152742
Four identical resistors when connected in series dissipate $5 \mathrm{~W}$ power. If they are connected in parallel, the power dissipated will be:
1 $80 \mathrm{~W}$
2 $60 \mathrm{~W}$
3 $40 \mathrm{~W}$
4 $20 \mathrm{~W}$
Explanation:
A We know that, Resistance of all four series connected $\left(\mathrm{R}_{\mathrm{S}}\right)=4 \mathrm{R}$ Power dissipated in all the four resistance, $\mathrm{P}_{\mathrm{S}}=\frac{\mathrm{V}^{2}}{\mathrm{R}_{\mathrm{S}}}=\frac{\mathrm{V}^{2}}{4 \mathrm{R}}$ $5=\frac{\mathrm{V}^{2}}{4 \mathrm{R}}$ $\frac{\mathrm{V}^{2}}{\mathrm{R}}=20$ For parallel, $\mathrm{R}_{\mathrm{P}}=\frac{\mathrm{R}}{4}$ Then, $\quad \mathrm{P}_{\mathrm{P}}=\frac{\mathrm{V}^{2}}{\frac{\mathrm{R}}{4}}=4 \frac{\mathrm{V}^{2}}{\mathrm{R}}$ $\mathrm{P}_{\mathrm{P}}=4 \times 20=80 \mathrm{~W}$
Current Electricity
152743
Two identical batteries, each of emf $2 \mathrm{~V}$ and internal resistance $1 \Omega$ pass a current through external resistance $R=0.5 \Omega$. The maximum power that can be developed across $R$ using these batteries is :
1 $3.2 \mathrm{~W}$
2 $8.2 \mathrm{~W}$
3 $4 \mathrm{~W}$
4 $2 \mathrm{~W}$
Explanation:
D $E_{e q}=\frac{E_{1} r_{2}+E_{2} r_{1}}{r_{1}+r_{2}}$ $=\frac{2 \times 1+2 \times 1}{1+1}=2 V$ By maximum power transfer theorem, $\mathrm{P}_{\text {max }}=\frac{\mathrm{V}_{\text {th }}^{2}}{4 \mathrm{R}_{\mathrm{L}}}=\frac{(2)^{2}}{4 \times 0.5}$ $\mathrm{P}_{\text {max }}=2 \mathrm{~W}$
JCECE-2003
Current Electricity
152745
Several lamps of $50 \mathrm{~W}$ and $100 \mathrm{~V}$ rating are available. How many of them can be connected in parallel across a battery of a $120 \mathrm{~V}$ of internal resistance $10 \Omega$, so that all bulbs glow in full power?
1 2
2 4
3 6
4 8
Explanation:
B Resistance of bulb, $\mathrm{R}=\frac{\mathrm{V}^{2}}{\mathrm{P}}=\frac{100 \times 100}{5}=200 \Omega$ 'n' lamps connected is parallel, $\therefore$ Total resistance of $\mathrm{n}$ lamps $=\frac{200}{\mathrm{n}}$ Current through each lamp $=\frac{\mathrm{P}}{\mathrm{V}}=\frac{50}{100}=\frac{1}{2} \mathrm{~A}$ Current through $\mathrm{n}$ lamp $=\frac{\mathrm{n}}{2}$ Total resistance of circuit $=10+\frac{200}{n}$ $\because$ Current, $I=\frac{\mathrm{emf}}{\text { Total resistance }}$ $\frac{\mathrm{n}}{2}=\frac{120}{10+\frac{200}{\mathrm{n}}} \Rightarrow \frac{\mathrm{n}}{2}=\frac{120 \mathrm{n}}{10 \mathrm{n}+200}$ $10 \mathrm{n}+200=240 \Rightarrow 10 \mathrm{n}=40$ $\mathrm{n}=4$
COMEDK 2015
Current Electricity
152746
