm{x}}-\mathrm{V}_{\mathrm{y}} =\mathrm{E}-\frac{\mathrm{E}}{\mathrm{r}} \times \mathrm{r}$ $=0$
Manipal UGET-2010
Current Electricity
152583
Calculate the value $E$, for given circuit, when value of $2 \mathrm{~A}$ current is either flowing in clockwise or anti- clockwise direction.
1 $32 \mathrm{~V}, 8 \mathrm{~V}$
2 $38 \mathrm{~V}, 2 \mathrm{~V}$
3 $32 \mathrm{~V}, 2 \mathrm{~V}$
4 $30 \mathrm{~V}, 8 \mathrm{~V}$
Explanation:
A Given, Current (I) = 2 Amp Case I, If current flow in clockwise Then, $6 \times 2+20-\mathrm{E}=0$ $\mathrm{E}=32 \mathrm{~V}$ Case II, If current flow anticlockwise Then, $6 \times 2+\mathrm{E}-20=0 \Rightarrow \mathrm{E}=8$
CG PET- 2008
Current Electricity
152584
Emf of generator is $6 \mathrm{~V}$ and internal resistance is $0.5 \mathrm{k} \Omega$. If internal resistance of voltmeter is $2.5 \mathrm{k} \Omega$ then reading of voltmeter must be
1 $10^{-3} \mathrm{~V}$
2 $1 \mathrm{~V}$
3 $5 \mathrm{~V}$
4 $10 \mathrm{~V}$
Explanation:
C Given that, emf of generator, $\mathrm{E}=6 \mathrm{~V}$ $\mathrm{r}=0.5 \mathrm{k} \Omega=0.5 \times 10^{3} \Omega$ Reading voltmeter, $=$ ? We know that, Current flowing through the circuit, $I=\frac{V}{R+G}$ $I=\frac{6}{(2.5+0.5) \times 10^{3}}=\frac{1}{500} \mathrm{~A}$ $\therefore \quad \mathrm{V}=\mathrm{I} \times \mathrm{G}$ $\mathrm{V}=\frac{1}{500} \times 2.5 \times 10^{3}$ $\mathrm{~V}=5 \text { Volt }$
CG PET- 2007
Current Electricity
152585
Thermo emf of a couple is $3 \mu \mathrm{V} /{ }^{\circ} \mathrm{C}$ if temperature of cold junction is $20^{\circ} \mathrm{C}$ and thermo emf is increased to $0.3 \mathrm{mV}$ then temperature of hot junction
1 $80^{\circ} \mathrm{C}$
2 $100^{\circ} \mathrm{C}$
3 $120^{\circ} \mathrm{C}$
4 $140^{\circ} \mathrm{C}$
Explanation:
C Given, $\mathrm{a} =3 \mu \mathrm{V} /{ }^{\circ} \mathrm{C}$ $\mathrm{a} =3 \times 10^{-6} \mathrm{~V}$ $\mathrm{e} =0.3 \mathrm{mV}=0.3 \times 10^{-3} \mathrm{~V}$ $=3 \times 10^{-4} \mathrm{~V}$ We know that, $\mathrm{e}=\mathrm{a} \times(\mathrm{T}-20)$ $\mathrm{T}-20=\frac{3 \times 10^{-4}}{3 \times 10^{-6}}=100^{\circ} \mathrm{C}$ The temperature at the hot junction. $\mathrm{T}-20=100$ $\mathrm{~T}=120^{\circ} \mathrm{C}$
152582
Two similar accumulators each of emf $E$ and internal resistance $r$ are connected as shown in the following figure. Then, the potential difference between $x$ and $y$ is
1 $2 \mathrm{E}$
2 $\mathrm{E}$
3 zero
4 None of these
Explanation:
C Let, 'I' be the current in the above given circuit, then $E=E=I r+I r=2 I r$ $\text { Current, } I=\frac{E}{r}$ Potential difference $\left(\mathrm{V}_{\mathrm{x}}-\mathrm{V}_{\mathrm{y}}\right)=\mathrm{E}-\mathrm{Ir}$ $\mathrm{V}_{\mathrm{x}}-\mathrm{V}_{\mathrm{y}} =\mathrm{E}-\frac{\mathrm{E}}{\mathrm{r}} \times \mathrm{r}$ $=0$
Manipal UGET-2010
Current Electricity
152583
Calculate the value $E$, for given circuit, when value of $2 \mathrm{~A}$ current is either flowing in clockwise or anti- clockwise direction.
