357137
Two bulbs marked \(60\,\;W,220\,\;V\) and \(100\;\,W,220\,\;V\) are connected in series and the series combination is now connected across a \(220\;\,V\) main supply. The power dissipiated in the circuit is
1 \(37.5\;\,W\)
2 \(75\;\,W\)
3 \(80\;\,W\)
4 \(40\;\,W\)
Explanation:
When connected in series, power dissipated is \({P_S} = \frac{{{P_1}{P_2}}}{{{P_1} + {P_2}}} = \frac{{100 \times 60}}{{100 + 60}}\) \( = \frac{{6000}}{{160}} = 37.5\,\;W\)
PHXII03:CURRENT ELECTRICITY
357138
An electric heater of resistance \(6\,\Omega \) is run for \(10 \mathrm{~min}\) on a \(120\;V\) line. The energy liberated in this period of time is
357139
Heat produced in a resistance if 1.5 \(A\) current is passed through it is 500\(J\) in 20 sec. If the current is 3\(A\), the heat produced in the same resistor in the same duration will be
1 500 \(J\)
2 1000 \(J\)
3 1500 \(J\)
4 2000 \(J\)
Explanation:
Heat Produced \( = H = {I^2}Rt\) \(\therefore \;\frac{{{H_1}}}{{{H_2}}} = {\left( {\frac{{{I_1}}}{{{I_2}}}} \right)^2}\) (\(\because \) time \(t\) is same in both cases) \( \Rightarrow {H_2} = 500 \times {\left( {\frac{3}{{1.5}}} \right)^2} = 2000\,J\)
JEE - 2021
PHXII03:CURRENT ELECTRICITY
357140
Find the power supplied by 10 \(V\) battery in the circuit shown in figure
1 \(8\,W\)
2 \(15\,W\)
3 \(20\,W\)
4 \(3\,W\)
Explanation:
Let us distribute the potentials at different junctions of the circuit as shown, Writing KCL equation for \(x, \dfrac{x}{5}+\dfrac{x-10}{10}+\dfrac{x+30}{10}+\dfrac{x+5}{5}=0\) \(\frac{{2x + x - 10 + x + 30 + 2x + 10}}{{10}} = 0\) \( \Rightarrow x = - 5\;V\) Current through the batteries and power supplied by batteries,\(I_{10 {~V}}=\dfrac{x}{5}+\dfrac{x+30}{10}=-1+2.5=1.5 {~A}\) and \(P_{\text {supplied } 10 {~V}}=\varepsilon I=10 \times 1.5=15 {~W}\)
357137
Two bulbs marked \(60\,\;W,220\,\;V\) and \(100\;\,W,220\,\;V\) are connected in series and the series combination is now connected across a \(220\;\,V\) main supply. The power dissipiated in the circuit is
1 \(37.5\;\,W\)
2 \(75\;\,W\)
3 \(80\;\,W\)
4 \(40\;\,W\)
Explanation:
When connected in series, power dissipated is \({P_S} = \frac{{{P_1}{P_2}}}{{{P_1} + {P_2}}} = \frac{{100 \times 60}}{{100 + 60}}\) \( = \frac{{6000}}{{160}} = 37.5\,\;W\)
PHXII03:CURRENT ELECTRICITY
357138
An electric heater of resistance \(6\,\Omega \) is run for \(10 \mathrm{~min}\) on a \(120\;V\) line. The energy liberated in this period of time is
357139
Heat produced in a resistance if 1.5 \(A\) current is passed through it is 500\(J\) in 20 sec. If the current is 3\(A\), the heat produced in the same resistor in the same duration will be
1 500 \(J\)
2 1000 \(J\)
3 1500 \(J\)
4 2000 \(J\)
Explanation:
Heat Produced \( = H = {I^2}Rt\) \(\therefore \;\frac{{{H_1}}}{{{H_2}}} = {\left( {\frac{{{I_1}}}{{{I_2}}}} \right)^2}\) (\(\because \) time \(t\) is same in both cases) \( \Rightarrow {H_2} = 500 \times {\left( {\frac{3}{{1.5}}} \right)^2} = 2000\,J\)
JEE - 2021
PHXII03:CURRENT ELECTRICITY
357140
Find the power supplied by 10 \(V\) battery in the circuit shown in figure
1 \(8\,W\)
2 \(15\,W\)
3 \(20\,W\)
4 \(3\,W\)
Explanation:
Let us distribute the potentials at different junctions of the circuit as shown, Writing KCL equation for \(x, \dfrac{x}{5}+\dfrac{x-10}{10}+\dfrac{x+30}{10}+\dfrac{x+5}{5}=0\) \(\frac{{2x + x - 10 + x + 30 + 2x + 10}}{{10}} = 0\) \( \Rightarrow x = - 5\;V\) Current through the batteries and power supplied by batteries,\(I_{10 {~V}}=\dfrac{x}{5}+\dfrac{x+30}{10}=-1+2.5=1.5 {~A}\) and \(P_{\text {supplied } 10 {~V}}=\varepsilon I=10 \times 1.5=15 {~W}\)
