152736
If $R_{1}$ and $R_{2}$ be the resistance of the filaments of $200 \mathrm{~W}$ and $100 \mathrm{~W}$ electric bulbs operating at $220 \mathrm{~V}$, then $\left(\frac{R_{1}}{R_{2}}\right)$ is
1 1
2 2
3 0.5
4 4
5 0.25
Explanation:
C Given, $\mathrm{P}_{1}=200 \mathrm{~W}, \mathrm{P}_{2}=100 \mathrm{~W}$ We know that, $\because \quad \text { Power, } \mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ $\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{100}{200}=\frac{1}{2}=0.5$
Kerala CEE 2007
Current Electricity
152739
In the adjoining figure, if 10 calorie heat is produced per second in $5 \Omega$ resister due to the flow of current through it, then the heat produced in $6 \Omega$ resistor is
1 $1 \mathrm{cal} / \mathrm{s}$
2 $2 \mathrm{cal} / \mathrm{s}$
3 $3 \mathrm{cal} / \mathrm{s}$
4 $4 \mathrm{cal} / \mathrm{s}$
Explanation:
C We know that, For $5 \Omega$ $\frac{\mathrm{H}}{\mathrm{t}}=10 \mathrm{cal} / \mathrm{s}$ Heat produced $(\mathrm{H})=\mathrm{I}^{2} \mathrm{Rt}$ $\frac{\mathrm{H}}{\mathrm{t}} =\mathrm{I}^{2} \mathrm{R}$ $10 =\mathrm{I}_{1}^{2} \mathrm{R}$ $\mathrm{I}_{1}^{2} =\frac{10}{5}=2$ $\mathrm{I}_{1} =\sqrt{2} \mathrm{~A}$ Voltage develop in $5 \Omega$ resistance $=$ Voltage across $(6+$ 4) $\Omega$ registance $\mathrm{I}_{1} 5=\mathrm{I}_{2}(4+6)$ $\mathrm{I}_{2}=\frac{5 \mathrm{I}_{1}}{10}=\frac{5 \times \sqrt{2}}{10}=\frac{1}{\sqrt{2}} \mathrm{~A}$ $\frac{\mathrm{H}}{\mathrm{t}}=\mathrm{I}_{2}^{2} R$ Heat produced in $6 \Omega$ resistance $\frac{\mathrm{H}}{\mathrm{t}}=\left(\frac{1}{\sqrt{2}}\right)^{2} \times 6=\frac{1}{2} \times 6=3 \mathrm{cal} / \mathrm{s}$
JCECE-2013
Current Electricity
152740
An electric heater boils $1 \mathrm{~kg}$ of water in a time $t_{1}$. Another heater boils the same amount of water in a time $t_{2}$. When the two heater are connected in parallel, the time required by them together to boil the same amount of water is:
1 $t_{1}+t_{2}$
2 $t_{1} t_{2}$
3 $\frac{t_{1}+t_{2}}{2}$
4 $\frac{t_{1} t_{2}}{t_{1}+t_{2}}$
Explanation:
D We know that, $R_{1}, R_{2}$ be the resistance of two heaters of $H$ the required $1 \mathrm{~kg}$ of water if $\mathrm{V}$ is the voltage applied. $\mathrm{H}=\frac{\mathrm{V}^{2}}{\mathrm{R}_{1}} \mathrm{t}_{1}=\frac{\mathrm{V}^{2}}{\mathrm{R}_{2}} \mathrm{t}_{2}$ $\frac{\mathrm{t}_{1}}{\mathrm{R}_{1}}=\frac{\mathrm{t}_{2}}{\mathrm{R}_{2}}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{\mathrm{t}_{1}}{\mathrm{t}_{2}}$ When two heaters are in parallel and $t$ is time taken to boil the given water $H=\left(\frac{V^{2}}{R_{1}}+\frac{V^{2}}{R_{2}}\right) t=\frac{V^{2}}{R_{1}} t_{1}$ $t=\frac{t_{1}}{R_{1}}\left(\frac{1}{R_{1}}+\frac{1}{R_{2}}\right)^{-1}$ $t=\frac{t_{1}}{1+\frac{R_{1}}{R_{2}}}$ $t=\frac{t_{1}}{1+\frac{t_{1}}{t_{2}}}=\frac{t_{1} t_{2}}{t_{1}+t_{2}}$
JCECE-2005]
Current Electricity
152741
An electric bulb is marked $100 \mathrm{~W}, 230 \mathrm{~V}$. If the supply voltage drops to $115 \mathrm{~V}$, what is the total energy produced by the bulb in $10 \mathrm{~min}$ ?
