357124
Two resistors of resistances \(100\,\Omega \) and \(200\,\Omega \) are connected in parallel in an electrical circuit. The ratio of the thermal energy developed in \(100\,\Omega \) to that in \(200\,\Omega \) in a given time is
1 \(2:1\)
2 \(1:4\)
3 \(4:1\)
4 \(1:2\)
Explanation:
As both resistors are in parallel combination so potential drop (\(V\)) across both are same. \(P = \frac{{{V^2}}}{R} \Rightarrow P \propto \frac{1}{R}\) We know Power \(P = \frac{{{\rm{Energy}}}}{{{\rm{time}}}} = \frac{E}{t}\) Time(t) is same for both \({R_1},{R_2}\) \(\frac{{{E_1}}}{{{E_2}}} = \frac{{{P_1}}}{{{P_2}}} = \frac{{{R_2}}}{{{R_1}}} = \frac{{200}}{{100}} = \frac{2}{1}\) \( = 2:1\) Note: We are asked to find ratio of thermal energies
NEET - 2022
PHXII03:CURRENT ELECTRICITY
357125
There are \(25\;W - 220\;V\) bulb and a \(100\;W\) - \(220\;V\) line. Which electric bulb will glow more brightly?
1 \(25\;W\) bulb
2 \(100\;W\) bulb
3 Both will have equal incandescence
4 Neither \(25\;W\) nor \(100\;W\) bulb will give light
Explanation:
Since power is given by \(P=\dfrac{V^{2}}{R}\), \(\text { so } R=\dfrac{V^{2}}{P}\) For the first bulb, \(R_{1}=\left(\dfrac{V^{2}}{P_{1}}\right)=\left[\dfrac{(220)^{2}}{25}\right]=1936\, \Omega\) For the second bulb, \(R_{2}=\left(\dfrac{V^{2}}{P_{2}}\right)=\left[\dfrac{(220)^{2}}{100}\right]=484\, \Omega\) Current in series combination is the same in the two bulbs and current \(I\) is given by \(I = \frac{V}{{{R_1} + {R_2}}} = \frac{{220}}{{1936 + 484}} = \frac{{220}}{{2420}} = \frac{1}{{11}}\;A\) If the actual powers in the bulbs be \(P_{1}\) and \(P_{2}\) then \(P_1^\prime {\rm{ }} = {I^2}{R_1} = {\left( {\frac{1}{{11}}} \right)^2} \times 1936 = 16\;W\) and \(P_2^\prime {\rm{ }} = {I^2}{R_2} = {\left( {\frac{1}{{11}}} \right)^2} \times 484 = 4\;W\) Since \({P^\prime }_1 > {P^\prime }_2,25\;W\) bulb will glow more brightly.
PHXII03:CURRENT ELECTRICITY
357126
Assertion : Fuse wire must have high resistance and low melting point. Reason : Fuse is used for small current flow only.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Fuse wires need high resistance to generate more heat when current exceeds the limit, ensuring they melt quickly and break the circuit for protection. The low melting point facilitates rapid melting with a temperature increase. So correct option is (3).
