152761
The electric current passes through a metallic wire produces heat because of- 1001. Two wires have resistances $R$ and $2 R$. When both are joining in series and in parallel, then ratio of heats generated in these situations on applying the same voltage, is :
1 collisions of conduction electrons with each other
2 collisions of the atoms of the metal with each other
3 the energy released in the ionization of the atoms of the metal
4 collisions of the conduction electrons with the atoms of the metallic wire
Explanation:
D When a current passes through a wire, a part of the electrical energy is converted into heat energy as a result of some resistance experienced by it and as a result the wire gets heated. This is known as heating effect of electric current. The electric current passes through a metallic wire produces heat because of collisions of the conduction electrons with the atoms of the metallic wire. (a) $2: 1$ (b) $1: 2$ (c) $2: 9$ (d) $9: 2$ Ans: c : Given, Resistance $\left(\mathrm{R}_{1}\right)=\mathrm{R}$ $R_{2}=2 R$ When two wires are joining in series. The net resistance $\left(R^{\prime}\right)=R_{1}+R_{2}$ $\mathrm{R}^{\prime}=\mathrm{R}+2 \mathrm{R}$ $\mathrm{R}^{\prime}=3 \mathrm{R}$ Heat produced in $\mathrm{R}^{\prime}$, $\mathrm{H}_{\mathrm{S}}=\frac{\mathrm{V}^{2}}{\mathrm{R}^{\prime}}=\frac{\mathrm{V}^{2}}{3 \mathrm{R}}$ When two wires are joining in parallel. The net resistance $\left(\frac{1}{\mathrm{R}^{\prime \prime}}\right)=\frac{1}{\mathrm{R}}+\frac{1}{2 \mathrm{R}}$ $\mathrm{R}^{\prime \prime}=\frac{2 \mathrm{R}}{3}$ Heat produced in R", $\mathrm{H}^{\prime \prime}=\frac{\mathrm{V}^{2}}{\mathrm{R}}, \quad \mathrm{H}^{\prime \prime}=\frac{\mathrm{V}^{2}}{\frac{2 \mathrm{R}}{3}}$ $\mathrm{H}^{\prime \prime}=\frac{3}{2} \frac{\mathrm{V}^{2}}{\mathrm{R}}$ On dividing equation (i) by equation (ii) $\frac{\mathrm{H}^{\prime}}{\mathrm{H}^{\prime \prime}} =\frac{\frac{\mathrm{V}^{2}}{3 \mathrm{R}}}{\frac{3 \mathrm{~V}^{2}}{2 \mathrm{R}}}$ $\frac{\mathrm{H}^{\prime}}{\mathrm{H}^{\prime \prime}} =\frac{2}{9}$ $\therefore \quad \mathrm{H}^{\prime}: \mathrm{H}^{\prime \prime} =2: 9$ Current Electricity
BCECE-2004
Current Electricity
152762
The electric current passes through a metallic wire produces heat because of- 1001. Two wires have resistances $R$ and $2 R$. When both are joining in series and in parallel, then ratio of heats generated in these situations on applying the same voltage, is :
1 collisions of conduction electrons with each other
2 collisions of the atoms of the metal with each other
3 the energy released in the ionization of the atoms of the metal
4 collisions of the conduction electrons with the atoms of the metallic wire
Explanation:
D When a current passes through a wire, a part of the electrical energy is converted into heat energy as a result of some resistance experienced by it and as a result the wire gets heated. This is known as heating effect of electric current. The electric current passes through a metallic wire produces heat because of collisions of the conduction electrons with the atoms of the metallic wire. (a) $2: 1$ (b) $1: 2$ (c) $2: 9$ (d) $9: 2$ Ans: c : Given, Resistance $\left(\mathrm{R}_{1}\right)=\mathrm{R}$ $R_{2}=2 R$ When two wires are joining in series. The net resistance $\left(R^{\prime}\right)=R_{1}+R_{2}$ $\mathrm{R}^{\prime}=\mathrm{R}+2 \mathrm{R}$ $\mathrm{R}^{\prime}=3 \mathrm{R}$ Heat produced in $\mathrm{R}^{\prime}$, $\mathrm{H}_{\mathrm{S}}=\frac{\mathrm{V}^{2}}{\mathrm{R}^{\prime}}=\frac{\mathrm{V}^{2}}{3 \mathrm{R}}$ When two wires are joining in parallel. The net resistance $\left(\frac{1}{\mathrm{R}^{\prime \prime}}\right)=\frac{1}{\mathrm{R}}+\frac{1}{2 \mathrm{R}}$ $\mathrm{R}^{\prime \prime}=\frac{2 \mathrm{R}}{3}$ Heat produced in R", $\mathrm{H}^{\prime \prime}=\frac{\mathrm{V}^{2}}{\mathrm{R}}, \quad \mathrm{H}^{\prime \prime}=\frac{\mathrm{V}^{2}}{\frac{2 \mathrm{R}}{3}}$ $\mathrm{H}^{\prime \prime}=\frac{3}{2} \frac{\mathrm{V}^{2}}{\mathrm{R}}$ On dividing equation (i) by equation (ii) $\frac{\mathrm{H}^{\prime}}{\mathrm{H}^{\prime \prime}} =\frac{\frac{\mathrm{V}^{2}}{3 \mathrm{R}}}{\frac{3 \mathrm{~V}^{2}}{2 \mathrm{R}}}$ $\frac{\mathrm{H}^{\prime}}{\mathrm{H}^{\prime \prime}} =\frac{2}{9}$ $\therefore \quad \mathrm{H}^{\prime}: \mathrm{H}^{\prime \prime} =2: 9$ Current Electricity