1 $30 \mathrm{~kJ}$
2 $20 \mathrm{~kJ}$
3 $15 \mathrm{~kJ}$
4 $10 \mathrm{~kJ}$
Explanation:
C We have that, Power of the bulb, $\mathrm{P}=100 \mathrm{~W}$ $\mathrm{V}=230 \mathrm{~V}$ $\mathrm{R}$ be the resistance of the bulb $\mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}} \quad\left[\because \mathrm{R}=\frac{\mathrm{V}^{2}}{\mathrm{P}}\right]$ $\mathrm{R}=\frac{230 \times 230}{100}=529 \Omega$ Changed supply voltage $\mathrm{V}=115 \mathrm{~V}$ Heat and light energy produced by the bulb in $10 \mathrm{~min}$ $\mathrm{H}=\frac{\mathrm{V}^{2} \mathrm{t}}{\mathrm{R}}=\frac{115 \times 115 \times 10 \times 60}{529}=15 \mathrm{~kJ}$
152736
If $R_{1}$ and $R_{2}$ be the resistance of the filaments of $200 \mathrm{~W}$ and $100 \mathrm{~W}$ electric bulbs operating at $220 \mathrm{~V}$, then $\left(\frac{R_{1}}{R_{2}}\right)$ is
1 1
2 2
3 0.5
4 4
5 0.25
Explanation:
C Given, $\mathrm{P}_{1}=200 \mathrm{~W}, \mathrm{P}_{2}=100 \mathrm{~W}$ We know that, $\because \quad \text { Power, } \mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ $\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{100}{200}=\frac{1}{2}=0.5$
Kerala CEE 2007
Current Electricity
152739
In the adjoining figure, if 10 calorie heat is produced per second in $5 \Omega$ resister due to the flow of current through it, then the heat produced in $6 \Omega$ resistor is
1 $1 \mathrm{cal} / \mathrm{s}$
2 $2 \mathrm{cal} / \mathrm{s}$
3 $3 \mathrm{cal} / \mathrm{s}$
4 $4 \mathrm{cal} / \mathrm{s}$
Explanation:
C We know that, For $5 \Omega$ $\frac{\mathrm{H}}{\mathrm{t}}=10 \mathrm{cal} / \mathrm{s}$ Heat produced $(\mathrm{H})=\mathrm{I}^{2} \mathrm{Rt}$ $\frac{\mathrm{H}}{\mathrm{t}} =\mathrm{I}^{2} \mathrm{R}$ $10 =\mathrm{I}_{1}^{2} \mathrm{R}$ $\mathrm{I}_{1}^{2} =\frac{10}{5}=2$ $\mathrm{I}_{1} =\sqrt{2} \mathrm{~A}$ Voltage develop in $5 \Omega$ resistance $=$ Voltage across $(6+$ 4) $\Omega$ registance $\mathrm{I}_{1} 5=\mathrm{I}_{2}(4+6)$ $\mathrm{I}_{2}=\frac{5 \mathrm{I}_{1}}{10}=\frac{5 \times \sqrt{2}}{10}=\frac{1}{\sqrt{2}} \mathrm{~A}$ $\frac{\mathrm{H}}{\mathrm{t}}=\mathrm{I}_{2}^{2} R$ Heat produced in $6 \Omega$ resistance $\frac{\mathrm{H}}{\mathrm{t}}=\left(\frac{1}{\sqrt{2}}\right)^{2} \times 6=\frac{1}{2} \times 6=3 \mathrm{cal} / \mathrm{s}$
JCECE-2013
Current Electricity
152740
An electric heater boils $1 \mathrm{~kg}$ of water in a time $t_{1}$. Another heater boils the same amount of water in a time $t_{2}$. When the two heater are connected in parallel, the time required by them together to boil the same amount of water is:
1 $t_{1}+t_{2}$
2 $t_{1} t_{2}$
3 $\frac{t_{1}+t_{2}}{2}$
4 $\frac{t_{1} t_{2}}{t_{1}+t_{2}}$
Explanation:
D We know that, $R_{1}, R_{2}$ be the resistance of two heaters of $H$ the required $1 \mathrm{~kg}$ of water if $\mathrm{V}$ is the voltage applied. $\mathrm{H}=\frac{\mathrm{V}^{2}}{\mathrm{R}_{1}} \mathrm{t}_{1}=\frac{\mathrm{V}^{2}}{\mathrm{R}_{2}} \mathrm{t}_{2}$ $\frac{\mathrm{t}_{1}}{\mathrm{R}_{1}}=\frac{\mathrm{t}_{2}}{\mathrm{R}_{2}}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{\mathrm{t}_{1}}{\mathrm{t}_{2}}$ When two heaters are in parallel and $t$ is time taken to boil the given water $H=\left(\frac{V^{2}}{R_{1}}+\frac{V^{2}}{R_{2}}\right) t=\frac{V^{2}}{R_{1}} t_{1}$ $t=\frac{t_{1}}{R_{1}}\left(\frac{1}{R_{1}}+\frac{1}{R_{2}}\right)^{-1}$ $t=\frac{t_{1}}{1+\frac{R_{1}}{R_{2}}}$ $t=\frac{t_{1}}{1+\frac{t_{1}}{t_{2}}}=\frac{t_{1} t_{2}}{t_{1}+t_{2}}$
JCECE-2005]
Current Electricity
152741
An electric bulb is marked $100 \mathrm{~W}, 230 \mathrm{~V}$. If the supply voltage drops to $115 \mathrm{~V}$, what is the total energy produced by the bulb in $10 \mathrm{~min}$ ?
