357172
If voltage across a bulb rated \(200\;V,\,\,100\;W\) drops by \(2.5 \%\) its rated value, the percentage of the rated value, by which the power would decrease is
1 \(20 \%\)
2 \(2.5 \%\)
3 \(5 \%\)
4 \(10 \%\)
Explanation:
Power: \(P=\dfrac{V^{2}}{R}\) For small variation in the value of power after differentiation: \(\dfrac{\Delta P}{P} \times 100=\dfrac{2 \times \Delta V}{V} \times 100 \%=2 \times 2.5=5 \%\) \(\therefore\) Power decreased by \(5 \%\).
PHXII03:CURRENT ELECTRICITY
357173
Which of the following acts as a circuit protection device?
1 Conductor
2 Inductor
3 Switch
4 Fuse
Explanation:
Fuse wire has less melting point so when excess current flows, it melts
NEET - 2019
PHXII03:CURRENT ELECTRICITY
357174
In this circuit, when certain current flows, the heat produced in \(5\Omega \) is 4.05 \(J\) in a time \(t\). The heat produced in \(2\Omega \) coil in the same time interval is
1 \(5.76\,J\)
2 \(1.44J\)
3 \(2.88J\)
4 \(2.02J\)
Explanation:
Current through the \(5\,\Omega \) resistor: \({I_1} = \frac{{(6 + 9)}}{{(6 + 9) + 5}}I = \frac{3}{4}I\) Ratio of heat generated at \(2\,\Omega \) and \(5\,\Omega \) resistors (\(Q\) and \({Q_1}\) respectively) at same time \(t\): \(\frac{Q}{{{Q_1}}} = \frac{{{I^2} \cdot 2 \cdot t}}{{I_1^2 \cdot 5 \cdot t}} = {\left( {\frac{4}{3}} \right)^2} \times \frac{2}{5} = \frac{{32}}{{45}}\) Given \({Q_1} = 4.05\,J,\) \(Q = \frac{{32}}{{45}} \times 4.05 = 2.88\,\,J\)
KCET - 2012
PHXII03:CURRENT ELECTRICITY
357175
Two lamps, each with a resistance of \(50\,\Omega \), are connected in series. The lamps will fuse if a power of more than 200 \(W\) is dissipated in it. What is the maximum voltage that can be applied to the circuit ?
1 \(140V\)
2 \(100V\)
3 \(200V\)
4 None of these
Explanation:
Let \({V_0}\) be the maximum voltage The current through a resistor is \(i = \frac{{{V_0}}}{{100}}\) The power through a resistor is \(P = {i^2}R\) \(200 = {\left( {\frac{{{V_0}}}{{100}}} \right)^2}50 \Rightarrow {V_0} = 200V\)
357172
If voltage across a bulb rated \(200\;V,\,\,100\;W\) drops by \(2.5 \%\) its rated value, the percentage of the rated value, by which the power would decrease is
1 \(20 \%\)
2 \(2.5 \%\)
3 \(5 \%\)
4 \(10 \%\)
Explanation:
Power: \(P=\dfrac{V^{2}}{R}\) For small variation in the value of power after differentiation: \(\dfrac{\Delta P}{P} \times 100=\dfrac{2 \times \Delta V}{V} \times 100 \%=2 \times 2.5=5 \%\) \(\therefore\) Power decreased by \(5 \%\).
PHXII03:CURRENT ELECTRICITY
357173
Which of the following acts as a circuit protection device?
