152681
There are a $25 \mathrm{~W}-220 \mathrm{~V}$ bulb and a $100 \mathrm{~W}$ $220 \mathrm{~V}$ line. Which electric bulb will glow more brightly?
1 $25 \mathrm{~W}$ bulb
2 $100 \mathrm{~W}$ bulb
3 Both will have equal incadescene
4 Neither $25 \mathrm{~W}$ nor $100 \mathrm{~W}$ bulb will give light
Explanation:
A Given, For $1^{\text {st }}$ bulb - Power $\left(\mathrm{P}_{1}\right)=25 \mathrm{~W}$ Voltage $\left(\mathrm{V}_{1}\right)=220 \mathrm{~V}$ Now, Resistance of $25 \mathrm{~W}$ bulb $\left(\mathrm{R}_{25}\right)=\frac{\mathrm{V}_{1}^{2}}{\mathrm{P}_{1}}=\frac{220^{2}}{25}=1936 \Omega$ For bulb $2^{\text {nd }}$ $\text { Power }\left(\mathrm{P}_{2}\right)=100 \mathrm{~W}$ $\text { Voltage }\left(\mathrm{V}_{2}\right)=220 \mathrm{~V}$ Resistance of $100 \mathrm{~W}$ bulb $\left(\mathrm{R}_{100}\right)=\frac{\mathrm{V}_{2}^{2}}{\mathrm{P}_{2}}=\frac{220^{2}}{100}=484 \Omega$ Therefore, both the bulb are connected in series, current through then will be the same. Now, $\text { Power, } \mathrm{P}=\mathrm{I}^{2} \mathrm{R}$ $\therefore$ Current same but resistance of $25 \mathrm{~W}$ bulb is more than $100 \mathrm{~W}$ bulb. So, $25 \mathrm{~W}$ bulb will glow more brightly.
VITEEE-2010
Current Electricity
152682
Two resistances $R$ and $2 R$ are connected in parallel in an electric circuit. The thermal energy developed in $R$ and $2 R$ in the ratio of
1 $1: 2$
2 $2: 1$
3 $1: 4$
4 $4: 1$
Explanation:
B Given, Resistance of $1^{\text {st }}$ resistor $=\mathrm{R}$ Resistance of $2^{\text {nd }}$ resistor $=2 \mathrm{R}$ We know that, Thermal energy $(P)=\frac{V^{2} t}{R}$ Therefore, $\quad \mathrm{H} \propto \frac{1}{\mathrm{R}}$ $\mathrm{H}_{1} \mathrm{R}_{1}=\mathrm{H}_{2} \mathrm{R}_{2}$ $\frac{\mathrm{H}_{1}}{\mathrm{H}_{2}}=\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}$ $\frac{\mathrm{H}_{1}}{\mathrm{H}_{2}}=\frac{2 \mathrm{R}}{\mathrm{R}}$ $\frac{\mathrm{H}_{1}}{\mathrm{H}_{2}}=\frac{2}{1}$ So, $\quad \mathrm{H}_{1}: \mathrm{H}_{2}=2: 1$
UPSEE - 2008
Current Electricity
152683
What will happen when a $40 \mathrm{~W}, 220 \mathrm{~V}$ lamp and $100 \mathrm{~W}, 220 \mathrm{~V}$ lamp are connected in series across $440 \mathrm{~V}$ supply?
1 $40 \mathrm{~W}$ lamp will fuse.
2 $100 \mathrm{~W}$ lamp will fuse.
3 Both the lamp will fuse.
4 Neither lamp will fuse.
Explanation:
A Case of lamp (I), $\mathrm{P}_{1}=40 \mathrm{~W}$ $\mathrm{V}_{1}=220 \mathrm{~V}$ Current, $\mathrm{I}=$ ? Current $\left(\mathrm{I}_{1}\right)=\frac{\mathrm{P}}{\mathrm{V}}=\frac{40}{220}=0.18 \mathrm{~A}$ $\mathrm{R}_{1}=\frac{\mathrm{V}^{2}}{\mathrm{P}}=\frac{(220)^{2}}{40}=1210 \Omega$ Case of lamp (II), $\mathrm{P}_{2}=100 \mathrm{~W}, \mathrm{~V}_{2}=220 \mathrm{~V}$ $\mathrm{I}_{2}=\frac{100}{220}=0.45 \mathrm{~A}$ $\mathrm{R}_{2}=\frac{(220)^{2}}{100}=484 \Omega$ Both are connected in series $440 \mathrm{~V}$ supply total current, $I=\frac{440}{R_{1}+R_{2}}=\frac{440}{1210+484}=\frac{440}{1694}=0.26 \mathrm{~A}$ This current is more than the rated current of lamp 40 W. So, $40 \mathrm{~W}$ lamp will fuse.