PHXII03:CURRENT ELECTRICITY
357127
Two bulbs of 500 \(W\) and 200 \(W\) are manufactured to operate on 220 \(V\) line. The ratio of heat produced in 500 \(W\) and 200 \(W\), in two cases, when firstly they are connected in parallel and secondly in series will be
1 \(\frac{5}{2}:\frac{5}{2}\)
2 \(\frac{5}{2}:\frac{2}{5}\)
3 \(\frac{2}{5}:\frac{2}{5}\)
4 \(\frac{2}{5}:\frac{5}{2}\)
Explanation:
In parallel, \(\frac{{{H_1}}}{{{H_2}}} = \frac{{{P_1}t}}{{{P_2}t}} = \frac{{{P_1}}}{{{P_2}}} = \frac{{500}}{{200}} = \frac{5}{2}\) In series \(\frac{{{H_1}}}{{{H_2}}} = \frac{{{I^2}{R_1}t}}{{{I^2}{R_2}t}} = \frac{{{R_1}}}{{{R_2}}} = \frac{{{V^2}/{P_1}}}{{{V^2}/{P_2}}} = \frac{{{P_2}}}{{{P_1}}} = \frac{{200}}{{500}} = \frac{2}{5}\)
PHXII03:CURRENT ELECTRICITY
357128
In a large building, there are \(12\) bulbs of \(40\;W\), \(4\) bulbs of \(100\;W,7\) fans of \(60\;W\) and \(1\) heater of \(1\;kW\). The voltage of the electric mains is \(230{\rm{ }}V\). The minimum capacity of the main fuse of the building will be
357124
Two resistors of resistances \(100\,\Omega \) and \(200\,\Omega \) are connected in parallel in an electrical circuit. The ratio of the thermal energy developed in \(100\,\Omega \) to that in \(200\,\Omega \) in a given time is
1 \(2:1\)
2 \(1:4\)
3 \(4:1\)
4 \(1:2\)
Explanation:
As both resistors are in parallel combination so potential drop (\(V\)) across both are same. \(P = \frac{{{V^2}}}{R} \Rightarrow P \propto \frac{1}{R}\) We know Power \(P = \frac{{{\rm{Energy}}}}{{{\rm{time}}}} = \frac{E}{t}\) Time(t) is same for both \({R_1},{R_2}\) \(\frac{{{E_1}}}{{{E_2}}} = \frac{{{P_1}}}{{{P_2}}} = \frac{{{R_2}}}{{{R_1}}} = \frac{{200}}{{100}} = \frac{2}{1}\) \( = 2:1\) Note: We are asked to find ratio of thermal energies
NEET - 2022
PHXII03:CURRENT ELECTRICITY
357125
There are \(25\;W - 220\;V\) bulb and a \(100\;W\) - \(220\;V\) line. Which electric bulb will glow more brightly?
1 \(25\;W\) bulb
2 \(100\;W\) bulb
3 Both will have equal incandescence
4 Neither \(25\;W\) nor \(100\;W\) bulb will give light
Explanation:
Since power is given by \(P=\dfrac{V^{2}}{R}\), \(\text { so } R=\dfrac{V^{2}}{P}\) For the first bulb, \(R_{1}=\left(\dfrac{V^{2}}{P_{1}}\right)=\left[\dfrac{(220)^{2}}{25}\right]=1936\, \Omega\) For the second bulb, \(R_{2}=\left(\dfrac{V^{2}}{P_{2}}\right)=\left[\dfrac{(220)^{2}}{100}\right]=484\, \Omega\) Current in series combination is the same in the two bulbs and current \(I\) is given by \(I = \frac{V}{{{R_1} + {R_2}}} = \frac{{220}}{{1936 + 484}} = \frac{{220}}{{2420}} = \frac{1}{{11}}\;A\) If the actual powers in the bulbs be \(P_{1}\) and \(P_{2}\) then \(P_1^\prime {\rm{ }} = {I^2}{R_1} = {\left( {\frac{1}{{11}}} \right)^2} \times 1936 = 16\;W\) and \(P_2^\prime {\rm{ }} = {I^2}{R_2} = {\left( {\frac{1}{{11}}} \right)^2} \times 484 = 4\;W\) Since \({P^\prime }_1 > {P^\prime }_2,25\;W\) bulb will glow more brightly.
PHXII03:CURRENT ELECTRICITY
357126
Assertion : Fuse wire must have high resistance and low melting point. Reason : Fuse is used for small current flow only.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Fuse wires need high resistance to generate more heat when current exceeds the limit, ensuring they melt quickly and break the circuit for protection. The low melting point facilitates rapid melting with a temperature increase. So correct option is (3).