1 Conductor
2 Inductor
3 Switch
4 Fuse
Explanation:
Fuse wire has less melting point so when excess current flows, it melts
NEET - 2019
PHXII03:CURRENT ELECTRICITY
357174
In this circuit, when certain current flows, the heat produced in \(5\Omega \) is 4.05 \(J\) in a time \(t\). The heat produced in \(2\Omega \) coil in the same time interval is
1 \(5.76\,J\)
2 \(1.44J\)
3 \(2.88J\)
4 \(2.02J\)
Explanation:
Current through the \(5\,\Omega \) resistor: \({I_1} = \frac{{(6 + 9)}}{{(6 + 9) + 5}}I = \frac{3}{4}I\) Ratio of heat generated at \(2\,\Omega \) and \(5\,\Omega \) resistors (\(Q\) and \({Q_1}\) respectively) at same time \(t\): \(\frac{Q}{{{Q_1}}} = \frac{{{I^2} \cdot 2 \cdot t}}{{I_1^2 \cdot 5 \cdot t}} = {\left( {\frac{4}{3}} \right)^2} \times \frac{2}{5} = \frac{{32}}{{45}}\) Given \({Q_1} = 4.05\,J,\) \(Q = \frac{{32}}{{45}} \times 4.05 = 2.88\,\,J\)
KCET - 2012
PHXII03:CURRENT ELECTRICITY
357175
Two lamps, each with a resistance of \(50\,\Omega \), are connected in series. The lamps will fuse if a power of more than 200 \(W\) is dissipated in it. What is the maximum voltage that can be applied to the circuit ?
1 \(140V\)
2 \(100V\)
3 \(200V\)
4 None of these
Explanation:
Let \({V_0}\) be the maximum voltage The current through a resistor is \(i = \frac{{{V_0}}}{{100}}\) The power through a resistor is \(P = {i^2}R\) \(200 = {\left( {\frac{{{V_0}}}{{100}}} \right)^2}50 \Rightarrow {V_0} = 200V\)
357172
If voltage across a bulb rated \(200\;V,\,\,100\;W\) drops by \(2.5 \%\) its rated value, the percentage of the rated value, by which the power would decrease is
1 \(20 \%\)
2 \(2.5 \%\)
3 \(5 \%\)
4 \(10 \%\)
Explanation:
Power: \(P=\dfrac{V^{2}}{R}\) For small variation in the value of power after differentiation: \(\dfrac{\Delta P}{P} \times 100=\dfrac{2 \times \Delta V}{V} \times 100 \%=2 \times 2.5=5 \%\) \(\therefore\) Power decreased by \(5 \%\).
PHXII03:CURRENT ELECTRICITY
357173
Which of the following acts as a circuit protection device?
1 Conductor
2 Inductor
3 Switch
4 Fuse
Explanation:
Fuse wire has less melting point so when excess current flows, it melts
NEET - 2019
PHXII03:CURRENT ELECTRICITY
357174
In this circuit, when certain current flows, the heat produced in \(5\Omega \) is 4.05 \(J\) in a time \(t\). The heat produced in \(2\Omega \) coil in the same time interval is
1 \(5.76\,J\)
2 \(1.44J\)
3 \(2.88J\)
4 \(2.02J\)
Explanation:
Current through the \(5\,\Omega \) resistor: \({I_1} = \frac{{(6 + 9)}}{{(6 + 9) + 5}}I = \frac{3}{4}I\) Ratio of heat generated at \(2\,\Omega \) and \(5\,\Omega \) resistors (\(Q\) and \({Q_1}\) respectively) at same time \(t\): \(\frac{Q}{{{Q_1}}} = \frac{{{I^2} \cdot 2 \cdot t}}{{I_1^2 \cdot 5 \cdot t}} = {\left( {\frac{4}{3}} \right)^2} \times \frac{2}{5} = \frac{{32}}{{45}}\) Given \({Q_1} = 4.05\,J,\) \(Q = \frac{{32}}{{45}} \times 4.05 = 2.88\,\,J\)
KCET - 2012
PHXII03:CURRENT ELECTRICITY
357175
Two lamps, each with a resistance of \(50\,\Omega \), are connected in series. The lamps will fuse if a power of more than 200 \(W\) is dissipated in it. What is the maximum voltage that can be applied to the circuit ?