SCRA-2012
Current Electricity
152684
An electric heater rated $220 \mathrm{~V}$ and $550 \mathrm{~W}$ is connected to AC mains. The current drawn by it is :
1 $0.8 \mathrm{~A}$
2 $2.5 \mathrm{~A}$
3 $0.4 \mathrm{~A}$
4 $1.25 \mathrm{~A}$
Explanation:
B Given, Voltage of electric heater $(\mathrm{V})=220 \mathrm{~V}$ Power $(\mathrm{P})=550 \mathrm{~W}$ We know that, $\text { Power }=\text { Voltage } \times \text { Current }$ $\mathrm{I}=\frac{\mathrm{P}}{\mathrm{V}}=\frac{550}{220}=2.5 \mathrm{~A}$
152681
There are a $25 \mathrm{~W}-220 \mathrm{~V}$ bulb and a $100 \mathrm{~W}$ $220 \mathrm{~V}$ line. Which electric bulb will glow more brightly?
1 $25 \mathrm{~W}$ bulb
2 $100 \mathrm{~W}$ bulb
3 Both will have equal incadescene
4 Neither $25 \mathrm{~W}$ nor $100 \mathrm{~W}$ bulb will give light
Explanation:
A Given, For $1^{\text {st }}$ bulb - Power $\left(\mathrm{P}_{1}\right)=25 \mathrm{~W}$ Voltage $\left(\mathrm{V}_{1}\right)=220 \mathrm{~V}$ Now, Resistance of $25 \mathrm{~W}$ bulb $\left(\mathrm{R}_{25}\right)=\frac{\mathrm{V}_{1}^{2}}{\mathrm{P}_{1}}=\frac{220^{2}}{25}=1936 \Omega$ For bulb $2^{\text {nd }}$ $\text { Power }\left(\mathrm{P}_{2}\right)=100 \mathrm{~W}$ $\text { Voltage }\left(\mathrm{V}_{2}\right)=220 \mathrm{~V}$ Resistance of $100 \mathrm{~W}$ bulb $\left(\mathrm{R}_{100}\right)=\frac{\mathrm{V}_{2}^{2}}{\mathrm{P}_{2}}=\frac{220^{2}}{100}=484 \Omega$ Therefore, both the bulb are connected in series, current through then will be the same. Now, $\text { Power, } \mathrm{P}=\mathrm{I}^{2} \mathrm{R}$ $\therefore$ Current same but resistance of $25 \mathrm{~W}$ bulb is more than $100 \mathrm{~W}$ bulb. So, $25 \mathrm{~W}$ bulb will glow more brightly.
VITEEE-2010
Current Electricity
152682
Two resistances $R$ and $2 R$ are connected in parallel in an electric circuit. The thermal energy developed in $R$ and $2 R$ in the ratio of
1 $1: 2$
2 $2: 1$
3 $1: 4$
4 $4: 1$
Explanation:
B Given, Resistance of $1^{\text {st }}$ resistor $=\mathrm{R}$ Resistance of $2^{\text {nd }}$ resistor $=2 \mathrm{R}$ We know that, Thermal energy $(P)=\frac{V^{2} t}{R}$ Therefore, $\quad \mathrm{H} \propto \frac{1}{\mathrm{R}}$ $\mathrm{H}_{1} \mathrm{R}_{1}=\mathrm{H}_{2} \mathrm{R}_{2}$ $\frac{\mathrm{H}_{1}}{\mathrm{H}_{2}}=\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}$ $\frac{\mathrm{H}_{1}}{\mathrm{H}_{2}}=\frac{2 \mathrm{R}}{\mathrm{R}}$ $\frac{\mathrm{H}_{1}}{\mathrm{H}_{2}}=\frac{2}{1}$ So, $\quad \mathrm{H}_{1}: \mathrm{H}_{2}=2: 1$
UPSEE - 2008
Current Electricity
152683
What will happen when a $40 \mathrm{~W}, 220 \mathrm{~V}$ lamp and $100 \mathrm{~W}, 220 \mathrm{~V}$ lamp are connected in series across $440 \mathrm{~V}$ supply?