1 $30 \mathrm{~kJ}$
2 $20 \mathrm{~kJ}$
3 $15 \mathrm{~kJ}$
4 $10 \mathrm{~kJ}$
Explanation:
C We have that, Power of the bulb, $\mathrm{P}=100 \mathrm{~W}$ $\mathrm{V}=230 \mathrm{~V}$ $\mathrm{R}$ be the resistance of the bulb $\mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}} \quad\left[\because \mathrm{R}=\frac{\mathrm{V}^{2}}{\mathrm{P}}\right]$ $\mathrm{R}=\frac{230 \times 230}{100}=529 \Omega$ Changed supply voltage $\mathrm{V}=115 \mathrm{~V}$ Heat and light energy produced by the bulb in $10 \mathrm{~min}$ $\mathrm{H}=\frac{\mathrm{V}^{2} \mathrm{t}}{\mathrm{R}}=\frac{115 \times 115 \times 10 \times 60}{529}=15 \mathrm{~kJ}$
NEET Test Series from KOTA - 10 Papers In MS WORD
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Current Electricity
152736
If $R_{1}$ and $R_{2}$ be the resistance of the filaments of $200 \mathrm{~W}$ and $100 \mathrm{~W}$ electric bulbs operating at $220 \mathrm{~V}$, then $\left(\frac{R_{1}}{R_{2}}\right)$ is
1 1
2 2
3 0.5
4 4
5 0.25
Explanation:
C Given, $\mathrm{P}_{1}=200 \mathrm{~W}, \mathrm{P}_{2}=100 \mathrm{~W}$ We know that, $\because \quad \text { Power, } \mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ $\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{100}{200}=\frac{1}{2}=0.5$
Kerala CEE 2007
Current Electricity
152739
In the adjoining figure, if 10 calorie heat is produced per second in $5 \Omega$ resister due to the flow of current through it, then the heat produced in $6 \Omega$ resistor is
1 $1 \mathrm{cal} / \mathrm{s}$
2 $2 \mathrm{cal} / \mathrm{s}$
3 $3 \mathrm{cal} / \mathrm{s}$
4 $4 \mathrm{cal} / \mathrm{s}$
Explanation:
C We know that, For $5 \Omega$ $\frac{\mathrm{H}}{\mathrm{t}}=10 \mathrm{cal} / \mathrm{s}$ Heat produced $(\mathrm{H})=\mathrm{I}^{2} \mathrm{Rt}$ $\frac{\mathrm{H}}{\mathrm{t}} =\mathrm{I}^{2} \mathrm{R}$ $10 =\mathrm{I}_{1}^{2} \mathrm{R}$ $\mathrm{I}_{1}^{2} =\frac{10}{5}=2$ $\mathrm{I}_{1} =\sqrt{2} \mathrm{~A}$ Voltage develop in $5 \Omega$ resistance $=$ Voltage across $(6+$ 4) $\Omega$ registance $\mathrm{I}_{1} 5=\mathrm{I}_{2}(4+6)$ $\mathrm{I}_{2}=\frac{5 \mathrm{I}_{1}}{10}=\frac{5 \times \sqrt{2}}{10}=\frac{1}{\sqrt{2}} \mathrm{~A}$ $\frac{\mathrm{H}}{\mathrm{t}}=\mathrm{I}_{2}^{2} R$ Heat produced in $6 \Omega$ resistance $\frac{\mathrm{H}}{\mathrm{t}}=\left(\frac{1}{\sqrt{2}}\right)^{2} \times 6=\frac{1}{2} \times 6=3 \mathrm{cal} / \mathrm{s}$
JCECE-2013
Current Electricity
152740
An electric heater boils $1 \mathrm{~kg}$ of water in a time $t_{1}$. Another heater boils the same amount of water in a time $t_{2}$. When the two heater are connected in parallel, the time required by them together to boil the same amount of water is:
1 $t_{1}+t_{2}$