1 $32 \mathrm{~V}, 8 \mathrm{~V}$
2 $38 \mathrm{~V}, 2 \mathrm{~V}$
3 $32 \mathrm{~V}, 2 \mathrm{~V}$
4 $30 \mathrm{~V}, 8 \mathrm{~V}$
Explanation:
A Given, Current (I) = 2 Amp Case I, If current flow in clockwise Then, $6 \times 2+20-\mathrm{E}=0$ $\mathrm{E}=32 \mathrm{~V}$ Case II, If current flow anticlockwise Then, $6 \times 2+\mathrm{E}-20=0 \Rightarrow \mathrm{E}=8$
CG PET- 2008
Current Electricity
152584
Emf of generator is $6 \mathrm{~V}$ and internal resistance is $0.5 \mathrm{k} \Omega$. If internal resistance of voltmeter is $2.5 \mathrm{k} \Omega$ then reading of voltmeter must be
1 $10^{-3} \mathrm{~V}$
2 $1 \mathrm{~V}$
3 $5 \mathrm{~V}$
4 $10 \mathrm{~V}$
Explanation:
C Given that, emf of generator, $\mathrm{E}=6 \mathrm{~V}$ $\mathrm{r}=0.5 \mathrm{k} \Omega=0.5 \times 10^{3} \Omega$ Reading voltmeter, $=$ ? We know that, Current flowing through the circuit, $I=\frac{V}{R+G}$ $I=\frac{6}{(2.5+0.5) \times 10^{3}}=\frac{1}{500} \mathrm{~A}$ $\therefore \quad \mathrm{V}=\mathrm{I} \times \mathrm{G}$ $\mathrm{V}=\frac{1}{500} \times 2.5 \times 10^{3}$ $\mathrm{~V}=5 \text { Volt }$
CG PET- 2007
Current Electricity
152585
Thermo emf of a couple is $3 \mu \mathrm{V} /{ }^{\circ} \mathrm{C}$ if temperature of cold junction is $20^{\circ} \mathrm{C}$ and thermo emf is increased to $0.3 \mathrm{mV}$ then temperature of hot junction
1 $80^{\circ} \mathrm{C}$
2 $100^{\circ} \mathrm{C}$
3 $120^{\circ} \mathrm{C}$
4 $140^{\circ} \mathrm{C}$
Explanation:
C Given, $\mathrm{a} =3 \mu \mathrm{V} /{ }^{\circ} \mathrm{C}$ $\mathrm{a} =3 \times 10^{-6} \mathrm{~V}$ $\mathrm{e} =0.3 \mathrm{mV}=0.3 \times 10^{-3} \mathrm{~V}$ $=3 \times 10^{-4} \mathrm{~V}$ We know that, $\mathrm{e}=\mathrm{a} \times(\mathrm{T}-20)$ $\mathrm{T}-20=\frac{3 \times 10^{-4}}{3 \times 10^{-6}}=100^{\circ} \mathrm{C}$ The temperature at the hot junction. $\mathrm{T}-20=100$ $\mathrm{~T}=120^{\circ} \mathrm{C}$
152582
Two similar accumulators each of emf $E$ and internal resistance $r$ are connected as shown in the following figure. Then, the potential difference between $x$ and $y$ is
1 $2 \mathrm{E}$
2 $\mathrm{E}$
3 zero
4 None of these
Explanation:
C Let, 'I' be the current in the above given circuit, then $E=E=I r+I r=2 I r$ $\text { Current, } I=\frac{E}{r}$ Potential difference $\left(\mathrm{V}_{\mathrm{x}}-\mathrm{V}_{\mathrm{y}}\right)=\mathrm{E}-\mathrm{Ir}$ $\mathrm{V}_{\mathrm{x}}-\mathrm{V}_{\mathrm{y}} =\mathrm{E}-\frac{\mathrm{E}}{\mathrm{r}} \times \mathrm{r}$ $=0$
Manipal UGET-2010
Current Electricity
152583
Calculate the value $E$, for given circuit, when value of $2 \mathrm{~A}$ current is either flowing in clockwise or anti- clockwise direction.