1 $40 \mathrm{~W}$ lamp will fuse.
2 $100 \mathrm{~W}$ lamp will fuse.
3 Both the lamp will fuse.
4 Neither lamp will fuse.
Explanation:
A Case of lamp (I), $\mathrm{P}_{1}=40 \mathrm{~W}$ $\mathrm{V}_{1}=220 \mathrm{~V}$ Current, $\mathrm{I}=$ ? Current $\left(\mathrm{I}_{1}\right)=\frac{\mathrm{P}}{\mathrm{V}}=\frac{40}{220}=0.18 \mathrm{~A}$ $\mathrm{R}_{1}=\frac{\mathrm{V}^{2}}{\mathrm{P}}=\frac{(220)^{2}}{40}=1210 \Omega$ Case of lamp (II), $\mathrm{P}_{2}=100 \mathrm{~W}, \mathrm{~V}_{2}=220 \mathrm{~V}$ $\mathrm{I}_{2}=\frac{100}{220}=0.45 \mathrm{~A}$ $\mathrm{R}_{2}=\frac{(220)^{2}}{100}=484 \Omega$ Both are connected in series $440 \mathrm{~V}$ supply total current, $I=\frac{440}{R_{1}+R_{2}}=\frac{440}{1210+484}=\frac{440}{1694}=0.26 \mathrm{~A}$ This current is more than the rated current of lamp 40 W. So, $40 \mathrm{~W}$ lamp will fuse.
SCRA-2012
Current Electricity
152684
An electric heater rated $220 \mathrm{~V}$ and $550 \mathrm{~W}$ is connected to AC mains. The current drawn by it is :
1 $0.8 \mathrm{~A}$
2 $2.5 \mathrm{~A}$
3 $0.4 \mathrm{~A}$
4 $1.25 \mathrm{~A}$
Explanation:
B Given, Voltage of electric heater $(\mathrm{V})=220 \mathrm{~V}$ Power $(\mathrm{P})=550 \mathrm{~W}$ We know that, $\text { Power }=\text { Voltage } \times \text { Current }$ $\mathrm{I}=\frac{\mathrm{P}}{\mathrm{V}}=\frac{550}{220}=2.5 \mathrm{~A}$
152681
There are a $25 \mathrm{~W}-220 \mathrm{~V}$ bulb and a $100 \mathrm{~W}$ $220 \mathrm{~V}$ line. Which electric bulb will glow more brightly?
1 $25 \mathrm{~W}$ bulb
2 $100 \mathrm{~W}$ bulb
3 Both will have equal incadescene
4 Neither $25 \mathrm{~W}$ nor $100 \mathrm{~W}$ bulb will give light
Explanation:
A Given, For $1^{\text {st }}$ bulb - Power $\left(\mathrm{P}_{1}\right)=25 \mathrm{~W}$ Voltage $\left(\mathrm{V}_{1}\right)=220 \mathrm{~V}$ Now, Resistance of $25 \mathrm{~W}$ bulb $\left(\mathrm{R}_{25}\right)=\frac{\mathrm{V}_{1}^{2}}{\mathrm{P}_{1}}=\frac{220^{2}}{25}=1936 \Omega$ For bulb $2^{\text {nd }}$ $\text { Power }\left(\mathrm{P}_{2}\right)=100 \mathrm{~W}$ $\text { Voltage }\left(\mathrm{V}_{2}\right)=220 \mathrm{~V}$ Resistance of $100 \mathrm{~W}$ bulb $\left(\mathrm{R}_{100}\right)=\frac{\mathrm{V}_{2}^{2}}{\mathrm{P}_{2}}=\frac{220^{2}}{100}=484 \Omega$ Therefore, both the bulb are connected in series, current through then will be the same. Now, $\text { Power, } \mathrm{P}=\mathrm{I}^{2} \mathrm{R}$ $\therefore$ Current same but resistance of $25 \mathrm{~W}$ bulb is more than $100 \mathrm{~W}$ bulb. So, $25 \mathrm{~W}$ bulb will glow more brightly.