PHXII03:CURRENT ELECTRICITY
357127
Two bulbs of 500 \(W\) and 200 \(W\) are manufactured to operate on 220 \(V\) line. The ratio of heat produced in 500 \(W\) and 200 \(W\), in two cases, when firstly they are connected in parallel and secondly in series will be
1 \(\frac{5}{2}:\frac{5}{2}\)
2 \(\frac{5}{2}:\frac{2}{5}\)
3 \(\frac{2}{5}:\frac{2}{5}\)
4 \(\frac{2}{5}:\frac{5}{2}\)
Explanation:
In parallel, \(\frac{{{H_1}}}{{{H_2}}} = \frac{{{P_1}t}}{{{P_2}t}} = \frac{{{P_1}}}{{{P_2}}} = \frac{{500}}{{200}} = \frac{5}{2}\) In series \(\frac{{{H_1}}}{{{H_2}}} = \frac{{{I^2}{R_1}t}}{{{I^2}{R_2}t}} = \frac{{{R_1}}}{{{R_2}}} = \frac{{{V^2}/{P_1}}}{{{V^2}/{P_2}}} = \frac{{{P_2}}}{{{P_1}}} = \frac{{200}}{{500}} = \frac{2}{5}\)
PHXII03:CURRENT ELECTRICITY
357128
In a large building, there are \(12\) bulbs of \(40\;W\), \(4\) bulbs of \(100\;W,7\) fans of \(60\;W\) and \(1\) heater of \(1\;kW\). The voltage of the electric mains is \(230{\rm{ }}V\). The minimum capacity of the main fuse of the building will be
357124
Two resistors of resistances \(100\,\Omega \) and \(200\,\Omega \) are connected in parallel in an electrical circuit. The ratio of the thermal energy developed in \(100\,\Omega \) to that in \(200\,\Omega \) in a given time is
1 \(2:1\)
2 \(1:4\)
3 \(4:1\)
4 \(1:2\)
Explanation:
As both resistors are in parallel combination so potential drop (\(V\)) across both are same. \(P = \frac{{{V^2}}}{R} \Rightarrow P \propto \frac{1}{R}\) We know Power \(P = \frac{{{\rm{Energy}}}}{{{\rm{time}}}} = \frac{E}{t}\) Time(t) is same for both \({R_1},{R_2}\) \(\frac{{{E_1}}}{{{E_2}}} = \frac{{{P_1}}}{{{P_2}}} = \frac{{{R_2}}}{{{R_1}}} = \frac{{200}}{{100}} = \frac{2}{1}\) \( = 2:1\) Note: We are asked to find ratio of thermal energies
NEET - 2022
PHXII03:CURRENT ELECTRICITY
357125
There are \(25\;W - 220\;V\) bulb and a \(100\;W\) - \(220\;V\) line. Which electric bulb will glow more brightly?
1 \(25\;W\) bulb
2 \(100\;W\) bulb
3 Both will have equal incandescence
4 Neither \(25\;W\) nor \(100\;W\) bulb will give light
Explanation:
Since power is given by \(P=\dfrac{V^{2}}{R}\), \(\text { so } R=\dfrac{V^{2}}{P}\) For the first bulb, \(R_{1}=\left(\dfrac{V^{2}}{P_{1}}\right)=\left[\dfrac{(220)^{2}}{25}\right]=1936\, \Omega\) For the second bulb, \(R_{2}=\left(\dfrac{V^{2}}{P_{2}}\right)=\left[\dfrac{(220)^{2}}{100}\right]=484\, \Omega\) Current in series combination is the same in the two bulbs and current \(I\) is given by \(I = \frac{V}{{{R_1} + {R_2}}} = \frac{{220}}{{1936 + 484}} = \frac{{220}}{{2420}} = \frac{1}{{11}}\;A\) If the actual powers in the bulbs be \(P_{1}\) and \(P_{2}\) then \(P_1^\prime {\rm{ }} = {I^2}{R_1} = {\left( {\frac{1}{{11}}} \right)^2} \times 1936 = 16\;W\) and \(P_2^\prime {\rm{ }} = {I^2}{R_2} = {\left( {\frac{1}{{11}}} \right)^2} \times 484 = 4\;W\) Since \({P^\prime }_1 > {P^\prime }_2,25\;W\) bulb will glow more brightly.