A wire when connected to $220 \mathrm{~V}$ mains supply has power dissipation $P_{1}$. Now the wire is cut into two equal pieces which are connected in parallel to the same supply. Power dissipation in this case is $P_{2}$. Then $P_{2}: P_{1}$ is
1 1
2 4
3 2
4 3
Explanation:
B Case I : \(P_1=\frac{V^2}{R}\) Case II : The wire is cut into two equal pieces. Therefore the resistance of the individual wire is $R / 2$. These are connected in parallel, $\because \quad \mathrm{R}_{\text {eq }}=\frac{\mathrm{R}}{4}$ $\because \quad \mathrm{P}_{2}=\frac{\mathrm{V}^{2}}{\frac{\mathrm{R}}{4}}=4\left(\frac{\mathrm{V}^{2}}{\mathrm{R}}\right)=4 \mathrm{P}_{1} \Rightarrow \frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=4$ $\therefore \quad$ Ratio of $\mathrm{P}_{2}: \mathrm{P}_{1}=4: 1$
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Current Electricity
152742
Four identical resistors when connected in series dissipate $5 \mathrm{~W}$ power. If they are connected in parallel, the power dissipated will be:
1 $80 \mathrm{~W}$
2 $60 \mathrm{~W}$
3 $40 \mathrm{~W}$
4 $20 \mathrm{~W}$
Explanation:
A We know that, Resistance of all four series connected $\left(\mathrm{R}_{\mathrm{S}}\right)=4 \mathrm{R}$ Power dissipated in all the four resistance, $\mathrm{P}_{\mathrm{S}}=\frac{\mathrm{V}^{2}}{\mathrm{R}_{\mathrm{S}}}=\frac{\mathrm{V}^{2}}{4 \mathrm{R}}$ $5=\frac{\mathrm{V}^{2}}{4 \mathrm{R}}$ $\frac{\mathrm{V}^{2}}{\mathrm{R}}=20$ For parallel, $\mathrm{R}_{\mathrm{P}}=\frac{\mathrm{R}}{4}$ Then, $\quad \mathrm{P}_{\mathrm{P}}=\frac{\mathrm{V}^{2}}{\frac{\mathrm{R}}{4}}=4 \frac{\mathrm{V}^{2}}{\mathrm{R}}$ $\mathrm{P}_{\mathrm{P}}=4 \times 20=80 \mathrm{~W}$
Current Electricity
152743
Two identical batteries, each of emf $2 \mathrm{~V}$ and internal resistance $1 \Omega$ pass a current through external resistance $R=0.5 \Omega$. The maximum power that can be developed across $R$ using these batteries is :
1 $3.2 \mathrm{~W}$
2 $8.2 \mathrm{~W}$
3 $4 \mathrm{~W}$
4 $2 \mathrm{~W}$
Explanation:
D $E_{e q}=\frac{E_{1} r_{2}+E_{2} r_{1}}{r_{1}+r_{2}}$ $=\frac{2 \times 1+2 \times 1}{1+1}=2 V$ By maximum power transfer theorem, $\mathrm{P}_{\text {max }}=\frac{\mathrm{V}_{\text {th }}^{2}}{4 \mathrm{R}_{\mathrm{L}}}=\frac{(2)^{2}}{4 \times 0.5}$ $\mathrm{P}_{\text {max }}=2 \mathrm{~W}$
JCECE-2003
Current Electricity
152745
Several lamps of $50 \mathrm{~W}$ and $100 \mathrm{~V}$ rating are available. How many of them can be connected in parallel across a battery of a $120 \mathrm{~V}$ of internal resistance $10 \Omega$, so that all bulbs glow in full power?
