152775
One filament takes $10 \mathrm{~min}$ to heat a kettle and another takes $15 \mathrm{~min}$ if connected in parallel. They combindly take...... minute to heat the same kettle.
1 6
2 12.5
3 25
4 7.5
Explanation:
A We know that, Heat, $\quad \mathrm{H}=\frac{\mathrm{V}^{2}}{\mathrm{R}} \mathrm{t} \Rightarrow \mathrm{R}=\frac{\mathrm{V}^{2}}{\mathrm{H}} \mathrm{t}$ In first filament $\mathrm{R}_{1}=\frac{\mathrm{V}^{2}}{\mathrm{H}} \mathrm{t}_{1}$ In second filament $\mathrm{R}_{2}=\frac{\mathrm{V}^{2}}{\mathrm{H}} \mathrm{t}_{2}$ When filament in parallel $\mathrm{H}=\frac{\mathrm{V}^{2}}{\mathrm{R}_{\mathrm{eq}}} \mathrm{t}_{\mathrm{eq}}$ $\mathrm{R}_{\mathrm{eq}}=\frac{\mathrm{V}^{2}}{\mathrm{H}} \mathrm{t}_{\mathrm{eq}}$ From equation (i), (ii) and (iii), we get- $\frac{1}{\mathrm{R}_{\text {eq }}}=\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}$ $\frac{\mathrm{H}}{\mathrm{V}^{2} \mathrm{t}_{\text {eq }}}=\frac{\mathrm{H}}{\mathrm{V}^{2} \mathrm{t}_{1}}+\frac{\mathrm{H}}{\mathrm{V}^{2} \mathrm{t}_{2}}$ $\frac{1}{\mathrm{t}_{\mathrm{eq}}}=\frac{1}{10}+\frac{1}{15}=\frac{3+2}{30}=\frac{5}{30}$ $\mathrm{t}_{\mathrm{eq}}=\frac{30}{5}$ $\mathrm{t}_{\mathrm{eq}}=6 \mathrm{~mm}$
UP CPMT-2003
Current Electricity
152702
A 5 A fuse wire can withstand a maximum power of $1 \mathrm{~W}$ in circuit. The resistance of the fuse wire is
1 $0.2 \Omega$
2 $5 \Omega$
3 $0.4 \Omega$
4 $0.04 \Omega$
Explanation:
D Given, Power $(\mathrm{P})=1 \mathrm{~W}$, current $(\mathrm{I})=5 \mathrm{~A}$, resistance $=$ ? We know that, $\mathrm{P}=\mathrm{I}^{2} \mathrm{R}$ $\mathrm{R}=\frac{\mathrm{P}}{\mathrm{I}^{2}}=\frac{1}{25}=0.04 \Omega$
AIPMT-2005
Current Electricity
152707
Four bulbs each marked $40 \mathrm{~W}, 250 \mathrm{~V}$ are connected in series with $250 \mathrm{~V}$ source. The total power output is
1 $10 \mathrm{~W}$
2 $40 \mathrm{~W}$
3 $160 \mathrm{~W}$
4 $320 \mathrm{~W}$
Explanation:
A Four bulbs each marked $40 \mathrm{~W}, 250 \mathrm{~V}$ are connected in series. Voltage $(\mathrm{V})=250 \mathrm{~V}$ $\mathrm{R}=\frac{\mathrm{V}^{2}}{\mathrm{P}}=\frac{(250)^{2}}{40} \Omega$ $\mathrm{R}_{\text {eq }}=\mathrm{R}+\mathrm{R}+\mathrm{R}+\mathrm{R}=4 \mathrm{R}=4 \times \frac{(250)^{2}}{40}=\frac{(250)^{2}}{10} \Omega$ $\therefore \mathrm{P}_{\text {output }}=\frac{\mathrm{V}^{2}}{\mathrm{R}_{\text {eq }}}=\frac{(250)^{2} \times 10}{(250)^{2}}=10 \mathrm{~W}$
EAMCET 1995
Current Electricity
152708
3 lamps of $250 \mathrm{~W}, 500 \mathrm{~W}, 1000 \mathrm{~W}$ are connected in series across the mains. The one which will light the brightest is
1 $250 \mathrm{~W}$
2 $500 \mathrm{~W}$
3 $1000 \mathrm{~W}$
4 all three will light with equal brightness
Explanation:
A When (n) lamps are connected is series across the mains. Which power is lowest in all it will light the brightest. $\mathrm{P} \propto \frac{1}{\mathrm{R}} $So, $250 \mathrm{~W}, 500 \mathrm{~W}$ and $1000 \mathrm{~W}$ are Hence, $250 \mathrm{~W}$ is brightest in all.