VITEEE-2010
Current Electricity
152682
Two resistances $R$ and $2 R$ are connected in parallel in an electric circuit. The thermal energy developed in $R$ and $2 R$ in the ratio of
1 $1: 2$
2 $2: 1$
3 $1: 4$
4 $4: 1$
Explanation:
B Given, Resistance of $1^{\text {st }}$ resistor $=\mathrm{R}$ Resistance of $2^{\text {nd }}$ resistor $=2 \mathrm{R}$ We know that, Thermal energy $(P)=\frac{V^{2} t}{R}$ Therefore, $\quad \mathrm{H} \propto \frac{1}{\mathrm{R}}$ $\mathrm{H}_{1} \mathrm{R}_{1}=\mathrm{H}_{2} \mathrm{R}_{2}$ $\frac{\mathrm{H}_{1}}{\mathrm{H}_{2}}=\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}$ $\frac{\mathrm{H}_{1}}{\mathrm{H}_{2}}=\frac{2 \mathrm{R}}{\mathrm{R}}$ $\frac{\mathrm{H}_{1}}{\mathrm{H}_{2}}=\frac{2}{1}$ So, $\quad \mathrm{H}_{1}: \mathrm{H}_{2}=2: 1$
UPSEE - 2008
Current Electricity
152683
What will happen when a $40 \mathrm{~W}, 220 \mathrm{~V}$ lamp and $100 \mathrm{~W}, 220 \mathrm{~V}$ lamp are connected in series across $440 \mathrm{~V}$ supply?
1 $40 \mathrm{~W}$ lamp will fuse.
2 $100 \mathrm{~W}$ lamp will fuse.
3 Both the lamp will fuse.
4 Neither lamp will fuse.
Explanation:
A Case of lamp (I), $\mathrm{P}_{1}=40 \mathrm{~W}$ $\mathrm{V}_{1}=220 \mathrm{~V}$ Current, $\mathrm{I}=$ ? Current $\left(\mathrm{I}_{1}\right)=\frac{\mathrm{P}}{\mathrm{V}}=\frac{40}{220}=0.18 \mathrm{~A}$ $\mathrm{R}_{1}=\frac{\mathrm{V}^{2}}{\mathrm{P}}=\frac{(220)^{2}}{40}=1210 \Omega$ Case of lamp (II), $\mathrm{P}_{2}=100 \mathrm{~W}, \mathrm{~V}_{2}=220 \mathrm{~V}$ $\mathrm{I}_{2}=\frac{100}{220}=0.45 \mathrm{~A}$ $\mathrm{R}_{2}=\frac{(220)^{2}}{100}=484 \Omega$ Both are connected in series $440 \mathrm{~V}$ supply total current, $I=\frac{440}{R_{1}+R_{2}}=\frac{440}{1210+484}=\frac{440}{1694}=0.26 \mathrm{~A}$ This current is more than the rated current of lamp 40 W. So, $40 \mathrm{~W}$ lamp will fuse.
SCRA-2012
Current Electricity
152684
An electric heater rated $220 \mathrm{~V}$ and $550 \mathrm{~W}$ is connected to AC mains. The current drawn by it is :
1 $0.8 \mathrm{~A}$
2 $2.5 \mathrm{~A}$
3 $0.4 \mathrm{~A}$
4 $1.25 \mathrm{~A}$
Explanation:
B Given, Voltage of electric heater $(\mathrm{V})=220 \mathrm{~V}$ Power $(\mathrm{P})=550 \mathrm{~W}$ We know that, $\text { Power }=\text { Voltage } \times \text { Current }$ $\mathrm{I}=\frac{\mathrm{P}}{\mathrm{V}}=\frac{550}{220}=2.5 \mathrm{~A}$
152681
There are a $25 \mathrm{~W}-220 \mathrm{~V}$ bulb and a $100 \mathrm{~W}$ $220 \mathrm{~V}$ line. Which electric bulb will glow more brightly?
1 $25 \mathrm{~W}$ bulb
2 $100 \mathrm{~W}$ bulb
3 Both will have equal incadescene
4 Neither $25 \mathrm{~W}$ nor $100 \mathrm{~W}$ bulb will give light
Explanation:
A Given, For $1^{\text {st }}$ bulb - Power $\left(\mathrm{P}_{1}\right)=25 \mathrm{~W}$ Voltage $\left(\mathrm{V}_{1}\right)=220 \mathrm{~V}$ Now, Resistance of $25 \mathrm{~W}$ bulb $\left(\mathrm{R}_{25}\right)=\frac{\mathrm{V}_{1}^{2}}{\mathrm{P}_{1}}=\frac{220^{2}}{25}=1936 \Omega$ For bulb $2^{\text {nd }}$ $\text { Power }\left(\mathrm{P}_{2}\right)=100 \mathrm{~W}$ $\text { Voltage }\left(\mathrm{V}_{2}\right)=220 \mathrm{~V}$ Resistance of $100 \mathrm{~W}$ bulb $\left(\mathrm{R}_{100}\right)=\frac{\mathrm{V}_{2}^{2}}{\mathrm{P}_{2}}=\frac{220^{2}}{100}=484 \Omega$ Therefore, both the bulb are connected in series, current through then will be the same. Now, $\text { Power, } \mathrm{P}=\mathrm{I}^{2} \mathrm{R}$ $\therefore$ Current same but resistance of $25 \mathrm{~W}$ bulb is more than $100 \mathrm{~W}$ bulb. So, $25 \mathrm{~W}$ bulb will glow more brightly.