BCECE-2004
Current Electricity
152763
The heat produced by a $100 \mathrm{~W}$ heater in $2 \mathrm{~min}$ will be equal to :
1 $12 \times 10^{3} \mathrm{~J}$
2 $10 \times 10^{3} \mathrm{~J}$
3 $6 \times 10^{3} \mathrm{~J}$
4 $3 \times 10^{3} \mathrm{~J}$
Explanation:
A Given, $\text { Power }(P)=100 \mathrm{~W}$ $\text { Time }(t)=2 \min =2 \times 60=120 \mathrm{sec}$ We know that, $\mathrm{P} =\frac{\mathrm{W}}{\mathrm{t}}$ $\mathrm{W} =\mathrm{P} \times \mathrm{t}$ $\mathrm{W} =100 \times 120$ $\mathrm{~W} =12000 \mathrm{~J}$ $\mathrm{~W} =12 \times 10^{3} \mathrm{~J}$
BCECE-2004
Current Electricity
152765
An electric bulb marked as $50 \mathrm{~W}-200 \mathrm{~V}$ is connected across a $100 \mathrm{~V}$ supply. The present power of the bulb is
1 $37.5 \mathrm{~W}$
2 $25 \mathrm{~W}$
3 $12.5 \mathrm{~W}$
4 $10 \mathrm{~W}$
Explanation:
C Given, $\mathrm{P}_{1}=50 \mathrm{~W}$ $\mathrm{~V}_{1}=200 \mathrm{~V}$ $\mathrm{~V}_{2}=100 \mathrm{~V}$ Let, Resistance of the bulb $=\mathrm{R}$ We know that, $\text { Power, } P_{1}=\frac{V_{1}^{2}}{R}$ $R=\frac{V_{1}^{2}}{P_{1}}$ $R=\frac{(200)^{2}}{50}$ $R=800 \Omega$ Then, $\text { Power, } \mathrm{P}_{2}=\frac{\mathrm{V}_{2}^{2}}{\mathrm{R}}$ $\mathrm{P}_{2}=\frac{(100)^{2}}{800}$ $\mathrm{P}_{2}=12.5 \mathrm{~W}$ Therefore, the present power of the bulb is $12.5 \mathrm{~W}$.
WB JEE 2012
Current Electricity
152766
An electrical cable having a resistance of $0.2 \Omega$ delivers $10 \mathrm{~kW}$ at $200 \mathrm{~V}$ D.C. to a factory. What is the efficiency of transmission?
1 $65 \%$
2 $75 \%$
3 $85 \%$
4 $95 \%$
Explanation:
D Given, $\mathrm{P}=10 \mathrm{~kW}=10 \times 10^{3}$ $\mathrm{V}=200 \mathrm{~V}$ Power, $\mathrm{P}=\mathrm{VI}$ $I=\frac{P}{V}$ $I=\frac{10 \times 10^{3}}{200}=50 \mathrm{~A}$ Power loss $=\mathrm{I}^{2} \mathrm{R}$ $=(50)^{2} \times(0.2)$ $=500 \mathrm{~W}$ $\because \quad$ efficiency of transmission $=\frac{10000}{(10000+500)} \times 100=95 \%$
152761
The electric current passes through a metallic wire produces heat because of- 1001. Two wires have resistances $R$ and $2 R$. When both are joining in series and in parallel, then ratio of heats generated in these situations on applying the same voltage, is :
1 collisions of conduction electrons with each other
2 collisions of the atoms of the metal with each other
3 the energy released in the ionization of the atoms of the metal
4 collisions of the conduction electrons with the atoms of the metallic wire
Explanation:
D When a current passes through a wire, a part of the electrical energy is converted into heat energy as a result of some resistance experienced by it and as a result the wire gets heated. This is known as heating effect of electric current. The electric current passes through a metallic wire produces heat because of collisions of the conduction electrons with the atoms of the metallic wire. (a) $2: 1$ (b) $1: 2$ (c) $2: 9$ (d) $9: 2$ Ans: c : Given, Resistance $\left(\mathrm{R}_{1}\right)=\mathrm{R}$ $R_{2}=2 R$ When two wires are joining in series. The net resistance $\left(R^{\prime}\right)=R_{1}+R_{2}$ $\mathrm{R}^{\prime}=\mathrm{R}+2 \mathrm{R}$ $\mathrm{R}^{\prime}=3 \mathrm{R}$ Heat produced in $\mathrm{R}^{\prime}$, $\mathrm{H}_{\mathrm{S}}=\frac{\mathrm{V}^{2}}{\mathrm{R}^{\prime}}=\frac{\mathrm{V}^{2}}{3 \mathrm{R}}$ When two wires are joining in parallel. The net resistance $\left(\frac{1}{\mathrm{R}^{\prime \prime}}\right)=\frac{1}{\mathrm{R}}+\frac{1}{2 \mathrm{R}}$ $\mathrm{R}^{\prime \prime}=\frac{2 \mathrm{R}}{3}$ Heat produced in R", $\mathrm{H}^{\prime \prime}=\frac{\mathrm{V}^{2}}{\mathrm{R}}, \quad \mathrm{H}^{\prime \prime}=\frac{\mathrm{V}^{2}}{\frac{2 \mathrm{R}}{3}}$ $\mathrm{H}^{\prime \prime}=\frac{3}{2} \frac{\mathrm{V}^{2}}{\mathrm{R}}$ On dividing equation (i) by equation (ii) $\frac{\mathrm{H}^{\prime}}{\mathrm{H}^{\prime \prime}} =\frac{\frac{\mathrm{V}^{2}}{3 \mathrm{R}}}{\frac{3 \mathrm{~V}^{2}}{2 \mathrm{R}}}$ $\frac{\mathrm{H}^{\prime}}{\mathrm{H}^{\prime \prime}} =\frac{2}{9}$ $\therefore \quad \mathrm{H}^{\prime}: \mathrm{H}^{\prime \prime} =2: 9$ Current Electricity