1 $32 \mathrm{~V}, 8 \mathrm{~V}$
2 $38 \mathrm{~V}, 2 \mathrm{~V}$
3 $32 \mathrm{~V}, 2 \mathrm{~V}$
4 $30 \mathrm{~V}, 8 \mathrm{~V}$
Explanation:
A Given, Current (I) = 2 Amp Case I, If current flow in clockwise Then, $6 \times 2+20-\mathrm{E}=0$ $\mathrm{E}=32 \mathrm{~V}$ Case II, If current flow anticlockwise Then, $6 \times 2+\mathrm{E}-20=0 \Rightarrow \mathrm{E}=8$
CG PET- 2008
Current Electricity
152584
Emf of generator is $6 \mathrm{~V}$ and internal resistance is $0.5 \mathrm{k} \Omega$. If internal resistance of voltmeter is $2.5 \mathrm{k} \Omega$ then reading of voltmeter must be
1 $10^{-3} \mathrm{~V}$
2 $1 \mathrm{~V}$
3 $5 \mathrm{~V}$
4 $10 \mathrm{~V}$
Explanation:
C Given that, emf of generator, $\mathrm{E}=6 \mathrm{~V}$ $\mathrm{r}=0.5 \mathrm{k} \Omega=0.5 \times 10^{3} \Omega$ Reading voltmeter, $=$ ? We know that, Current flowing through the circuit, $I=\frac{V}{R+G}$ $I=\frac{6}{(2.5+0.5) \times 10^{3}}=\frac{1}{500} \mathrm{~A}$ $\therefore \quad \mathrm{V}=\mathrm{I} \times \mathrm{G}$ $\mathrm{V}=\frac{1}{500} \times 2.5 \times 10^{3}$ $\mathrm{~V}=5 \text { Volt }$
CG PET- 2007
Current Electricity
152585
Thermo emf of a couple is $3 \mu \mathrm{V} /{ }^{\circ} \mathrm{C}$ if temperature of cold junction is $20^{\circ} \mathrm{C}$ and thermo emf is increased to $0.3 \mathrm{mV}$ then temperature of hot junction
1 $80^{\circ} \mathrm{C}$
2 $100^{\circ} \mathrm{C}$
3 $120^{\circ} \mathrm{C}$
4 $140^{\circ} \mathrm{C}$
Explanation:
C Given, $\mathrm{a} =3 \mu \mathrm{V} /{ }^{\circ} \mathrm{C}$ $\mathrm{a} =3 \times 10^{-6} \mathrm{~V}$ $\mathrm{e} =0.3 \mathrm{mV}=0.3 \times 10^{-3} \mathrm{~V}$ $=3 \times 10^{-4} \mathrm{~V}$ We know that, $\mathrm{e}=\mathrm{a} \times(\mathrm{T}-20)$ $\mathrm{T}-20=\frac{3 \times 10^{-4}}{3 \times 10^{-6}}=100^{\circ} \mathrm{C}$ The temperature at the hot junction. $\mathrm{T}-20=100$ $\mathrm{~T}=120^{\circ} \mathrm{C}$
152582
Two similar accumulators each of emf $E$ and internal resistance $r$ are connected as shown in the following figure. Then, the potential difference between $x$ and $y$ is
1 $2 \mathrm{E}$
2 $\mathrm{E}$
3 zero
4 None of these
Explanation:
C Let, 'I' be the current in the above given circuit, then $E=E=I r+I r=2 I r$ $\text { Current, } I=\frac{E}{r}$ Potential difference $\left(\mathrm{V}_{\mathrm{x}}-\mathrm{V}_{\mathrm{y}}\right)=\mathrm{E}-\mathrm{Ir}$ $\mathrm{V}_{\mathrm{x}}-\mathrm{V}_{\mathrm{y}} =\mathrm{E}-\frac{\mathrm{E}}{\mathrm{r}} \times \mathrm{r}$ $=0$
Manipal UGET-2010
Current Electricity
152583
Calculate the value $E$, for given circuit, when value of $2 \mathrm{~A}$ current is either flowing in clockwise or anti- clockwise direction.
