152694
A $100 \mathrm{~W} 200 \mathrm{~V}$ bulb is connected to a $160 \mathrm{~V}$ power supply. The power consumption would be
1 $125 \mathrm{~W}$
2 $100 \mathrm{~W}$
3 $80 \mathrm{~W}$
4 $64 \mathrm{~W}$
Explanation:
D Given, Power $(\mathrm{P})=100 \mathrm{~W}$ Bulb voltage $(\mathrm{V})=200$ Volt Supply power $\left(\mathrm{V}_{1}\right)=160$ Volt We know that, Resistance of the bulb $(\mathrm{R})=\frac{\mathrm{V}^{2}}{\mathrm{P}}$ $\therefore \quad \mathrm{R}=\frac{200^{2}}{100}=400 \Omega$ Now, Actual power consumption $(\mathrm{P})=\frac{\mathrm{V}_{1}^{2}}{\mathrm{R}}$ $\therefore \quad \mathrm{P}^{\prime}=\frac{160^{2}}{400}=64 \mathrm{~W}$
AIPMT-1997
Current Electricity
152695
A heating coil is labelled $100 \mathrm{~W}, 220 \mathrm{~V}$. The coil is cut in two equal halves and the two pieces are joined in parallel to the same source. The energy now liberated per second is
1 $25 \mathrm{~J}$
2 $50 \mathrm{~J}$
3 $200 \mathrm{~J}$
4 $400 \mathrm{~J}$
Explanation:
D Given, Power $(\mathrm{P})=100 \mathrm{~W}$ Voltage $(\mathrm{V})=220 \mathrm{~V}$ Let resistance of heating coil $=\mathrm{R}$ $\mathrm{R}=\frac{(220)^{2}}{100}=484$ We know that, The resistance of each part is $\mathrm{R} / 2$ and connect parallel. $\therefore$ Equivalent resistance $=\mathrm{R}_{\mathrm{eq}}$ $\frac{1}{\mathrm{R}_{\mathrm{eq}}} =\frac{1}{\mathrm{R} / 2}+\frac{1}{\mathrm{R} / 2}$ $=\frac{\mathrm{R} / 2+\mathrm{R} / 2}{(\mathrm{R} / 2)(\mathrm{R} / 2)}$ $\therefore \quad \mathrm{R}_{\mathrm{eq}}=\frac{(\mathrm{R} / 2)(\mathrm{R} / 2)}{(\mathrm{R} / 2)+(\mathrm{R} / 2)}=\frac{\mathrm{R}}{4}$ $\therefore$ Dissipated power, $\mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}_{\mathrm{eq}}}=\frac{(220)^{2}}{\mathrm{R} / 4}$ $=\frac{(220)^{2}}{\frac{484}{4}}=400 \mathrm{~J}$
AIPMT-1995
Current Electricity
152696
A $4 \mu \mathrm{F}$ conductor is charged to $400 \mathrm{~V}$ and then its plates are joined through a resistance of $1 \mathrm{k} \Omega$. The heat produced in the resistance is
1 $0.16 \mathrm{~J}$
2 $1.28 \mathrm{~J}$
3 $0.64 \mathrm{~J}$
4 $0.32 \mathrm{~J}$ [APIMT-1994]
Explanation:
D Given,} Capacitance of capacitor $(\mathrm{C})=4 \mu \mathrm{F}=4 \times 10^{-6} \mathrm{~F}$ Voltage applied $(\mathrm{V})=400$ Volt We know that, Energy stored $=$ Heat energy produced through resistance $\mathrm{U} =\frac{1}{2} \mathrm{CV}^{2}=\frac{1}{2} \times 4 \times 10^{-6} \times 400^{2}$ $=32 \times 10^{-2} \text { Joule }$ $\therefore \quad \mathrm{U} =0.32 \text { Joule }$
Current Electricity
152697
40 electric bulbs are connected in series across a $220 \mathrm{~V}$ supply. After one bulb is fused the remaining 39 are connected again in series across the same supply. The illumination will be
1 more with 40 bulbs than with 39
2 more with 39 bulbs than with 40
3 equal in both the cases
4 in the ratio $40^{2}: 39^{2}$
Explanation:
B Case -I: Total resistance of 40 bulbs $=40 \mathrm{R}$ Circuit current $\left(\mathrm{I}_{1}\right)=\frac{220}{40 \mathrm{R}}$ Power dissipated by 40 bulbs, $\mathrm{P}_{1}=40\left[\mathrm{I}_{1}^{2} \mathrm{R}\right]=40\left(\frac{220}{40 \mathrm{R}}\right)^{2} \times \mathrm{R}$ $\mathrm{P}_{1}=\frac{(220)^{2}}{40 \mathrm{R}} \mathrm{W}$ Case-II: Total resistance of 39 bulbs $=39 \mathrm{R}$ Circuit current $\left(\mathrm{I}_{2}\right)=\frac{220}{39 \mathrm{R}}$ Power dissipated by 39 bulbs, $\mathrm{P}_{2}=39\left[\mathrm{I}_{2}^{2} \mathrm{R}\right]=39\left(\frac{220}{39 \mathrm{R}}\right)^{2} \times \mathrm{R}$ $\mathrm{P}_{2}=\frac{(220)^{2}}{39 \mathrm{R}} \mathrm{W}$ A look at equation (i) and (ii) shows that $P_{2}>P_{1}$. So, illumination of 39 bulbs more than 40 bulbs .