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Current Electricity
152775
One filament takes $10 \mathrm{~min}$ to heat a kettle and another takes $15 \mathrm{~min}$ if connected in parallel. They combindly take...... minute to heat the same kettle.
1 6
2 12.5
3 25
4 7.5
Explanation:
A We know that, Heat, $\quad \mathrm{H}=\frac{\mathrm{V}^{2}}{\mathrm{R}} \mathrm{t} \Rightarrow \mathrm{R}=\frac{\mathrm{V}^{2}}{\mathrm{H}} \mathrm{t}$ In first filament $\mathrm{R}_{1}=\frac{\mathrm{V}^{2}}{\mathrm{H}} \mathrm{t}_{1}$ In second filament $\mathrm{R}_{2}=\frac{\mathrm{V}^{2}}{\mathrm{H}} \mathrm{t}_{2}$ When filament in parallel $\mathrm{H}=\frac{\mathrm{V}^{2}}{\mathrm{R}_{\mathrm{eq}}} \mathrm{t}_{\mathrm{eq}}$ $\mathrm{R}_{\mathrm{eq}}=\frac{\mathrm{V}^{2}}{\mathrm{H}} \mathrm{t}_{\mathrm{eq}}$ From equation (i), (ii) and (iii), we get- $\frac{1}{\mathrm{R}_{\text {eq }}}=\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}$ $\frac{\mathrm{H}}{\mathrm{V}^{2} \mathrm{t}_{\text {eq }}}=\frac{\mathrm{H}}{\mathrm{V}^{2} \mathrm{t}_{1}}+\frac{\mathrm{H}}{\mathrm{V}^{2} \mathrm{t}_{2}}$ $\frac{1}{\mathrm{t}_{\mathrm{eq}}}=\frac{1}{10}+\frac{1}{15}=\frac{3+2}{30}=\frac{5}{30}$ $\mathrm{t}_{\mathrm{eq}}=\frac{30}{5}$ $\mathrm{t}_{\mathrm{eq}}=6 \mathrm{~mm}$
UP CPMT-2003
Current Electricity
152702
A 5 A fuse wire can withstand a maximum power of $1 \mathrm{~W}$ in circuit. The resistance of the fuse wire is
1 $0.2 \Omega$
2 $5 \Omega$
3 $0.4 \Omega$
4 $0.04 \Omega$
Explanation:
D Given, Power $(\mathrm{P})=1 \mathrm{~W}$, current $(\mathrm{I})=5 \mathrm{~A}$, resistance $=$ ? We know that, $\mathrm{P}=\mathrm{I}^{2} \mathrm{R}$ $\mathrm{R}=\frac{\mathrm{P}}{\mathrm{I}^{2}}=\frac{1}{25}=0.04 \Omega$
AIPMT-2005
Current Electricity
152707
Four bulbs each marked $40 \mathrm{~W}, 250 \mathrm{~V}$ are connected in series with $250 \mathrm{~V}$ source. The total power output is
1 $10 \mathrm{~W}$
2 $40 \mathrm{~W}$
3 $160 \mathrm{~W}$
4 $320 \mathrm{~W}$
Explanation:
A Four bulbs each marked $40 \mathrm{~W}, 250 \mathrm{~V}$ are connected in series. Voltage $(\mathrm{V})=250 \mathrm{~V}$ $\mathrm{R}=\frac{\mathrm{V}^{2}}{\mathrm{P}}=\frac{(250)^{2}}{40} \Omega$ $\mathrm{R}_{\text {eq }}=\mathrm{R}+\mathrm{R}+\mathrm{R}+\mathrm{R}=4 \mathrm{R}=4 \times \frac{(250)^{2}}{40}=\frac{(250)^{2}}{10} \Omega$ $\therefore \mathrm{P}_{\text {output }}=\frac{\mathrm{V}^{2}}{\mathrm{R}_{\text {eq }}}=\frac{(250)^{2} \times 10}{(250)^{2}}=10 \mathrm{~W}$
EAMCET 1995
Current Electricity
152708
3 lamps of $250 \mathrm{~W}, 500 \mathrm{~W}, 1000 \mathrm{~W}$ are connected in series across the mains. The one which will light the brightest is
1 $250 \mathrm{~W}$
2 $500 \mathrm{~W}$
3 $1000 \mathrm{~W}$
4 all three will light with equal brightness
Explanation:
A When (n) lamps are connected is series across the mains. Which power is lowest in all it will light the brightest. $\mathrm{P} \propto \frac{1}{\mathrm{R}} $So, $250 \mathrm{~W}, 500 \mathrm{~W}$ and $1000 \mathrm{~W}$ are Hence, $250 \mathrm{~W}$ is brightest in all.