BCECE-2004
Current Electricity
152762
The electric current passes through a metallic wire produces heat because of- 1001. Two wires have resistances $R$ and $2 R$. When both are joining in series and in parallel, then ratio of heats generated in these situations on applying the same voltage, is :
1 collisions of conduction electrons with each other
2 collisions of the atoms of the metal with each other
3 the energy released in the ionization of the atoms of the metal
4 collisions of the conduction electrons with the atoms of the metallic wire
Explanation:
D When a current passes through a wire, a part of the electrical energy is converted into heat energy as a result of some resistance experienced by it and as a result the wire gets heated. This is known as heating effect of electric current. The electric current passes through a metallic wire produces heat because of collisions of the conduction electrons with the atoms of the metallic wire. (a) $2: 1$ (b) $1: 2$ (c) $2: 9$ (d) $9: 2$ Ans: c : Given, Resistance $\left(\mathrm{R}_{1}\right)=\mathrm{R}$ $R_{2}=2 R$ When two wires are joining in series. The net resistance $\left(R^{\prime}\right)=R_{1}+R_{2}$ $\mathrm{R}^{\prime}=\mathrm{R}+2 \mathrm{R}$ $\mathrm{R}^{\prime}=3 \mathrm{R}$ Heat produced in $\mathrm{R}^{\prime}$, $\mathrm{H}_{\mathrm{S}}=\frac{\mathrm{V}^{2}}{\mathrm{R}^{\prime}}=\frac{\mathrm{V}^{2}}{3 \mathrm{R}}$ When two wires are joining in parallel. The net resistance $\left(\frac{1}{\mathrm{R}^{\prime \prime}}\right)=\frac{1}{\mathrm{R}}+\frac{1}{2 \mathrm{R}}$ $\mathrm{R}^{\prime \prime}=\frac{2 \mathrm{R}}{3}$ Heat produced in R", $\mathrm{H}^{\prime \prime}=\frac{\mathrm{V}^{2}}{\mathrm{R}}, \quad \mathrm{H}^{\prime \prime}=\frac{\mathrm{V}^{2}}{\frac{2 \mathrm{R}}{3}}$ $\mathrm{H}^{\prime \prime}=\frac{3}{2} \frac{\mathrm{V}^{2}}{\mathrm{R}}$ On dividing equation (i) by equation (ii) $\frac{\mathrm{H}^{\prime}}{\mathrm{H}^{\prime \prime}} =\frac{\frac{\mathrm{V}^{2}}{3 \mathrm{R}}}{\frac{3 \mathrm{~V}^{2}}{2 \mathrm{R}}}$ $\frac{\mathrm{H}^{\prime}}{\mathrm{H}^{\prime \prime}} =\frac{2}{9}$ $\therefore \quad \mathrm{H}^{\prime}: \mathrm{H}^{\prime \prime} =2: 9$ Current Electricity
BCECE-2004
Current Electricity
152763
The heat produced by a $100 \mathrm{~W}$ heater in $2 \mathrm{~min}$ will be equal to :
1 $12 \times 10^{3} \mathrm{~J}$
2 $10 \times 10^{3} \mathrm{~J}$
3 $6 \times 10^{3} \mathrm{~J}$
4 $3 \times 10^{3} \mathrm{~J}$
Explanation:
A Given, $\text { Power }(P)=100 \mathrm{~W}$ $\text { Time }(t)=2 \min =2 \times 60=120 \mathrm{sec}$ We know that, $\mathrm{P} =\frac{\mathrm{W}}{\mathrm{t}}$ $\mathrm{W} =\mathrm{P} \times \mathrm{t}$ $\mathrm{W} =100 \times 120$ $\mathrm{~W} =12000 \mathrm{~J}$ $\mathrm{~W} =12 \times 10^{3} \mathrm{~J}$
BCECE-2004
Current Electricity
152765
An electric bulb marked as $50 \mathrm{~W}-200 \mathrm{~V}$ is connected across a $100 \mathrm{~V}$ supply. The present power of the bulb is
1 $37.5 \mathrm{~W}$
2 $25 \mathrm{~W}$
3 $12.5 \mathrm{~W}$
4 $10 \mathrm{~W}$
Explanation:
C Given, $\mathrm{P}_{1}=50 \mathrm{~W}$ $\mathrm{~V}_{1}=200 \mathrm{~V}$ $\mathrm{~V}_{2}=100 \mathrm{~V}$ Let, Resistance of the bulb $=\mathrm{R}$ We know that, $\text { Power, } P_{1}=\frac{V_{1}^{2}}{R}$ $R=\frac{V_{1}^{2}}{P_{1}}$ $R=\frac{(200)^{2}}{50}$ $R=800 \Omega$ Then, $\text { Power, } \mathrm{P}_{2}=\frac{\mathrm{V}_{2}^{2}}{\mathrm{R}}$ $\mathrm{P}_{2}=\frac{(100)^{2}}{800}$ $\mathrm{P}_{2}=12.5 \mathrm{~W}$ Therefore, the present power of the bulb is $12.5 \mathrm{~W}$.