1 2
2 4
3 6
4 8
Explanation:
B Resistance of bulb, $\mathrm{R}=\frac{\mathrm{V}^{2}}{\mathrm{P}}=\frac{100 \times 100}{5}=200 \Omega$ 'n' lamps connected is parallel, $\therefore$ Total resistance of $\mathrm{n}$ lamps $=\frac{200}{\mathrm{n}}$ Current through each lamp $=\frac{\mathrm{P}}{\mathrm{V}}=\frac{50}{100}=\frac{1}{2} \mathrm{~A}$ Current through $\mathrm{n}$ lamp $=\frac{\mathrm{n}}{2}$ Total resistance of circuit $=10+\frac{200}{n}$ $\because$ Current, $I=\frac{\mathrm{emf}}{\text { Total resistance }}$ $\frac{\mathrm{n}}{2}=\frac{120}{10+\frac{200}{\mathrm{n}}} \Rightarrow \frac{\mathrm{n}}{2}=\frac{120 \mathrm{n}}{10 \mathrm{n}+200}$ $10 \mathrm{n}+200=240 \Rightarrow 10 \mathrm{n}=40$ $\mathrm{n}=4$
COMEDK 2015
Current Electricity
152746
A wire when connected to $220 \mathrm{~V}$ mains supply has power dissipation $P_{1}$. Now the wire is cut into two equal pieces which are connected in parallel to the same supply. Power dissipation in this case is $P_{2}$. Then $P_{2}: P_{1}$ is
1 1
2 4
3 2
4 3
Explanation:
B Case I : \(P_1=\frac{V^2}{R}\) Case II : The wire is cut into two equal pieces. Therefore the resistance of the individual wire is $R / 2$. These are connected in parallel, $\because \quad \mathrm{R}_{\text {eq }}=\frac{\mathrm{R}}{4}$ $\because \quad \mathrm{P}_{2}=\frac{\mathrm{V}^{2}}{\frac{\mathrm{R}}{4}}=4\left(\frac{\mathrm{V}^{2}}{\mathrm{R}}\right)=4 \mathrm{P}_{1} \Rightarrow \frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=4$ $\therefore \quad$ Ratio of $\mathrm{P}_{2}: \mathrm{P}_{1}=4: 1$
152742
Four identical resistors when connected in series dissipate $5 \mathrm{~W}$ power. If they are connected in parallel, the power dissipated will be:
1 $80 \mathrm{~W}$
2 $60 \mathrm{~W}$
3 $40 \mathrm{~W}$
4 $20 \mathrm{~W}$
Explanation:
A We know that, Resistance of all four series connected $\left(\mathrm{R}_{\mathrm{S}}\right)=4 \mathrm{R}$ Power dissipated in all the four resistance, $\mathrm{P}_{\mathrm{S}}=\frac{\mathrm{V}^{2}}{\mathrm{R}_{\mathrm{S}}}=\frac{\mathrm{V}^{2}}{4 \mathrm{R}}$ $5=\frac{\mathrm{V}^{2}}{4 \mathrm{R}}$ $\frac{\mathrm{V}^{2}}{\mathrm{R}}=20$ For parallel, $\mathrm{R}_{\mathrm{P}}=\frac{\mathrm{R}}{4}$ Then, $\quad \mathrm{P}_{\mathrm{P}}=\frac{\mathrm{V}^{2}}{\frac{\mathrm{R}}{4}}=4 \frac{\mathrm{V}^{2}}{\mathrm{R}}$ $\mathrm{P}_{\mathrm{P}}=4 \times 20=80 \mathrm{~W}$
Current Electricity
152743
Two identical batteries, each of emf $2 \mathrm{~V}$ and internal resistance $1 \Omega$ pass a current through external resistance $R=0.5 \Omega$. The maximum power that can be developed across $R$ using these batteries is :
1 $3.2 \mathrm{~W}$
2 $8.2 \mathrm{~W}$
3 $4 \mathrm{~W}$
4 $2 \mathrm{~W}$
Explanation:
D $E_{e q}=\frac{E_{1} r_{2}+E_{2} r_{1}}{r_{1}+r_{2}}$ $=\frac{2 \times 1+2 \times 1}{1+1}=2 V$ By maximum power transfer theorem, $\mathrm{P}_{\text {max }}=\frac{\mathrm{V}_{\text {th }}^{2}}{4 \mathrm{R}_{\mathrm{L}}}=\frac{(2)^{2}}{4 \times 0.5}$ $\mathrm{P}_{\text {max }}=2 \mathrm{~W}$
JCECE-2003
Current Electricity
152745
Several lamps of $50 \mathrm{~W}$ and $100 \mathrm{~V}$ rating are available. How many of them can be connected in parallel across a battery of a $120 \mathrm{~V}$ of internal resistance $10 \Omega$, so that all bulbs glow in full power?