VITEEE-2010
Current Electricity
152682
Two resistances $R$ and $2 R$ are connected in parallel in an electric circuit. The thermal energy developed in $R$ and $2 R$ in the ratio of
1 $1: 2$
2 $2: 1$
3 $1: 4$
4 $4: 1$
Explanation:
B Given, Resistance of $1^{\text {st }}$ resistor $=\mathrm{R}$ Resistance of $2^{\text {nd }}$ resistor $=2 \mathrm{R}$ We know that, Thermal energy $(P)=\frac{V^{2} t}{R}$ Therefore, $\quad \mathrm{H} \propto \frac{1}{\mathrm{R}}$ $\mathrm{H}_{1} \mathrm{R}_{1}=\mathrm{H}_{2} \mathrm{R}_{2}$ $\frac{\mathrm{H}_{1}}{\mathrm{H}_{2}}=\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}$ $\frac{\mathrm{H}_{1}}{\mathrm{H}_{2}}=\frac{2 \mathrm{R}}{\mathrm{R}}$ $\frac{\mathrm{H}_{1}}{\mathrm{H}_{2}}=\frac{2}{1}$ So, $\quad \mathrm{H}_{1}: \mathrm{H}_{2}=2: 1$
UPSEE - 2008
Current Electricity
152683
What will happen when a $40 \mathrm{~W}, 220 \mathrm{~V}$ lamp and $100 \mathrm{~W}, 220 \mathrm{~V}$ lamp are connected in series across $440 \mathrm{~V}$ supply?
1 $40 \mathrm{~W}$ lamp will fuse.
2 $100 \mathrm{~W}$ lamp will fuse.
3 Both the lamp will fuse.
4 Neither lamp will fuse.
Explanation:
A Case of lamp (I), $\mathrm{P}_{1}=40 \mathrm{~W}$ $\mathrm{V}_{1}=220 \mathrm{~V}$ Current, $\mathrm{I}=$ ? Current $\left(\mathrm{I}_{1}\right)=\frac{\mathrm{P}}{\mathrm{V}}=\frac{40}{220}=0.18 \mathrm{~A}$ $\mathrm{R}_{1}=\frac{\mathrm{V}^{2}}{\mathrm{P}}=\frac{(220)^{2}}{40}=1210 \Omega$ Case of lamp (II), $\mathrm{P}_{2}=100 \mathrm{~W}, \mathrm{~V}_{2}=220 \mathrm{~V}$ $\mathrm{I}_{2}=\frac{100}{220}=0.45 \mathrm{~A}$ $\mathrm{R}_{2}=\frac{(220)^{2}}{100}=484 \Omega$ Both are connected in series $440 \mathrm{~V}$ supply total current, $I=\frac{440}{R_{1}+R_{2}}=\frac{440}{1210+484}=\frac{440}{1694}=0.26 \mathrm{~A}$ This current is more than the rated current of lamp 40 W. So, $40 \mathrm{~W}$ lamp will fuse.
SCRA-2012
Current Electricity
152684
An electric heater rated $220 \mathrm{~V}$ and $550 \mathrm{~W}$ is connected to AC mains. The current drawn by it is :
1 $0.8 \mathrm{~A}$
2 $2.5 \mathrm{~A}$
3 $0.4 \mathrm{~A}$
4 $1.25 \mathrm{~A}$
Explanation:
B Given, Voltage of electric heater $(\mathrm{V})=220 \mathrm{~V}$ Power $(\mathrm{P})=550 \mathrm{~W}$ We know that, $\text { Power }=\text { Voltage } \times \text { Current }$ $\mathrm{I}=\frac{\mathrm{P}}{\mathrm{V}}=\frac{550}{220}=2.5 \mathrm{~A}$