357137
Two bulbs marked \(60\,\;W,220\,\;V\) and \(100\;\,W,220\,\;V\) are connected in series and the series combination is now connected across a \(220\;\,V\) main supply. The power dissipiated in the circuit is
1 \(37.5\;\,W\)
2 \(75\;\,W\)
3 \(80\;\,W\)
4 \(40\;\,W\)
Explanation:
When connected in series, power dissipated is \({P_S} = \frac{{{P_1}{P_2}}}{{{P_1} + {P_2}}} = \frac{{100 \times 60}}{{100 + 60}}\) \( = \frac{{6000}}{{160}} = 37.5\,\;W\)
PHXII03:CURRENT ELECTRICITY
357138
An electric heater of resistance \(6\,\Omega \) is run for \(10 \mathrm{~min}\) on a \(120\;V\) line. The energy liberated in this period of time is
357139
Heat produced in a resistance if 1.5 \(A\) current is passed through it is 500\(J\) in 20 sec. If the current is 3\(A\), the heat produced in the same resistor in the same duration will be
1 500 \(J\)
2 1000 \(J\)
3 1500 \(J\)
4 2000 \(J\)
Explanation:
Heat Produced \( = H = {I^2}Rt\) \(\therefore \;\frac{{{H_1}}}{{{H_2}}} = {\left( {\frac{{{I_1}}}{{{I_2}}}} \right)^2}\) (\(\because \) time \(t\) is same in both cases) \( \Rightarrow {H_2} = 500 \times {\left( {\frac{3}{{1.5}}} \right)^2} = 2000\,J\)
JEE - 2021
PHXII03:CURRENT ELECTRICITY
357140
Find the power supplied by 10 \(V\) battery in the circuit shown in figure
1 \(8\,W\)
2 \(15\,W\)
3 \(20\,W\)
4 \(3\,W\)
Explanation:
Let us distribute the potentials at different junctions of the circuit as shown, Writing KCL equation for \(x, \dfrac{x}{5}+\dfrac{x-10}{10}+\dfrac{x+30}{10}+\dfrac{x+5}{5}=0\) \(\frac{{2x + x - 10 + x + 30 + 2x + 10}}{{10}} = 0\) \( \Rightarrow x = - 5\;V\) Current through the batteries and power supplied by batteries,\(I_{10 {~V}}=\dfrac{x}{5}+\dfrac{x+30}{10}=-1+2.5=1.5 {~A}\) and \(P_{\text {supplied } 10 {~V}}=\varepsilon I=10 \times 1.5=15 {~W}\)
357137
Two bulbs marked \(60\,\;W,220\,\;V\) and \(100\;\,W,220\,\;V\) are connected in series and the series combination is now connected across a \(220\;\,V\) main supply. The power dissipiated in the circuit is
1 \(37.5\;\,W\)
2 \(75\;\,W\)
3 \(80\;\,W\)
4 \(40\;\,W\)
Explanation:
When connected in series, power dissipated is \({P_S} = \frac{{{P_1}{P_2}}}{{{P_1} + {P_2}}} = \frac{{100 \times 60}}{{100 + 60}}\) \( = \frac{{6000}}{{160}} = 37.5\,\;W\)
PHXII03:CURRENT ELECTRICITY
357138
An electric heater of resistance \(6\,\Omega \) is run for \(10 \mathrm{~min}\) on a \(120\;V\) line. The energy liberated in this period of time is
357139
Heat produced in a resistance if 1.5 \(A\) current is passed through it is 500\(J\) in 20 sec. If the current is 3\(A\), the heat produced in the same resistor in the same duration will be
1 500 \(J\)
2 1000 \(J\)
3 1500 \(J\)
4 2000 \(J\)
Explanation:
Heat Produced \( = H = {I^2}Rt\) \(\therefore \;\frac{{{H_1}}}{{{H_2}}} = {\left( {\frac{{{I_1}}}{{{I_2}}}} \right)^2}\) (\(\because \) time \(t\) is same in both cases) \( \Rightarrow {H_2} = 500 \times {\left( {\frac{3}{{1.5}}} \right)^2} = 2000\,J\)
JEE - 2021
PHXII03:CURRENT ELECTRICITY
357140
Find the power supplied by 10 \(V\) battery in the circuit shown in figure
1 \(8\,W\)
2 \(15\,W\)
3 \(20\,W\)
4 \(3\,W\)
Explanation:
Let us distribute the potentials at different junctions of the circuit as shown, Writing KCL equation for \(x, \dfrac{x}{5}+\dfrac{x-10}{10}+\dfrac{x+30}{10}+\dfrac{x+5}{5}=0\) \(\frac{{2x + x - 10 + x + 30 + 2x + 10}}{{10}} = 0\) \( \Rightarrow x = - 5\;V\) Current through the batteries and power supplied by batteries,\(I_{10 {~V}}=\dfrac{x}{5}+\dfrac{x+30}{10}=-1+2.5=1.5 {~A}\) and \(P_{\text {supplied } 10 {~V}}=\varepsilon I=10 \times 1.5=15 {~W}\)