1 $32 \mathrm{~V}, 8 \mathrm{~V}$
2 $38 \mathrm{~V}, 2 \mathrm{~V}$
3 $32 \mathrm{~V}, 2 \mathrm{~V}$
4 $30 \mathrm{~V}, 8 \mathrm{~V}$
Explanation:
A Given, Current (I) = 2 Amp Case I, If current flow in clockwise Then, $6 \times 2+20-\mathrm{E}=0$ $\mathrm{E}=32 \mathrm{~V}$ Case II, If current flow anticlockwise Then, $6 \times 2+\mathrm{E}-20=0 \Rightarrow \mathrm{E}=8$
CG PET- 2008
Current Electricity
152584
Emf of generator is $6 \mathrm{~V}$ and internal resistance is $0.5 \mathrm{k} \Omega$. If internal resistance of voltmeter is $2.5 \mathrm{k} \Omega$ then reading of voltmeter must be
1 $10^{-3} \mathrm{~V}$
2 $1 \mathrm{~V}$
3 $5 \mathrm{~V}$
4 $10 \mathrm{~V}$
Explanation:
C Given that, emf of generator, $\mathrm{E}=6 \mathrm{~V}$ $\mathrm{r}=0.5 \mathrm{k} \Omega=0.5 \times 10^{3} \Omega$ Reading voltmeter, $=$ ? We know that, Current flowing through the circuit, $I=\frac{V}{R+G}$ $I=\frac{6}{(2.5+0.5) \times 10^{3}}=\frac{1}{500} \mathrm{~A}$ $\therefore \quad \mathrm{V}=\mathrm{I} \times \mathrm{G}$ $\mathrm{V}=\frac{1}{500} \times 2.5 \times 10^{3}$ $\mathrm{~V}=5 \text { Volt }$
CG PET- 2007
Current Electricity
152585
Thermo emf of a couple is $3 \mu \mathrm{V} /{ }^{\circ} \mathrm{C}$ if temperature of cold junction is $20^{\circ} \mathrm{C}$ and thermo emf is increased to $0.3 \mathrm{mV}$ then temperature of hot junction
1 $80^{\circ} \mathrm{C}$
2 $100^{\circ} \mathrm{C}$
3 $120^{\circ} \mathrm{C}$
4 $140^{\circ} \mathrm{C}$
Explanation:
C Given, $\mathrm{a} =3 \mu \mathrm{V} /{ }^{\circ} \mathrm{C}$ $\mathrm{a} =3 \times 10^{-6} \mathrm{~V}$ $\mathrm{e} =0.3 \mathrm{mV}=0.3 \times 10^{-3} \mathrm{~V}$ $=3 \times 10^{-4} \mathrm{~V}$ We know that, $\mathrm{e}=\mathrm{a} \times(\mathrm{T}-20)$ $\mathrm{T}-20=\frac{3 \times 10^{-4}}{3 \times 10^{-6}}=100^{\circ} \mathrm{C}$ The temperature at the hot junction. $\mathrm{T}-20=100$ $\mathrm{~T}=120^{\circ} \mathrm{C}$