PHXII03:CURRENT ELECTRICITY
357126
Assertion : Fuse wire must have high resistance and low melting point. Reason : Fuse is used for small current flow only.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Fuse wires need high resistance to generate more heat when current exceeds the limit, ensuring they melt quickly and break the circuit for protection. The low melting point facilitates rapid melting with a temperature increase. So correct option is (3).
PHXII03:CURRENT ELECTRICITY
357127
Two bulbs of 500 \(W\) and 200 \(W\) are manufactured to operate on 220 \(V\) line. The ratio of heat produced in 500 \(W\) and 200 \(W\), in two cases, when firstly they are connected in parallel and secondly in series will be
1 \(\frac{5}{2}:\frac{5}{2}\)
2 \(\frac{5}{2}:\frac{2}{5}\)
3 \(\frac{2}{5}:\frac{2}{5}\)
4 \(\frac{2}{5}:\frac{5}{2}\)
Explanation:
In parallel, \(\frac{{{H_1}}}{{{H_2}}} = \frac{{{P_1}t}}{{{P_2}t}} = \frac{{{P_1}}}{{{P_2}}} = \frac{{500}}{{200}} = \frac{5}{2}\) In series \(\frac{{{H_1}}}{{{H_2}}} = \frac{{{I^2}{R_1}t}}{{{I^2}{R_2}t}} = \frac{{{R_1}}}{{{R_2}}} = \frac{{{V^2}/{P_1}}}{{{V^2}/{P_2}}} = \frac{{{P_2}}}{{{P_1}}} = \frac{{200}}{{500}} = \frac{2}{5}\)
PHXII03:CURRENT ELECTRICITY
357128
In a large building, there are \(12\) bulbs of \(40\;W\), \(4\) bulbs of \(100\;W,7\) fans of \(60\;W\) and \(1\) heater of \(1\;kW\). The voltage of the electric mains is \(230{\rm{ }}V\). The minimum capacity of the main fuse of the building will be
357124
Two resistors of resistances \(100\,\Omega \) and \(200\,\Omega \) are connected in parallel in an electrical circuit. The ratio of the thermal energy developed in \(100\,\Omega \) to that in \(200\,\Omega \) in a given time is
1 \(2:1\)
2 \(1:4\)
3 \(4:1\)
4 \(1:2\)
Explanation:
As both resistors are in parallel combination so potential drop (\(V\)) across both are same. \(P = \frac{{{V^2}}}{R} \Rightarrow P \propto \frac{1}{R}\) We know Power \(P = \frac{{{\rm{Energy}}}}{{{\rm{time}}}} = \frac{E}{t}\) Time(t) is same for both \({R_1},{R_2}\) \(\frac{{{E_1}}}{{{E_2}}} = \frac{{{P_1}}}{{{P_2}}} = \frac{{{R_2}}}{{{R_1}}} = \frac{{200}}{{100}} = \frac{2}{1}\) \( = 2:1\) Note: We are asked to find ratio of thermal energies
NEET - 2022
PHXII03:CURRENT ELECTRICITY
357125
There are \(25\;W - 220\;V\) bulb and a \(100\;W\) - \(220\;V\) line. Which electric bulb will glow more brightly?