2 $t_{1} t_{2}$
3 $\frac{t_{1}+t_{2}}{2}$
4 $\frac{t_{1} t_{2}}{t_{1}+t_{2}}$
Explanation:
D We know that, $R_{1}, R_{2}$ be the resistance of two heaters of $H$ the required $1 \mathrm{~kg}$ of water if $\mathrm{V}$ is the voltage applied. $\mathrm{H}=\frac{\mathrm{V}^{2}}{\mathrm{R}_{1}} \mathrm{t}_{1}=\frac{\mathrm{V}^{2}}{\mathrm{R}_{2}} \mathrm{t}_{2}$ $\frac{\mathrm{t}_{1}}{\mathrm{R}_{1}}=\frac{\mathrm{t}_{2}}{\mathrm{R}_{2}}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{\mathrm{t}_{1}}{\mathrm{t}_{2}}$ When two heaters are in parallel and $t$ is time taken to boil the given water $H=\left(\frac{V^{2}}{R_{1}}+\frac{V^{2}}{R_{2}}\right) t=\frac{V^{2}}{R_{1}} t_{1}$ $t=\frac{t_{1}}{R_{1}}\left(\frac{1}{R_{1}}+\frac{1}{R_{2}}\right)^{-1}$ $t=\frac{t_{1}}{1+\frac{R_{1}}{R_{2}}}$ $t=\frac{t_{1}}{1+\frac{t_{1}}{t_{2}}}=\frac{t_{1} t_{2}}{t_{1}+t_{2}}$
JCECE-2005]
Current Electricity
152741
An electric bulb is marked $100 \mathrm{~W}, 230 \mathrm{~V}$. If the supply voltage drops to $115 \mathrm{~V}$, what is the total energy produced by the bulb in $10 \mathrm{~min}$ ?
1 $30 \mathrm{~kJ}$
2 $20 \mathrm{~kJ}$
3 $15 \mathrm{~kJ}$
4 $10 \mathrm{~kJ}$
Explanation:
C We have that, Power of the bulb, $\mathrm{P}=100 \mathrm{~W}$ $\mathrm{V}=230 \mathrm{~V}$ $\mathrm{R}$ be the resistance of the bulb $\mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}} \quad\left[\because \mathrm{R}=\frac{\mathrm{V}^{2}}{\mathrm{P}}\right]$ $\mathrm{R}=\frac{230 \times 230}{100}=529 \Omega$ Changed supply voltage $\mathrm{V}=115 \mathrm{~V}$ Heat and light energy produced by the bulb in $10 \mathrm{~min}$ $\mathrm{H}=\frac{\mathrm{V}^{2} \mathrm{t}}{\mathrm{R}}=\frac{115 \times 115 \times 10 \times 60}{529}=15 \mathrm{~kJ}$
152736
If $R_{1}$ and $R_{2}$ be the resistance of the filaments of $200 \mathrm{~W}$ and $100 \mathrm{~W}$ electric bulbs operating at $220 \mathrm{~V}$, then $\left(\frac{R_{1}}{R_{2}}\right)$ is
1 1
2 2
3 0.5
4 4
5 0.25
Explanation:
C Given, $\mathrm{P}_{1}=200 \mathrm{~W}, \mathrm{P}_{2}=100 \mathrm{~W}$ We know that, $\because \quad \text { Power, } \mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ $\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{100}{200}=\frac{1}{2}=0.5$
Kerala CEE 2007
Current Electricity
152739
In the adjoining figure, if 10 calorie heat is produced per second in $5 \Omega$ resister due to the flow of current through it, then the heat produced in $6 \Omega$ resistor is
1 $1 \mathrm{cal} / \mathrm{s}$
2 $2 \mathrm{cal} / \mathrm{s}$
3 $3 \mathrm{cal} / \mathrm{s}$
4 $4 \mathrm{cal} / \mathrm{s}$
Explanation:
C We know that, For $5 \Omega$ $\frac{\mathrm{H}}{\mathrm{t}}=10 \mathrm{cal} / \mathrm{s}$ Heat produced $(\mathrm{H})=\mathrm{I}^{2} \mathrm{Rt}$ $\frac{\mathrm{H}}{\mathrm{t}} =\mathrm{I}^{2} \mathrm{R}$ $10 =\mathrm{I}_{1}^{2} \mathrm{R}$ $\mathrm{I}_{1}^{2} =\frac{10}{5}=2$ $\mathrm{I}_{1} =\sqrt{2} \mathrm{~A}$ Voltage develop in $5 \Omega$ resistance $=$ Voltage across $(6+$ 4) $\Omega$ registance $\mathrm{I}_{1} 5=\mathrm{I}_{2}(4+6)$ $\mathrm{I}_{2}=\frac{5 \mathrm{I}_{1}}{10}=\frac{5 \times \sqrt{2}}{10}=\frac{1}{\sqrt{2}} \mathrm{~A}$ $\frac{\mathrm{H}}{\mathrm{t}}=\mathrm{I}_{2}^{2} R$ Heat produced in $6 \Omega$ resistance $\frac{\mathrm{H}}{\mathrm{t}}=\left(\frac{1}{\sqrt{2}}\right)^{2} \times 6=\frac{1}{2} \times 6=3 \mathrm{cal} / \mathrm{s}$
JCECE-2013
Current Electricity
152740
An electric heater boils $1 \mathrm{~kg}$ of water in a time $t_{1}$. Another heater boils the same amount of water in a time $t_{2}$. When the two heater are connected in parallel, the time required by them together to boil the same amount of water is:
1 $t_{1}+t_{2}$
2 $t_{1} t_{2}$
3 $\frac{t_{1}+t_{2}}{2}$
4 $\frac{t_{1} t_{2}}{t_{1}+t_{2}}$
Explanation:
D We know that, $R_{1}, R_{2}$ be the resistance of two heaters of $H$ the required $1 \mathrm{~kg}$ of water if $\mathrm{V}$ is the voltage applied. $\mathrm{H}=\frac{\mathrm{V}^{2}}{\mathrm{R}_{1}} \mathrm{t}_{1}=\frac{\mathrm{V}^{2}}{\mathrm{R}_{2}} \mathrm{t}_{2}$ $\frac{\mathrm{t}_{1}}{\mathrm{R}_{1}}=\frac{\mathrm{t}_{2}}{\mathrm{R}_{2}}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{\mathrm{t}_{1}}{\mathrm{t}_{2}}$ When two heaters are in parallel and $t$ is time taken to boil the given water $H=\left(\frac{V^{2}}{R_{1}}+\frac{V^{2}}{R_{2}}\right) t=\frac{V^{2}}{R_{1}} t_{1}$ $t=\frac{t_{1}}{R_{1}}\left(\frac{1}{R_{1}}+\frac{1}{R_{2}}\right)^{-1}$ $t=\frac{t_{1}}{1+\frac{R_{1}}{R_{2}}}$ $t=\frac{t_{1}}{1+\frac{t_{1}}{t_{2}}}=\frac{t_{1} t_{2}}{t_{1}+t_{2}}$
JCECE-2005]
Current Electricity
152741
An electric bulb is marked $100 \mathrm{~W}, 230 \mathrm{~V}$. If the supply voltage drops to $115 \mathrm{~V}$, what is the total energy produced by the bulb in $10 \mathrm{~min}$ ?
1 $30 \mathrm{~kJ}$
2 $20 \mathrm{~kJ}$
3 $15 \mathrm{~kJ}$
4 $10 \mathrm{~kJ}$
Explanation:
C We have that, Power of the bulb, $\mathrm{P}=100 \mathrm{~W}$ $\mathrm{V}=230 \mathrm{~V}$ $\mathrm{R}$ be the resistance of the bulb $\mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}} \quad\left[\because \mathrm{R}=\frac{\mathrm{V}^{2}}{\mathrm{P}}\right]$ $\mathrm{R}=\frac{230 \times 230}{100}=529 \Omega$ Changed supply voltage $\mathrm{V}=115 \mathrm{~V}$ Heat and light energy produced by the bulb in $10 \mathrm{~min}$ $\mathrm{H}=\frac{\mathrm{V}^{2} \mathrm{t}}{\mathrm{R}}=\frac{115 \times 115 \times 10 \times 60}{529}=15 \mathrm{~kJ}$