152775
One filament takes $10 \mathrm{~min}$ to heat a kettle and another takes $15 \mathrm{~min}$ if connected in parallel. They combindly take...... minute to heat the same kettle.
1 6
2 12.5
3 25
4 7.5
Explanation:
A We know that, Heat, $\quad \mathrm{H}=\frac{\mathrm{V}^{2}}{\mathrm{R}} \mathrm{t} \Rightarrow \mathrm{R}=\frac{\mathrm{V}^{2}}{\mathrm{H}} \mathrm{t}$ In first filament $\mathrm{R}_{1}=\frac{\mathrm{V}^{2}}{\mathrm{H}} \mathrm{t}_{1}$ In second filament $\mathrm{R}_{2}=\frac{\mathrm{V}^{2}}{\mathrm{H}} \mathrm{t}_{2}$ When filament in parallel $\mathrm{H}=\frac{\mathrm{V}^{2}}{\mathrm{R}_{\mathrm{eq}}} \mathrm{t}_{\mathrm{eq}}$ $\mathrm{R}_{\mathrm{eq}}=\frac{\mathrm{V}^{2}}{\mathrm{H}} \mathrm{t}_{\mathrm{eq}}$ From equation (i), (ii) and (iii), we get- $\frac{1}{\mathrm{R}_{\text {eq }}}=\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}$ $\frac{\mathrm{H}}{\mathrm{V}^{2} \mathrm{t}_{\text {eq }}}=\frac{\mathrm{H}}{\mathrm{V}^{2} \mathrm{t}_{1}}+\frac{\mathrm{H}}{\mathrm{V}^{2} \mathrm{t}_{2}}$ $\frac{1}{\mathrm{t}_{\mathrm{eq}}}=\frac{1}{10}+\frac{1}{15}=\frac{3+2}{30}=\frac{5}{30}$ $\mathrm{t}_{\mathrm{eq}}=\frac{30}{5}$ $\mathrm{t}_{\mathrm{eq}}=6 \mathrm{~mm}$
UP CPMT-2003
Current Electricity
152702
A 5 A fuse wire can withstand a maximum power of $1 \mathrm{~W}$ in circuit. The resistance of the fuse wire is
1 $0.2 \Omega$
2 $5 \Omega$
3 $0.4 \Omega$
4 $0.04 \Omega$
Explanation:
D Given, Power $(\mathrm{P})=1 \mathrm{~W}$, current $(\mathrm{I})=5 \mathrm{~A}$, resistance $=$ ? We know that, $\mathrm{P}=\mathrm{I}^{2} \mathrm{R}$ $\mathrm{R}=\frac{\mathrm{P}}{\mathrm{I}^{2}}=\frac{1}{25}=0.04 \Omega$
AIPMT-2005
Current Electricity
152707
Four bulbs each marked $40 \mathrm{~W}, 250 \mathrm{~V}$ are connected in series with $250 \mathrm{~V}$ source. The total power output is
1 $10 \mathrm{~W}$
2 $40 \mathrm{~W}$
3 $160 \mathrm{~W}$
4 $320 \mathrm{~W}$
Explanation:
A Four bulbs each marked $40 \mathrm{~W}, 250 \mathrm{~V}$ are connected in series. Voltage $(\mathrm{V})=250 \mathrm{~V}$ $\mathrm{R}=\frac{\mathrm{V}^{2}}{\mathrm{P}}=\frac{(250)^{2}}{40} \Omega$ $\mathrm{R}_{\text {eq }}=\mathrm{R}+\mathrm{R}+\mathrm{R}+\mathrm{R}=4 \mathrm{R}=4 \times \frac{(250)^{2}}{40}=\frac{(250)^{2}}{10} \Omega$ $\therefore \mathrm{P}_{\text {output }}=\frac{\mathrm{V}^{2}}{\mathrm{R}_{\text {eq }}}=\frac{(250)^{2} \times 10}{(250)^{2}}=10 \mathrm{~W}$
EAMCET 1995
Current Electricity
152708
3 lamps of $250 \mathrm{~W}, 500 \mathrm{~W}, 1000 \mathrm{~W}$ are connected in series across the mains. The one which will light the brightest is
1 $250 \mathrm{~W}$
2 $500 \mathrm{~W}$
3 $1000 \mathrm{~W}$
4 all three will light with equal brightness
Explanation:
A When (n) lamps are connected is series across the mains. Which power is lowest in all it will light the brightest. $\mathrm{P} \propto \frac{1}{\mathrm{R}} $So, $250 \mathrm{~W}, 500 \mathrm{~W}$ and $1000 \mathrm{~W}$ are Hence, $250 \mathrm{~W}$ is brightest in all.