1 \(25\;W\) bulb
2 \(100\;W\) bulb
3 Both will have equal incandescence
4 Neither \(25\;W\) nor \(100\;W\) bulb will give light
Explanation:
Since power is given by \(P=\dfrac{V^{2}}{R}\), \(\text { so } R=\dfrac{V^{2}}{P}\) For the first bulb, \(R_{1}=\left(\dfrac{V^{2}}{P_{1}}\right)=\left[\dfrac{(220)^{2}}{25}\right]=1936\, \Omega\) For the second bulb, \(R_{2}=\left(\dfrac{V^{2}}{P_{2}}\right)=\left[\dfrac{(220)^{2}}{100}\right]=484\, \Omega\) Current in series combination is the same in the two bulbs and current \(I\) is given by \(I = \frac{V}{{{R_1} + {R_2}}} = \frac{{220}}{{1936 + 484}} = \frac{{220}}{{2420}} = \frac{1}{{11}}\;A\) If the actual powers in the bulbs be \(P_{1}\) and \(P_{2}\) then \(P_1^\prime {\rm{ }} = {I^2}{R_1} = {\left( {\frac{1}{{11}}} \right)^2} \times 1936 = 16\;W\) and \(P_2^\prime {\rm{ }} = {I^2}{R_2} = {\left( {\frac{1}{{11}}} \right)^2} \times 484 = 4\;W\) Since \({P^\prime }_1 > {P^\prime }_2,25\;W\) bulb will glow more brightly.
PHXII03:CURRENT ELECTRICITY
357126
Assertion : Fuse wire must have high resistance and low melting point. Reason : Fuse is used for small current flow only.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Fuse wires need high resistance to generate more heat when current exceeds the limit, ensuring they melt quickly and break the circuit for protection. The low melting point facilitates rapid melting with a temperature increase. So correct option is (3).
PHXII03:CURRENT ELECTRICITY
357127
Two bulbs of 500 \(W\) and 200 \(W\) are manufactured to operate on 220 \(V\) line. The ratio of heat produced in 500 \(W\) and 200 \(W\), in two cases, when firstly they are connected in parallel and secondly in series will be
1 \(\frac{5}{2}:\frac{5}{2}\)
2 \(\frac{5}{2}:\frac{2}{5}\)
3 \(\frac{2}{5}:\frac{2}{5}\)
4 \(\frac{2}{5}:\frac{5}{2}\)
Explanation:
In parallel, \(\frac{{{H_1}}}{{{H_2}}} = \frac{{{P_1}t}}{{{P_2}t}} = \frac{{{P_1}}}{{{P_2}}} = \frac{{500}}{{200}} = \frac{5}{2}\) In series \(\frac{{{H_1}}}{{{H_2}}} = \frac{{{I^2}{R_1}t}}{{{I^2}{R_2}t}} = \frac{{{R_1}}}{{{R_2}}} = \frac{{{V^2}/{P_1}}}{{{V^2}/{P_2}}} = \frac{{{P_2}}}{{{P_1}}} = \frac{{200}}{{500}} = \frac{2}{5}\)
PHXII03:CURRENT ELECTRICITY
357128
In a large building, there are \(12\) bulbs of \(40\;W\), \(4\) bulbs of \(100\;W,7\) fans of \(60\;W\) and \(1\) heater of \(1\;kW\). The voltage of the electric mains is \(230{\rm{ }}V\). The minimum capacity of the main fuse of the building will be
357124
Two resistors of resistances \(100\,\Omega \) and \(200\,\Omega \) are connected in parallel in an electrical circuit. The ratio of the thermal energy developed in \(100\,\Omega \) to that in \(200\,\Omega \) in a given time is
1 \(2:1\)
2 \(1:4\)
3 \(4:1\)
4 \(1:2\)
Explanation:
As both resistors are in parallel combination so potential drop (\(V\)) across both are same. \(P = \frac{{{V^2}}}{R} \Rightarrow P \propto \frac{1}{R}\) We know Power \(P = \frac{{{\rm{Energy}}}}{{{\rm{time}}}} = \frac{E}{t}\) Time(t) is same for both \({R_1},{R_2}\) \(\frac{{{E_1}}}{{{E_2}}} = \frac{{{P_1}}}{{{P_2}}} = \frac{{{R_2}}}{{{R_1}}} = \frac{{200}}{{100}} = \frac{2}{1}\) \( = 2:1\) Note: We are asked to find ratio of thermal energies
NEET - 2022
PHXII03:CURRENT ELECTRICITY
357125
There are \(25\;W - 220\;V\) bulb and a \(100\;W\) - \(220\;V\) line. Which electric bulb will glow more brightly?