1 2
2 4
3 6
4 8
Explanation:
B Resistance of bulb, $\mathrm{R}=\frac{\mathrm{V}^{2}}{\mathrm{P}}=\frac{100 \times 100}{5}=200 \Omega$ 'n' lamps connected is parallel, $\therefore$ Total resistance of $\mathrm{n}$ lamps $=\frac{200}{\mathrm{n}}$ Current through each lamp $=\frac{\mathrm{P}}{\mathrm{V}}=\frac{50}{100}=\frac{1}{2} \mathrm{~A}$ Current through $\mathrm{n}$ lamp $=\frac{\mathrm{n}}{2}$ Total resistance of circuit $=10+\frac{200}{n}$ $\because$ Current, $I=\frac{\mathrm{emf}}{\text { Total resistance }}$ $\frac{\mathrm{n}}{2}=\frac{120}{10+\frac{200}{\mathrm{n}}} \Rightarrow \frac{\mathrm{n}}{2}=\frac{120 \mathrm{n}}{10 \mathrm{n}+200}$ $10 \mathrm{n}+200=240 \Rightarrow 10 \mathrm{n}=40$ $\mathrm{n}=4$
COMEDK 2015
Current Electricity
152746
A wire when connected to $220 \mathrm{~V}$ mains supply has power dissipation $P_{1}$. Now the wire is cut into two equal pieces which are connected in parallel to the same supply. Power dissipation in this case is $P_{2}$. Then $P_{2}: P_{1}$ is
1 1
2 4
3 2
4 3
Explanation:
B Case I : \(P_1=\frac{V^2}{R}\) Case II : The wire is cut into two equal pieces. Therefore the resistance of the individual wire is $R / 2$. These are connected in parallel, $\because \quad \mathrm{R}_{\text {eq }}=\frac{\mathrm{R}}{4}$ $\because \quad \mathrm{P}_{2}=\frac{\mathrm{V}^{2}}{\frac{\mathrm{R}}{4}}=4\left(\frac{\mathrm{V}^{2}}{\mathrm{R}}\right)=4 \mathrm{P}_{1} \Rightarrow \frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=4$ $\therefore \quad$ Ratio of $\mathrm{P}_{2}: \mathrm{P}_{1}=4: 1$
152742
Four identical resistors when connected in series dissipate $5 \mathrm{~W}$ power. If they are connected in parallel, the power dissipated will be:
1 $80 \mathrm{~W}$
2 $60 \mathrm{~W}$
3 $40 \mathrm{~W}$
4 $20 \mathrm{~W}$
Explanation:
A We know that, Resistance of all four series connected $\left(\mathrm{R}_{\mathrm{S}}\right)=4 \mathrm{R}$ Power dissipated in all the four resistance, $\mathrm{P}_{\mathrm{S}}=\frac{\mathrm{V}^{2}}{\mathrm{R}_{\mathrm{S}}}=\frac{\mathrm{V}^{2}}{4 \mathrm{R}}$ $5=\frac{\mathrm{V}^{2}}{4 \mathrm{R}}$ $\frac{\mathrm{V}^{2}}{\mathrm{R}}=20$ For parallel, $\mathrm{R}_{\mathrm{P}}=\frac{\mathrm{R}}{4}$ Then, $\quad \mathrm{P}_{\mathrm{P}}=\frac{\mathrm{V}^{2}}{\frac{\mathrm{R}}{4}}=4 \frac{\mathrm{V}^{2}}{\mathrm{R}}$ $\mathrm{P}_{\mathrm{P}}=4 \times 20=80 \mathrm{~W}$
Current Electricity
152743