152775
One filament takes $10 \mathrm{~min}$ to heat a kettle and another takes $15 \mathrm{~min}$ if connected in parallel. They combindly take...... minute to heat the same kettle.
1 6
2 12.5
3 25
4 7.5
Explanation:
A We know that, Heat, $\quad \mathrm{H}=\frac{\mathrm{V}^{2}}{\mathrm{R}} \mathrm{t} \Rightarrow \mathrm{R}=\frac{\mathrm{V}^{2}}{\mathrm{H}} \mathrm{t}$ In first filament $\mathrm{R}_{1}=\frac{\mathrm{V}^{2}}{\mathrm{H}} \mathrm{t}_{1}$ In second filament $\mathrm{R}_{2}=\frac{\mathrm{V}^{2}}{\mathrm{H}} \mathrm{t}_{2}$ When filament in parallel $\mathrm{H}=\frac{\mathrm{V}^{2}}{\mathrm{R}_{\mathrm{eq}}} \mathrm{t}_{\mathrm{eq}}$ $\mathrm{R}_{\mathrm{eq}}=\frac{\mathrm{V}^{2}}{\mathrm{H}} \mathrm{t}_{\mathrm{eq}}$ From equation (i), (ii) and (iii), we get- $\frac{1}{\mathrm{R}_{\text {eq }}}=\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}$ $\frac{\mathrm{H}}{\mathrm{V}^{2} \mathrm{t}_{\text {eq }}}=\frac{\mathrm{H}}{\mathrm{V}^{2} \mathrm{t}_{1}}+\frac{\mathrm{H}}{\mathrm{V}^{2} \mathrm{t}_{2}}$ $\frac{1}{\mathrm{t}_{\mathrm{eq}}}=\frac{1}{10}+\frac{1}{15}=\frac{3+2}{30}=\frac{5}{30}$ $\mathrm{t}_{\mathrm{eq}}=\frac{30}{5}$ $\mathrm{t}_{\mathrm{eq}}=6 \mathrm{~mm}$
UP CPMT-2003
Current Electricity
152702
A 5 A fuse wire can withstand a maximum power of $1 \mathrm{~W}$ in circuit. The resistance of the fuse wire is
1 $0.2 \Omega$
2 $5 \Omega$
3 $0.4 \Omega$
4 $0.04 \Omega$
Explanation:
D Given, Power $(\mathrm{P})=1 \mathrm{~W}$, current $(\mathrm{I})=5 \mathrm{~A}$, resistance $=$ ? We know that, $\mathrm{P}=\mathrm{I}^{2} \mathrm{R}$ $\mathrm{R}=\frac{\mathrm{P}}{\mathrm{I}^{2}}=\frac{1}{25}=0.04 \Omega$
AIPMT-2005
Current Electricity
152707
Four bulbs each marked $40 \mathrm{~W}, 250 \mathrm{~V}$ are connected in series with $250 \mathrm{~V}$ source. The total power output is
1 $10 \mathrm{~W}$
2 $40 \mathrm{~W}$
3 $160 \mathrm{~W}$
4 $320 \mathrm{~W}$
Explanation:
A Four bulbs each marked $40 \mathrm{~W}, 250 \mathrm{~V}$ are connected in series. Voltage $(\mathrm{V})=250 \mathrm{~V}$ $\mathrm{R}=\frac{\mathrm{V}^{2}}{\mathrm{P}}=\frac{(250)^{2}}{40} \Omega$ $\mathrm{R}_{\text {eq }}=\mathrm{R}+\mathrm{R}+\mathrm{R}+\mathrm{R}=4 \mathrm{R}=4 \times \frac{(250)^{2}}{40}=\frac{(250)^{2}}{10} \Omega$ $\therefore \mathrm{P}_{\text {output }}=\frac{\mathrm{V}^{2}}{\mathrm{R}_{\text {eq }}}=\frac{(250)^{2} \times 10}{(250)^{2}}=10 \mathrm{~W}$
EAMCET 1995
Current Electricity
152708
3 lamps of $250 \mathrm{~W}, 500 \mathrm{~W}, 1000 \mathrm{~W}$ are connected in series across the mains. The one which will light the brightest is
1 $250 \mathrm{~W}$
2 $500 \mathrm{~W}$
3 $1000 \mathrm{~W}$
4 all three will light with equal brightness
Explanation:
A When (n) lamps are connected is series across the mains. Which power is lowest in all it will light the brightest. $\mathrm{P} \propto \frac{1}{\mathrm{R}} $So, $250 \mathrm{~W}, 500 \mathrm{~W}$ and $1000 \mathrm{~W}$ are Hence, $250 \mathrm{~W}$ is brightest in all.