152694
A $100 \mathrm{~W} 200 \mathrm{~V}$ bulb is connected to a $160 \mathrm{~V}$ power supply. The power consumption would be
1 $125 \mathrm{~W}$
2 $100 \mathrm{~W}$
3 $80 \mathrm{~W}$
4 $64 \mathrm{~W}$
Explanation:
D Given, Power $(\mathrm{P})=100 \mathrm{~W}$ Bulb voltage $(\mathrm{V})=200$ Volt Supply power $\left(\mathrm{V}_{1}\right)=160$ Volt We know that, Resistance of the bulb $(\mathrm{R})=\frac{\mathrm{V}^{2}}{\mathrm{P}}$ $\therefore \quad \mathrm{R}=\frac{200^{2}}{100}=400 \Omega$ Now, Actual power consumption $(\mathrm{P})=\frac{\mathrm{V}_{1}^{2}}{\mathrm{R}}$ $\therefore \quad \mathrm{P}^{\prime}=\frac{160^{2}}{400}=64 \mathrm{~W}$
AIPMT-1997
Current Electricity
152695
A heating coil is labelled $100 \mathrm{~W}, 220 \mathrm{~V}$. The coil is cut in two equal halves and the two pieces are joined in parallel to the same source. The energy now liberated per second is
1 $25 \mathrm{~J}$
2 $50 \mathrm{~J}$
3 $200 \mathrm{~J}$
4 $400 \mathrm{~J}$
Explanation:
D Given, Power $(\mathrm{P})=100 \mathrm{~W}$ Voltage $(\mathrm{V})=220 \mathrm{~V}$ Let resistance of heating coil $=\mathrm{R}$ $\mathrm{R}=\frac{(220)^{2}}{100}=484$ We know that, The resistance of each part is $\mathrm{R} / 2$ and connect parallel. $\therefore$ Equivalent resistance $=\mathrm{R}_{\mathrm{eq}}$ $\frac{1}{\mathrm{R}_{\mathrm{eq}}} =\frac{1}{\mathrm{R} / 2}+\frac{1}{\mathrm{R} / 2}$ $=\frac{\mathrm{R} / 2+\mathrm{R} / 2}{(\mathrm{R} / 2)(\mathrm{R} / 2)}$ $\therefore \quad \mathrm{R}_{\mathrm{eq}}=\frac{(\mathrm{R} / 2)(\mathrm{R} / 2)}{(\mathrm{R} / 2)+(\mathrm{R} / 2)}=\frac{\mathrm{R}}{4}$ $\therefore$ Dissipated power, $\mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}_{\mathrm{eq}}}=\frac{(220)^{2}}{\mathrm{R} / 4}$ $=\frac{(220)^{2}}{\frac{484}{4}}=400 \mathrm{~J}$
AIPMT-1995
Current Electricity
152696
A $4 \mu \mathrm{F}$ conductor is charged to $400 \mathrm{~V}$ and then its plates are joined through a resistance of $1 \mathrm{k} \Omega$. The heat produced in the resistance is
1 $0.16 \mathrm{~J}$
2 $1.28 \mathrm{~J}$
3 $0.64 \mathrm{~J}$
4 $0.32 \mathrm{~J}$ [APIMT-1994]
Explanation:
D Given,} Capacitance of capacitor $(\mathrm{C})=4 \mu \mathrm{F}=4 \times 10^{-6} \mathrm{~F}$ Voltage applied $(\mathrm{V})=400$ Volt We know that, Energy stored $=$ Heat energy produced through resistance $\mathrm{U} =\frac{1}{2} \mathrm{CV}^{2}=\frac{1}{2} \times 4 \times 10^{-6} \times 400^{2}$ $=32 \times 10^{-2} \text { Joule }$ $\therefore \quad \mathrm{U} =0.32 \text { Joule }$