1 \(25\;W\) bulb
2 \(100\;W\) bulb
3 Both will have equal incandescence
4 Neither \(25\;W\) nor \(100\;W\) bulb will give light
Explanation:
Since power is given by \(P=\dfrac{V^{2}}{R}\), \(\text { so } R=\dfrac{V^{2}}{P}\) For the first bulb, \(R_{1}=\left(\dfrac{V^{2}}{P_{1}}\right)=\left[\dfrac{(220)^{2}}{25}\right]=1936\, \Omega\) For the second bulb, \(R_{2}=\left(\dfrac{V^{2}}{P_{2}}\right)=\left[\dfrac{(220)^{2}}{100}\right]=484\, \Omega\) Current in series combination is the same in the two bulbs and current \(I\) is given by \(I = \frac{V}{{{R_1} + {R_2}}} = \frac{{220}}{{1936 + 484}} = \frac{{220}}{{2420}} = \frac{1}{{11}}\;A\) If the actual powers in the bulbs be \(P_{1}\) and \(P_{2}\) then \(P_1^\prime {\rm{ }} = {I^2}{R_1} = {\left( {\frac{1}{{11}}} \right)^2} \times 1936 = 16\;W\) and \(P_2^\prime {\rm{ }} = {I^2}{R_2} = {\left( {\frac{1}{{11}}} \right)^2} \times 484 = 4\;W\) Since \({P^\prime }_1 > {P^\prime }_2,25\;W\) bulb will glow more brightly.
PHXII03:CURRENT ELECTRICITY
357126
Assertion : Fuse wire must have high resistance and low melting point. Reason : Fuse is used for small current flow only.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Fuse wires need high resistance to generate more heat when current exceeds the limit, ensuring they melt quickly and break the circuit for protection. The low melting point facilitates rapid melting with a temperature increase. So correct option is (3).
PHXII03:CURRENT ELECTRICITY
357127
Two bulbs of 500 \(W\) and 200 \(W\) are manufactured to operate on 220 \(V\) line. The ratio of heat produced in 500 \(W\) and 200 \(W\), in two cases, when firstly they are connected in parallel and secondly in series will be
1 \(\frac{5}{2}:\frac{5}{2}\)
2 \(\frac{5}{2}:\frac{2}{5}\)
3 \(\frac{2}{5}:\frac{2}{5}\)
4 \(\frac{2}{5}:\frac{5}{2}\)
Explanation:
In parallel, \(\frac{{{H_1}}}{{{H_2}}} = \frac{{{P_1}t}}{{{P_2}t}} = \frac{{{P_1}}}{{{P_2}}} = \frac{{500}}{{200}} = \frac{5}{2}\) In series \(\frac{{{H_1}}}{{{H_2}}} = \frac{{{I^2}{R_1}t}}{{{I^2}{R_2}t}} = \frac{{{R_1}}}{{{R_2}}} = \frac{{{V^2}/{P_1}}}{{{V^2}/{P_2}}} = \frac{{{P_2}}}{{{P_1}}} = \frac{{200}}{{500}} = \frac{2}{5}\)
PHXII03:CURRENT ELECTRICITY
357128
In a large building, there are \(12\) bulbs of \(40\;W\), \(4\) bulbs of \(100\;W,7\) fans of \(60\;W\) and \(1\) heater of \(1\;kW\). The voltage of the electric mains is \(230{\rm{ }}V\). The minimum capacity of the main fuse of the building will be