Two identical batteries, each of emf $2 \mathrm{~V}$ and internal resistance $1 \Omega$ pass a current through external resistance $R=0.5 \Omega$. The maximum power that can be developed across $R$ using these batteries is :
1 $3.2 \mathrm{~W}$
2 $8.2 \mathrm{~W}$
3 $4 \mathrm{~W}$
4 $2 \mathrm{~W}$
Explanation:
D $E_{e q}=\frac{E_{1} r_{2}+E_{2} r_{1}}{r_{1}+r_{2}}$ $=\frac{2 \times 1+2 \times 1}{1+1}=2 V$ By maximum power transfer theorem, $\mathrm{P}_{\text {max }}=\frac{\mathrm{V}_{\text {th }}^{2}}{4 \mathrm{R}_{\mathrm{L}}}=\frac{(2)^{2}}{4 \times 0.5}$ $\mathrm{P}_{\text {max }}=2 \mathrm{~W}$
JCECE-2003
Current Electricity
152745
Several lamps of $50 \mathrm{~W}$ and $100 \mathrm{~V}$ rating are available. How many of them can be connected in parallel across a battery of a $120 \mathrm{~V}$ of internal resistance $10 \Omega$, so that all bulbs glow in full power?
1 2
2 4
3 6
4 8
Explanation:
B Resistance of bulb, $\mathrm{R}=\frac{\mathrm{V}^{2}}{\mathrm{P}}=\frac{100 \times 100}{5}=200 \Omega$ 'n' lamps connected is parallel, $\therefore$ Total resistance of $\mathrm{n}$ lamps $=\frac{200}{\mathrm{n}}$ Current through each lamp $=\frac{\mathrm{P}}{\mathrm{V}}=\frac{50}{100}=\frac{1}{2} \mathrm{~A}$ Current through $\mathrm{n}$ lamp $=\frac{\mathrm{n}}{2}$ Total resistance of circuit $=10+\frac{200}{n}$ $\because$ Current, $I=\frac{\mathrm{emf}}{\text { Total resistance }}$ $\frac{\mathrm{n}}{2}=\frac{120}{10+\frac{200}{\mathrm{n}}} \Rightarrow \frac{\mathrm{n}}{2}=\frac{120 \mathrm{n}}{10 \mathrm{n}+200}$ $10 \mathrm{n}+200=240 \Rightarrow 10 \mathrm{n}=40$ $\mathrm{n}=4$
COMEDK 2015
Current Electricity
152746
A wire when connected to $220 \mathrm{~V}$ mains supply has power dissipation $P_{1}$. Now the wire is cut into two equal pieces which are connected in parallel to the same supply. Power dissipation in this case is $P_{2}$. Then $P_{2}: P_{1}$ is
1 1
2 4
3 2
4 3
Explanation:
B Case I : \(P_1=\frac{V^2}{R}\) Case II : The wire is cut into two equal pieces. Therefore the resistance of the individual wire is $R / 2$. These are connected in parallel, $\because \quad \mathrm{R}_{\text {eq }}=\frac{\mathrm{R}}{4}$ $\because \quad \mathrm{P}_{2}=\frac{\mathrm{V}^{2}}{\frac{\mathrm{R}}{4}}=4\left(\frac{\mathrm{V}^{2}}{\mathrm{R}}\right)=4 \mathrm{P}_{1} \Rightarrow \frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=4$ $\therefore \quad$ Ratio of $\mathrm{P}_{2}: \mathrm{P}_{1}=4: 1$