Current Electricity
152697
40 electric bulbs are connected in series across a $220 \mathrm{~V}$ supply. After one bulb is fused the remaining 39 are connected again in series across the same supply. The illumination will be
1 more with 40 bulbs than with 39
2 more with 39 bulbs than with 40
3 equal in both the cases
4 in the ratio $40^{2}: 39^{2}$
Explanation:
B Case -I: Total resistance of 40 bulbs $=40 \mathrm{R}$ Circuit current $\left(\mathrm{I}_{1}\right)=\frac{220}{40 \mathrm{R}}$ Power dissipated by 40 bulbs, $\mathrm{P}_{1}=40\left[\mathrm{I}_{1}^{2} \mathrm{R}\right]=40\left(\frac{220}{40 \mathrm{R}}\right)^{2} \times \mathrm{R}$ $\mathrm{P}_{1}=\frac{(220)^{2}}{40 \mathrm{R}} \mathrm{W}$ Case-II: Total resistance of 39 bulbs $=39 \mathrm{R}$ Circuit current $\left(\mathrm{I}_{2}\right)=\frac{220}{39 \mathrm{R}}$ Power dissipated by 39 bulbs, $\mathrm{P}_{2}=39\left[\mathrm{I}_{2}^{2} \mathrm{R}\right]=39\left(\frac{220}{39 \mathrm{R}}\right)^{2} \times \mathrm{R}$ $\mathrm{P}_{2}=\frac{(220)^{2}}{39 \mathrm{R}} \mathrm{W}$ A look at equation (i) and (ii) shows that $P_{2}>P_{1}$. So, illumination of 39 bulbs more than 40 bulbs .
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Current Electricity
152694
A $100 \mathrm{~W} 200 \mathrm{~V}$ bulb is connected to a $160 \mathrm{~V}$ power supply. The power consumption would be
1 $125 \mathrm{~W}$
2 $100 \mathrm{~W}$
3 $80 \mathrm{~W}$
4 $64 \mathrm{~W}$
Explanation:
D Given, Power $(\mathrm{P})=100 \mathrm{~W}$ Bulb voltage $(\mathrm{V})=200$ Volt Supply power $\left(\mathrm{V}_{1}\right)=160$ Volt We know that, Resistance of the bulb $(\mathrm{R})=\frac{\mathrm{V}^{2}}{\mathrm{P}}$ $\therefore \quad \mathrm{R}=\frac{200^{2}}{100}=400 \Omega$ Now, Actual power consumption $(\mathrm{P})=\frac{\mathrm{V}_{1}^{2}}{\mathrm{R}}$ $\therefore \quad \mathrm{P}^{\prime}=\frac{160^{2}}{400}=64 \mathrm{~W}$
AIPMT-1997
Current Electricity
152695
A heating coil is labelled $100 \mathrm{~W}, 220 \mathrm{~V}$. The coil is cut in two equal halves and the two pieces are joined in parallel to the same source. The energy now liberated per second is
1 $25 \mathrm{~J}$
2 $50 \mathrm{~J}$
3 $200 \mathrm{~J}$
4 $400 \mathrm{~J}$
Explanation:
D Given, Power $(\mathrm{P})=100 \mathrm{~W}$ Voltage $(\mathrm{V})=220 \mathrm{~V}$ Let resistance of heating coil $=\mathrm{R}$ $\mathrm{R}=\frac{(220)^{2}}{100}=484$ We know that, The resistance of each part is $\mathrm{R} / 2$ and connect parallel. $\therefore$ Equivalent resistance $=\mathrm{R}_{\mathrm{eq}}$ $\frac{1}{\mathrm{R}_{\mathrm{eq}}} =\frac{1}{\mathrm{R} / 2}+\frac{1}{\mathrm{R} / 2}$ $=\frac{\mathrm{R} / 2+\mathrm{R} / 2}{(\mathrm{R} / 2)(\mathrm{R} / 2)}$ $\therefore \quad \mathrm{R}_{\mathrm{eq}}=\frac{(\mathrm{R} / 2)(\mathrm{R} / 2)}{(\mathrm{R} / 2)+(\mathrm{R} / 2)}=\frac{\mathrm{R}}{4}$ $\therefore$ Dissipated power, $\mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}_{\mathrm{eq}}}=\frac{(220)^{2}}{\mathrm{R} / 4}$ $=\frac{(220)^{2}}{\frac{484}{4}}=400 \mathrm{~J}$
AIPMT-1995
Current Electricity
152696
A $4 \mu \mathrm{F}$ conductor is charged to $400 \mathrm{~V}$ and then its plates are joined through a resistance of $1 \mathrm{k} \Omega$. The heat produced in the resistance is
1 $0.16 \mathrm{~J}$
2 $1.28 \mathrm{~J}$
3 $0.64 \mathrm{~J}$
4 $0.32 \mathrm{~J}$ [APIMT-1994]
Explanation:
D Given,} Capacitance of capacitor $(\mathrm{C})=4 \mu \mathrm{F}=4 \times 10^{-6} \mathrm{~F}$ Voltage applied $(\mathrm{V})=400$ Volt We know that, Energy stored $=$ Heat energy produced through resistance $\mathrm{U} =\frac{1}{2} \mathrm{CV}^{2}=\frac{1}{2} \times 4 \times 10^{-6} \times 400^{2}$ $=32 \times 10^{-2} \text { Joule }$ $\therefore \quad \mathrm{U} =0.32 \text { Joule }$
Current Electricity
152697
40 electric bulbs are connected in series across a $220 \mathrm{~V}$ supply. After one bulb is fused the remaining 39 are connected again in series across the same supply. The illumination will be
1 more with 40 bulbs than with 39
2 more with 39 bulbs than with 40
3 equal in both the cases
4 in the ratio $40^{2}: 39^{2}$
Explanation:
B Case -I: Total resistance of 40 bulbs $=40 \mathrm{R}$ Circuit current $\left(\mathrm{I}_{1}\right)=\frac{220}{40 \mathrm{R}}$ Power dissipated by 40 bulbs, $\mathrm{P}_{1}=40\left[\mathrm{I}_{1}^{2} \mathrm{R}\right]=40\left(\frac{220}{40 \mathrm{R}}\right)^{2} \times \mathrm{R}$ $\mathrm{P}_{1}=\frac{(220)^{2}}{40 \mathrm{R}} \mathrm{W}$ Case-II: Total resistance of 39 bulbs $=39 \mathrm{R}$ Circuit current $\left(\mathrm{I}_{2}\right)=\frac{220}{39 \mathrm{R}}$ Power dissipated by 39 bulbs, $\mathrm{P}_{2}=39\left[\mathrm{I}_{2}^{2} \mathrm{R}\right]=39\left(\frac{220}{39 \mathrm{R}}\right)^{2} \times \mathrm{R}$ $\mathrm{P}_{2}=\frac{(220)^{2}}{39 \mathrm{R}} \mathrm{W}$ A look at equation (i) and (ii) shows that $P_{2}>P_{1}$. So, illumination of 39 bulbs more than 40 bulbs .