WB JEE 2012
Current Electricity
152766
An electrical cable having a resistance of $0.2 \Omega$ delivers $10 \mathrm{~kW}$ at $200 \mathrm{~V}$ D.C. to a factory. What is the efficiency of transmission?
1 $65 \%$
2 $75 \%$
3 $85 \%$
4 $95 \%$
Explanation:
D Given, $\mathrm{P}=10 \mathrm{~kW}=10 \times 10^{3}$ $\mathrm{V}=200 \mathrm{~V}$ Power, $\mathrm{P}=\mathrm{VI}$ $I=\frac{P}{V}$ $I=\frac{10 \times 10^{3}}{200}=50 \mathrm{~A}$ Power loss $=\mathrm{I}^{2} \mathrm{R}$ $=(50)^{2} \times(0.2)$ $=500 \mathrm{~W}$ $\because \quad$ efficiency of transmission $=\frac{10000}{(10000+500)} \times 100=95 \%$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Current Electricity
152761
The electric current passes through a metallic wire produces heat because of- 1001. Two wires have resistances $R$ and $2 R$. When both are joining in series and in parallel, then ratio of heats generated in these situations on applying the same voltage, is :
1 collisions of conduction electrons with each other
2 collisions of the atoms of the metal with each other
3 the energy released in the ionization of the atoms of the metal
4 collisions of the conduction electrons with the atoms of the metallic wire
Explanation:
D When a current passes through a wire, a part of the electrical energy is converted into heat energy as a result of some resistance experienced by it and as a result the wire gets heated. This is known as heating effect of electric current. The electric current passes through a metallic wire produces heat because of collisions of the conduction electrons with the atoms of the metallic wire. (a) $2: 1$ (b) $1: 2$ (c) $2: 9$ (d) $9: 2$ Ans: c : Given, Resistance $\left(\mathrm{R}_{1}\right)=\mathrm{R}$ $R_{2}=2 R$ When two wires are joining in series. The net resistance $\left(R^{\prime}\right)=R_{1}+R_{2}$ $\mathrm{R}^{\prime}=\mathrm{R}+2 \mathrm{R}$ $\mathrm{R}^{\prime}=3 \mathrm{R}$ Heat produced in $\mathrm{R}^{\prime}$, $\mathrm{H}_{\mathrm{S}}=\frac{\mathrm{V}^{2}}{\mathrm{R}^{\prime}}=\frac{\mathrm{V}^{2}}{3 \mathrm{R}}$ When two wires are joining in parallel. The net resistance $\left(\frac{1}{\mathrm{R}^{\prime \prime}}\right)=\frac{1}{\mathrm{R}}+\frac{1}{2 \mathrm{R}}$ $\mathrm{R}^{\prime \prime}=\frac{2 \mathrm{R}}{3}$ Heat produced in R", $\mathrm{H}^{\prime \prime}=\frac{\mathrm{V}^{2}}{\mathrm{R}}, \quad \mathrm{H}^{\prime \prime}=\frac{\mathrm{V}^{2}}{\frac{2 \mathrm{R}}{3}}$ $\mathrm{H}^{\prime \prime}=\frac{3}{2} \frac{\mathrm{V}^{2}}{\mathrm{R}}$ On dividing equation (i) by equation (ii) $\frac{\mathrm{H}^{\prime}}{\mathrm{H}^{\prime \prime}} =\frac{\frac{\mathrm{V}^{2}}{3 \mathrm{R}}}{\frac{3 \mathrm{~V}^{2}}{2 \mathrm{R}}}$ $\frac{\mathrm{H}^{\prime}}{\mathrm{H}^{\prime \prime}} =\frac{2}{9}$ $\therefore \quad \mathrm{H}^{\prime}: \mathrm{H}^{\prime \prime} =2: 9$ Current Electricity
BCECE-2004
Current Electricity
152762
The electric current passes through a metallic wire produces heat because of- 1001. Two wires have resistances $R$ and $2 R$. When both are joining in series and in parallel, then ratio of heats generated in these situations on applying the same voltage, is :
1 collisions of conduction electrons with each other
2 collisions of the atoms of the metal with each other
3 the energy released in the ionization of the atoms of the metal
4 collisions of the conduction electrons with the atoms of the metallic wire
Explanation:
D When a current passes through a wire, a part of the electrical energy is converted into heat energy as a result of some resistance experienced by it and as a result the wire gets heated. This is known as heating effect of electric current. The electric current passes through a metallic wire produces heat because of collisions of the conduction electrons with the atoms of the metallic wire. (a) $2: 1$ (b) $1: 2$ (c) $2: 9$ (d) $9: 2$ Ans: c : Given, Resistance $\left(\mathrm{R}_{1}\right)=\mathrm{R}$ $R_{2}=2 R$ When two wires are joining in series. The net resistance $\left(R^{\prime}\right)=R_{1}+R_{2}$ $\mathrm{R}^{\prime}=\mathrm{R}+2 \mathrm{R}$ $\mathrm{R}^{\prime}=3 \mathrm{R}$ Heat produced in $\mathrm{R}^{\prime}$, $\mathrm{H}_{\mathrm{S}}=\frac{\mathrm{V}^{2}}{\mathrm{R}^{\prime}}=\frac{\mathrm{V}^{2}}{3 \mathrm{R}}$ When two wires are joining in parallel. The net resistance $\left(\frac{1}{\mathrm{R}^{\prime \prime}}\right)=\frac{1}{\mathrm{R}}+\frac{1}{2 \mathrm{R}}$ $\mathrm{R}^{\prime \prime}=\frac{2 \mathrm{R}}{3}$ Heat produced in R", $\mathrm{H}^{\prime \prime}=\frac{\mathrm{V}^{2}}{\mathrm{R}}, \quad \mathrm{H}^{\prime \prime}=\frac{\mathrm{V}^{2}}{\frac{2 \mathrm{R}}{3}}$ $\mathrm{H}^{\prime \prime}=\frac{3}{2} \frac{\mathrm{V}^{2}}{\mathrm{R}}$ On dividing equation (i) by equation (ii) $\frac{\mathrm{H}^{\prime}}{\mathrm{H}^{\prime \prime}} =\frac{\frac{\mathrm{V}^{2}}{3 \mathrm{R}}}{\frac{3 \mathrm{~V}^{2}}{2 \mathrm{R}}}$ $\frac{\mathrm{H}^{\prime}}{\mathrm{H}^{\prime \prime}} =\frac{2}{9}$ $\therefore \quad \mathrm{H}^{\prime}: \mathrm{H}^{\prime \prime} =2: 9$ Current Electricity
BCECE-2004
Current Electricity
152763
The heat produced by a $100 \mathrm{~W}$ heater in $2 \mathrm{~min}$ will be equal to :
1 $12 \times 10^{3} \mathrm{~J}$
2 $10 \times 10^{3} \mathrm{~J}$
3 $6 \times 10^{3} \mathrm{~J}$
4 $3 \times 10^{3} \mathrm{~J}$
Explanation:
A Given, $\text { Power }(P)=100 \mathrm{~W}$ $\text { Time }(t)=2 \min =2 \times 60=120 \mathrm{sec}$ We know that, $\mathrm{P} =\frac{\mathrm{W}}{\mathrm{t}}$ $\mathrm{W} =\mathrm{P} \times \mathrm{t}$ $\mathrm{W} =100 \times 120$ $\mathrm{~W} =12000 \mathrm{~J}$ $\mathrm{~W} =12 \times 10^{3} \mathrm{~J}$
BCECE-2004
Current Electricity
152765
An electric bulb marked as $50 \mathrm{~W}-200 \mathrm{~V}$ is connected across a $100 \mathrm{~V}$ supply. The present power of the bulb is
1 $37.5 \mathrm{~W}$
2 $25 \mathrm{~W}$
3 $12.5 \mathrm{~W}$
4 $10 \mathrm{~W}$
Explanation:
C Given, $\mathrm{P}_{1}=50 \mathrm{~W}$ $\mathrm{~V}_{1}=200 \mathrm{~V}$ $\mathrm{~V}_{2}=100 \mathrm{~V}$ Let, Resistance of the bulb $=\mathrm{R}$ We know that, $\text { Power, } P_{1}=\frac{V_{1}^{2}}{R}$ $R=\frac{V_{1}^{2}}{P_{1}}$ $R=\frac{(200)^{2}}{50}$ $R=800 \Omega$ Then, $\text { Power, } \mathrm{P}_{2}=\frac{\mathrm{V}_{2}^{2}}{\mathrm{R}}$ $\mathrm{P}_{2}=\frac{(100)^{2}}{800}$ $\mathrm{P}_{2}=12.5 \mathrm{~W}$ Therefore, the present power of the bulb is $12.5 \mathrm{~W}$.
WB JEE 2012
Current Electricity
152766
An electrical cable having a resistance of $0.2 \Omega$ delivers $10 \mathrm{~kW}$ at $200 \mathrm{~V}$ D.C. to a factory. What is the efficiency of transmission?
1 $65 \%$
2 $75 \%$
3 $85 \%$
4 $95 \%$
Explanation:
D Given, $\mathrm{P}=10 \mathrm{~kW}=10 \times 10^{3}$ $\mathrm{V}=200 \mathrm{~V}$ Power, $\mathrm{P}=\mathrm{VI}$ $I=\frac{P}{V}$ $I=\frac{10 \times 10^{3}}{200}=50 \mathrm{~A}$ Power loss $=\mathrm{I}^{2} \mathrm{R}$ $=(50)^{2} \times(0.2)$ $=500 \mathrm{~W}$ $\because \quad$ efficiency of transmission $=\frac{10000}{(10000+500)} \times 100=95 \%$
152761
The electric current passes through a metallic wire produces heat because of- 1001. Two wires have resistances $R$ and $2 R$. When both are joining in series and in parallel, then ratio of heats generated in these situations on applying the same voltage, is :
1 collisions of conduction electrons with each other
2 collisions of the atoms of the metal with each other
3 the energy released in the ionization of the atoms of the metal
4 collisions of the conduction electrons with the atoms of the metallic wire
Explanation:
D When a current passes through a wire, a part of the electrical energy is converted into heat energy as a result of some resistance experienced by it and as a result the wire gets heated. This is known as heating effect of electric current. The electric current passes through a metallic wire produces heat because of collisions of the conduction electrons with the atoms of the metallic wire. (a) $2: 1$ (b) $1: 2$ (c) $2: 9$ (d) $9: 2$ Ans: c : Given, Resistance $\left(\mathrm{R}_{1}\right)=\mathrm{R}$ $R_{2}=2 R$ When two wires are joining in series. The net resistance $\left(R^{\prime}\right)=R_{1}+R_{2}$ $\mathrm{R}^{\prime}=\mathrm{R}+2 \mathrm{R}$ $\mathrm{R}^{\prime}=3 \mathrm{R}$ Heat produced in $\mathrm{R}^{\prime}$, $\mathrm{H}_{\mathrm{S}}=\frac{\mathrm{V}^{2}}{\mathrm{R}^{\prime}}=\frac{\mathrm{V}^{2}}{3 \mathrm{R}}$ When two wires are joining in parallel. The net resistance $\left(\frac{1}{\mathrm{R}^{\prime \prime}}\right)=\frac{1}{\mathrm{R}}+\frac{1}{2 \mathrm{R}}$ $\mathrm{R}^{\prime \prime}=\frac{2 \mathrm{R}}{3}$ Heat produced in R", $\mathrm{H}^{\prime \prime}=\frac{\mathrm{V}^{2}}{\mathrm{R}}, \quad \mathrm{H}^{\prime \prime}=\frac{\mathrm{V}^{2}}{\frac{2 \mathrm{R}}{3}}$ $\mathrm{H}^{\prime \prime}=\frac{3}{2} \frac{\mathrm{V}^{2}}{\mathrm{R}}$ On dividing equation (i) by equation (ii) $\frac{\mathrm{H}^{\prime}}{\mathrm{H}^{\prime \prime}} =\frac{\frac{\mathrm{V}^{2}}{3 \mathrm{R}}}{\frac{3 \mathrm{~V}^{2}}{2 \mathrm{R}}}$ $\frac{\mathrm{H}^{\prime}}{\mathrm{H}^{\prime \prime}} =\frac{2}{9}$ $\therefore \quad \mathrm{H}^{\prime}: \mathrm{H}^{\prime \prime} =2: 9$ Current Electricity
BCECE-2004
Current Electricity
152762
The electric current passes through a metallic wire produces heat because of- 1001. Two wires have resistances $R$ and $2 R$. When both are joining in series and in parallel, then ratio of heats generated in these situations on applying the same voltage, is :
1 collisions of conduction electrons with each other
2 collisions of the atoms of the metal with each other
3 the energy released in the ionization of the atoms of the metal
4 collisions of the conduction electrons with the atoms of the metallic wire
Explanation:
D When a current passes through a wire, a part of the electrical energy is converted into heat energy as a result of some resistance experienced by it and as a result the wire gets heated. This is known as heating effect of electric current. The electric current passes through a metallic wire produces heat because of collisions of the conduction electrons with the atoms of the metallic wire. (a) $2: 1$ (b) $1: 2$ (c) $2: 9$ (d) $9: 2$ Ans: c : Given, Resistance $\left(\mathrm{R}_{1}\right)=\mathrm{R}$ $R_{2}=2 R$ When two wires are joining in series. The net resistance $\left(R^{\prime}\right)=R_{1}+R_{2}$ $\mathrm{R}^{\prime}=\mathrm{R}+2 \mathrm{R}$ $\mathrm{R}^{\prime}=3 \mathrm{R}$ Heat produced in $\mathrm{R}^{\prime}$, $\mathrm{H}_{\mathrm{S}}=\frac{\mathrm{V}^{2}}{\mathrm{R}^{\prime}}=\frac{\mathrm{V}^{2}}{3 \mathrm{R}}$ When two wires are joining in parallel. The net resistance $\left(\frac{1}{\mathrm{R}^{\prime \prime}}\right)=\frac{1}{\mathrm{R}}+\frac{1}{2 \mathrm{R}}$ $\mathrm{R}^{\prime \prime}=\frac{2 \mathrm{R}}{3}$ Heat produced in R", $\mathrm{H}^{\prime \prime}=\frac{\mathrm{V}^{2}}{\mathrm{R}}, \quad \mathrm{H}^{\prime \prime}=\frac{\mathrm{V}^{2}}{\frac{2 \mathrm{R}}{3}}$ $\mathrm{H}^{\prime \prime}=\frac{3}{2} \frac{\mathrm{V}^{2}}{\mathrm{R}}$ On dividing equation (i) by equation (ii) $\frac{\mathrm{H}^{\prime}}{\mathrm{H}^{\prime \prime}} =\frac{\frac{\mathrm{V}^{2}}{3 \mathrm{R}}}{\frac{3 \mathrm{~V}^{2}}{2 \mathrm{R}}}$ $\frac{\mathrm{H}^{\prime}}{\mathrm{H}^{\prime \prime}} =\frac{2}{9}$ $\therefore \quad \mathrm{H}^{\prime}: \mathrm{H}^{\prime \prime} =2: 9$ Current Electricity
BCECE-2004
Current Electricity
152763
The heat produced by a $100 \mathrm{~W}$ heater in $2 \mathrm{~min}$ will be equal to :
1 $12 \times 10^{3} \mathrm{~J}$
2 $10 \times 10^{3} \mathrm{~J}$
3 $6 \times 10^{3} \mathrm{~J}$
4 $3 \times 10^{3} \mathrm{~J}$
Explanation:
A Given, $\text { Power }(P)=100 \mathrm{~W}$ $\text { Time }(t)=2 \min =2 \times 60=120 \mathrm{sec}$ We know that, $\mathrm{P} =\frac{\mathrm{W}}{\mathrm{t}}$ $\mathrm{W} =\mathrm{P} \times \mathrm{t}$ $\mathrm{W} =100 \times 120$ $\mathrm{~W} =12000 \mathrm{~J}$ $\mathrm{~W} =12 \times 10^{3} \mathrm{~J}$
BCECE-2004
Current Electricity
152765
An electric bulb marked as $50 \mathrm{~W}-200 \mathrm{~V}$ is connected across a $100 \mathrm{~V}$ supply. The present power of the bulb is
1 $37.5 \mathrm{~W}$
2 $25 \mathrm{~W}$
3 $12.5 \mathrm{~W}$
4 $10 \mathrm{~W}$
Explanation:
C Given, $\mathrm{P}_{1}=50 \mathrm{~W}$ $\mathrm{~V}_{1}=200 \mathrm{~V}$ $\mathrm{~V}_{2}=100 \mathrm{~V}$ Let, Resistance of the bulb $=\mathrm{R}$ We know that, $\text { Power, } P_{1}=\frac{V_{1}^{2}}{R}$ $R=\frac{V_{1}^{2}}{P_{1}}$ $R=\frac{(200)^{2}}{50}$ $R=800 \Omega$ Then, $\text { Power, } \mathrm{P}_{2}=\frac{\mathrm{V}_{2}^{2}}{\mathrm{R}}$ $\mathrm{P}_{2}=\frac{(100)^{2}}{800}$ $\mathrm{P}_{2}=12.5 \mathrm{~W}$ Therefore, the present power of the bulb is $12.5 \mathrm{~W}$.