152694
A $100 \mathrm{~W} 200 \mathrm{~V}$ bulb is connected to a $160 \mathrm{~V}$ power supply. The power consumption would be
1 $125 \mathrm{~W}$
2 $100 \mathrm{~W}$
3 $80 \mathrm{~W}$
4 $64 \mathrm{~W}$
Explanation:
D Given, Power $(\mathrm{P})=100 \mathrm{~W}$ Bulb voltage $(\mathrm{V})=200$ Volt Supply power $\left(\mathrm{V}_{1}\right)=160$ Volt We know that, Resistance of the bulb $(\mathrm{R})=\frac{\mathrm{V}^{2}}{\mathrm{P}}$ $\therefore \quad \mathrm{R}=\frac{200^{2}}{100}=400 \Omega$ Now, Actual power consumption $(\mathrm{P})=\frac{\mathrm{V}_{1}^{2}}{\mathrm{R}}$ $\therefore \quad \mathrm{P}^{\prime}=\frac{160^{2}}{400}=64 \mathrm{~W}$
AIPMT-1997
Current Electricity
152695
A heating coil is labelled $100 \mathrm{~W}, 220 \mathrm{~V}$. The coil is cut in two equal halves and the two pieces are joined in parallel to the same source. The energy now liberated per second is
1 $25 \mathrm{~J}$
2 $50 \mathrm{~J}$
3 $200 \mathrm{~J}$
4 $400 \mathrm{~J}$
Explanation:
D Given, Power $(\mathrm{P})=100 \mathrm{~W}$ Voltage $(\mathrm{V})=220 \mathrm{~V}$ Let resistance of heating coil $=\mathrm{R}$ $\mathrm{R}=\frac{(220)^{2}}{100}=484$ We know that, The resistance of each part is $\mathrm{R} / 2$ and connect parallel. $\therefore$ Equivalent resistance $=\mathrm{R}_{\mathrm{eq}}$ $\frac{1}{\mathrm{R}_{\mathrm{eq}}} =\frac{1}{\mathrm{R} / 2}+\frac{1}{\mathrm{R} / 2}$ $=\frac{\mathrm{R} / 2+\mathrm{R} / 2}{(\mathrm{R} / 2)(\mathrm{R} / 2)}$ $\therefore \quad \mathrm{R}_{\mathrm{eq}}=\frac{(\mathrm{R} / 2)(\mathrm{R} / 2)}{(\mathrm{R} / 2)+(\mathrm{R} / 2)}=\frac{\mathrm{R}}{4}$ $\therefore$ Dissipated power, $\mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}_{\mathrm{eq}}}=\frac{(220)^{2}}{\mathrm{R} / 4}$ $=\frac{(220)^{2}}{\frac{484}{4}}=400 \mathrm{~J}$
AIPMT-1995
Current Electricity
152696
A $4 \mu \mathrm{F}$ conductor is charged to $400 \mathrm{~V}$ and then its plates are joined through a resistance of $1 \mathrm{k} \Omega$. The heat produced in the resistance is
1 $0.16 \mathrm{~J}$
2 $1.28 \mathrm{~J}$
3 $0.64 \mathrm{~J}$
4 $0.32 \mathrm{~J}$ [APIMT-1994]
Explanation:
D Given,} Capacitance of capacitor $(\mathrm{C})=4 \mu \mathrm{F}=4 \times 10^{-6} \mathrm{~F}$ Voltage applied $(\mathrm{V})=400$ Volt We know that, Energy stored $=$ Heat energy produced through resistance $\mathrm{U} =\frac{1}{2} \mathrm{CV}^{2}=\frac{1}{2} \times 4 \times 10^{-6} \times 400^{2}$ $=32 \times 10^{-2} \text { Joule }$ $\therefore \quad \mathrm{U} =0.32 \text { Joule }$
Current Electricity
152697
40 electric bulbs are connected in series across a $220 \mathrm{~V}$ supply. After one bulb is fused the remaining 39 are connected again in series across the same supply. The illumination will be
1 more with 40 bulbs than with 39
2 more with 39 bulbs than with 40
3 equal in both the cases
4 in the ratio $40^{2}: 39^{2}$
Explanation:
B Case -I: Total resistance of 40 bulbs $=40 \mathrm{R}$ Circuit current $\left(\mathrm{I}_{1}\right)=\frac{220}{40 \mathrm{R}}$ Power dissipated by 40 bulbs, $\mathrm{P}_{1}=40\left[\mathrm{I}_{1}^{2} \mathrm{R}\right]=40\left(\frac{220}{40 \mathrm{R}}\right)^{2} \times \mathrm{R}$ $\mathrm{P}_{1}=\frac{(220)^{2}}{40 \mathrm{R}} \mathrm{W}$ Case-II: Total resistance of 39 bulbs $=39 \mathrm{R}$ Circuit current $\left(\mathrm{I}_{2}\right)=\frac{220}{39 \mathrm{R}}$ Power dissipated by 39 bulbs, $\mathrm{P}_{2}=39\left[\mathrm{I}_{2}^{2} \mathrm{R}\right]=39\left(\frac{220}{39 \mathrm{R}}\right)^{2} \times \mathrm{R}$ $\mathrm{P}_{2}=\frac{(220)^{2}}{39 \mathrm{R}} \mathrm{W}$ A look at equation (i) and (ii) shows that $P_{2}>P_{1}$. So, illumination of 39 bulbs more than 40 bulbs .