WB JEE 2012
Current Electricity
152766
An electrical cable having a resistance of $0.2 \Omega$ delivers $10 \mathrm{~kW}$ at $200 \mathrm{~V}$ D.C. to a factory. What is the efficiency of transmission?
1 $65 \%$
2 $75 \%$
3 $85 \%$
4 $95 \%$
Explanation:
D Given, $\mathrm{P}=10 \mathrm{~kW}=10 \times 10^{3}$ $\mathrm{V}=200 \mathrm{~V}$ Power, $\mathrm{P}=\mathrm{VI}$ $I=\frac{P}{V}$ $I=\frac{10 \times 10^{3}}{200}=50 \mathrm{~A}$ Power loss $=\mathrm{I}^{2} \mathrm{R}$ $=(50)^{2} \times(0.2)$ $=500 \mathrm{~W}$ $\because \quad$ efficiency of transmission $=\frac{10000}{(10000+500)} \times 100=95 \%$
152761
The electric current passes through a metallic wire produces heat because of- 1001. Two wires have resistances $R$ and $2 R$. When both are joining in series and in parallel, then ratio of heats generated in these situations on applying the same voltage, is :
1 collisions of conduction electrons with each other
2 collisions of the atoms of the metal with each other
3 the energy released in the ionization of the atoms of the metal
4 collisions of the conduction electrons with the atoms of the metallic wire
Explanation:
D When a current passes through a wire, a part of the electrical energy is converted into heat energy as a result of some resistance experienced by it and as a result the wire gets heated. This is known as heating effect of electric current. The electric current passes through a metallic wire produces heat because of collisions of the conduction electrons with the atoms of the metallic wire. (a) $2: 1$ (b) $1: 2$ (c) $2: 9$ (d) $9: 2$ Ans: c : Given, Resistance $\left(\mathrm{R}_{1}\right)=\mathrm{R}$ $R_{2}=2 R$ When two wires are joining in series. The net resistance $\left(R^{\prime}\right)=R_{1}+R_{2}$ $\mathrm{R}^{\prime}=\mathrm{R}+2 \mathrm{R}$ $\mathrm{R}^{\prime}=3 \mathrm{R}$ Heat produced in $\mathrm{R}^{\prime}$, $\mathrm{H}_{\mathrm{S}}=\frac{\mathrm{V}^{2}}{\mathrm{R}^{\prime}}=\frac{\mathrm{V}^{2}}{3 \mathrm{R}}$ When two wires are joining in parallel. The net resistance $\left(\frac{1}{\mathrm{R}^{\prime \prime}}\right)=\frac{1}{\mathrm{R}}+\frac{1}{2 \mathrm{R}}$ $\mathrm{R}^{\prime \prime}=\frac{2 \mathrm{R}}{3}$ Heat produced in R", $\mathrm{H}^{\prime \prime}=\frac{\mathrm{V}^{2}}{\mathrm{R}}, \quad \mathrm{H}^{\prime \prime}=\frac{\mathrm{V}^{2}}{\frac{2 \mathrm{R}}{3}}$ $\mathrm{H}^{\prime \prime}=\frac{3}{2} \frac{\mathrm{V}^{2}}{\mathrm{R}}$ On dividing equation (i) by equation (ii) $\frac{\mathrm{H}^{\prime}}{\mathrm{H}^{\prime \prime}} =\frac{\frac{\mathrm{V}^{2}}{3 \mathrm{R}}}{\frac{3 \mathrm{~V}^{2}}{2 \mathrm{R}}}$ $\frac{\mathrm{H}^{\prime}}{\mathrm{H}^{\prime \prime}} =\frac{2}{9}$ $\therefore \quad \mathrm{H}^{\prime}: \mathrm{H}^{\prime \prime} =2: 9$ Current Electricity
BCECE-2004
Current Electricity
152762
The electric current passes through a metallic wire produces heat because of- 1001. Two wires have resistances $R$ and $2 R$. When both are joining in series and in parallel, then ratio of heats generated in these situations on applying the same voltage, is :
1 collisions of conduction electrons with each other
2 collisions of the atoms of the metal with each other
3 the energy released in the ionization of the atoms of the metal
4 collisions of the conduction electrons with the atoms of the metallic wire
Explanation:
D When a current passes through a wire, a part of the electrical energy is converted into heat energy as a result of some resistance experienced by it and as a result the wire gets heated. This is known as heating effect of electric current. The electric current passes through a metallic wire produces heat because of collisions of the conduction electrons with the atoms of the metallic wire. (a) $2: 1$ (b) $1: 2$ (c) $2: 9$ (d) $9: 2$ Ans: c : Given, Resistance $\left(\mathrm{R}_{1}\right)=\mathrm{R}$ $R_{2}=2 R$ When two wires are joining in series. The net resistance $\left(R^{\prime}\right)=R_{1}+R_{2}$ $\mathrm{R}^{\prime}=\mathrm{R}+2 \mathrm{R}$ $\mathrm{R}^{\prime}=3 \mathrm{R}$ Heat produced in $\mathrm{R}^{\prime}$, $\mathrm{H}_{\mathrm{S}}=\frac{\mathrm{V}^{2}}{\mathrm{R}^{\prime}}=\frac{\mathrm{V}^{2}}{3 \mathrm{R}}$ When two wires are joining in parallel. The net resistance $\left(\frac{1}{\mathrm{R}^{\prime \prime}}\right)=\frac{1}{\mathrm{R}}+\frac{1}{2 \mathrm{R}}$ $\mathrm{R}^{\prime \prime}=\frac{2 \mathrm{R}}{3}$ Heat produced in R", $\mathrm{H}^{\prime \prime}=\frac{\mathrm{V}^{2}}{\mathrm{R}}, \quad \mathrm{H}^{\prime \prime}=\frac{\mathrm{V}^{2}}{\frac{2 \mathrm{R}}{3}}$ $\mathrm{H}^{\prime \prime}=\frac{3}{2} \frac{\mathrm{V}^{2}}{\mathrm{R}}$ On dividing equation (i) by equation (ii) $\frac{\mathrm{H}^{\prime}}{\mathrm{H}^{\prime \prime}} =\frac{\frac{\mathrm{V}^{2}}{3 \mathrm{R}}}{\frac{3 \mathrm{~V}^{2}}{2 \mathrm{R}}}$ $\frac{\mathrm{H}^{\prime}}{\mathrm{H}^{\prime \prime}} =\frac{2}{9}$ $\therefore \quad \mathrm{H}^{\prime}: \mathrm{H}^{\prime \prime} =2: 9$ Current Electricity
BCECE-2004
Current Electricity
152763
The heat produced by a $100 \mathrm{~W}$ heater in $2 \mathrm{~min}$ will be equal to :
1 $12 \times 10^{3} \mathrm{~J}$
2 $10 \times 10^{3} \mathrm{~J}$
3 $6 \times 10^{3} \mathrm{~J}$
4 $3 \times 10^{3} \mathrm{~J}$
Explanation:
A Given, $\text { Power }(P)=100 \mathrm{~W}$ $\text { Time }(t)=2 \min =2 \times 60=120 \mathrm{sec}$ We know that, $\mathrm{P} =\frac{\mathrm{W}}{\mathrm{t}}$ $\mathrm{W} =\mathrm{P} \times \mathrm{t}$ $\mathrm{W} =100 \times 120$ $\mathrm{~W} =12000 \mathrm{~J}$ $\mathrm{~W} =12 \times 10^{3} \mathrm{~J}$
BCECE-2004
Current Electricity
152765
An electric bulb marked as $50 \mathrm{~W}-200 \mathrm{~V}$ is connected across a $100 \mathrm{~V}$ supply. The present power of the bulb is
1 $37.5 \mathrm{~W}$
2 $25 \mathrm{~W}$
3 $12.5 \mathrm{~W}$
4 $10 \mathrm{~W}$
Explanation:
C Given, $\mathrm{P}_{1}=50 \mathrm{~W}$ $\mathrm{~V}_{1}=200 \mathrm{~V}$ $\mathrm{~V}_{2}=100 \mathrm{~V}$ Let, Resistance of the bulb $=\mathrm{R}$ We know that, $\text { Power, } P_{1}=\frac{V_{1}^{2}}{R}$ $R=\frac{V_{1}^{2}}{P_{1}}$ $R=\frac{(200)^{2}}{50}$ $R=800 \Omega$ Then, $\text { Power, } \mathrm{P}_{2}=\frac{\mathrm{V}_{2}^{2}}{\mathrm{R}}$ $\mathrm{P}_{2}=\frac{(100)^{2}}{800}$ $\mathrm{P}_{2}=12.5 \mathrm{~W}$ Therefore, the present power of the bulb is $12.5 \mathrm{~W}$.
WB JEE 2012
Current Electricity
152766
An electrical cable having a resistance of $0.2 \Omega$ delivers $10 \mathrm{~kW}$ at $200 \mathrm{~V}$ D.C. to a factory. What is the efficiency of transmission?
1 $65 \%$
2 $75 \%$
3 $85 \%$
4 $95 \%$
Explanation:
D Given, $\mathrm{P}=10 \mathrm{~kW}=10 \times 10^{3}$ $\mathrm{V}=200 \mathrm{~V}$ Power, $\mathrm{P}=\mathrm{VI}$ $I=\frac{P}{V}$ $I=\frac{10 \times 10^{3}}{200}=50 \mathrm{~A}$ Power loss $=\mathrm{I}^{2} \mathrm{R}$ $=(50)^{2} \times(0.2)$ $=500 \mathrm{~W}$ $\because \quad$ efficiency of transmission $=\frac{10000}{(10000+500)} \times 100=95 \%$