1 \(140V\)
2 \(100V\)
3 \(200V\)
4 None of these
Explanation:
Let \({V_0}\) be the maximum voltage The current through a resistor is \(i = \frac{{{V_0}}}{{100}}\) The power through a resistor is \(P = {i^2}R\) \(200 = {\left( {\frac{{{V_0}}}{{100}}} \right)^2}50 \Rightarrow {V_0} = 200V\)
357172
If voltage across a bulb rated \(200\;V,\,\,100\;W\) drops by \(2.5 \%\) its rated value, the percentage of the rated value, by which the power would decrease is
1 \(20 \%\)
2 \(2.5 \%\)
3 \(5 \%\)
4 \(10 \%\)
Explanation:
Power: \(P=\dfrac{V^{2}}{R}\) For small variation in the value of power after differentiation: \(\dfrac{\Delta P}{P} \times 100=\dfrac{2 \times \Delta V}{V} \times 100 \%=2 \times 2.5=5 \%\) \(\therefore\) Power decreased by \(5 \%\).
PHXII03:CURRENT ELECTRICITY
357173
Which of the following acts as a circuit protection device?
1 Conductor
2 Inductor
3 Switch
4 Fuse
Explanation:
Fuse wire has less melting point so when excess current flows, it melts
NEET - 2019
PHXII03:CURRENT ELECTRICITY
357174
In this circuit, when certain current flows, the heat produced in \(5\Omega \) is 4.05 \(J\) in a time \(t\). The heat produced in \(2\Omega \) coil in the same time interval is
1 \(5.76\,J\)
2 \(1.44J\)
3 \(2.88J\)
4 \(2.02J\)
Explanation:
Current through the \(5\,\Omega \) resistor: \({I_1} = \frac{{(6 + 9)}}{{(6 + 9) + 5}}I = \frac{3}{4}I\) Ratio of heat generated at \(2\,\Omega \) and \(5\,\Omega \) resistors (\(Q\) and \({Q_1}\) respectively) at same time \(t\): \(\frac{Q}{{{Q_1}}} = \frac{{{I^2} \cdot 2 \cdot t}}{{I_1^2 \cdot 5 \cdot t}} = {\left( {\frac{4}{3}} \right)^2} \times \frac{2}{5} = \frac{{32}}{{45}}\) Given \({Q_1} = 4.05\,J,\) \(Q = \frac{{32}}{{45}} \times 4.05 = 2.88\,\,J\)
KCET - 2012
PHXII03:CURRENT ELECTRICITY
357175
Two lamps, each with a resistance of \(50\,\Omega \), are connected in series. The lamps will fuse if a power of more than 200 \(W\) is dissipated in it. What is the maximum voltage that can be applied to the circuit ?
1 \(140V\)
2 \(100V\)
3 \(200V\)
4 None of these
Explanation:
Let \({V_0}\) be the maximum voltage The current through a resistor is \(i = \frac{{{V_0}}}{{100}}\) The power through a resistor is \(P = {i^2}R\) \(200 = {\left( {\frac{{{V_0}}}{{100}}} \right)^2}50 \Rightarrow {V_0} = 200V\)
357172
If voltage across a bulb rated \(200\;V,\,\,100\;W\) drops by \(2.5 \%\) its rated value, the percentage of the rated value, by which the power would decrease is
1 \(20 \%\)
2 \(2.5 \%\)
3 \(5 \%\)
4 \(10 \%\)
Explanation:
Power: \(P=\dfrac{V^{2}}{R}\) For small variation in the value of power after differentiation: \(\dfrac{\Delta P}{P} \times 100=\dfrac{2 \times \Delta V}{V} \times 100 \%=2 \times 2.5=5 \%\) \(\therefore\) Power decreased by \(5 \%\).
PHXII03:CURRENT ELECTRICITY
357173
Which of the following acts as a circuit protection device?
1 Conductor
2 Inductor
3 Switch
4 Fuse
Explanation:
Fuse wire has less melting point so when excess current flows, it melts
NEET - 2019
PHXII03:CURRENT ELECTRICITY
357174
In this circuit, when certain current flows, the heat produced in \(5\Omega \) is 4.05 \(J\) in a time \(t\). The heat produced in \(2\Omega \) coil in the same time interval is
1 \(5.76\,J\)
2 \(1.44J\)
3 \(2.88J\)
4 \(2.02J\)
Explanation:
Current through the \(5\,\Omega \) resistor: \({I_1} = \frac{{(6 + 9)}}{{(6 + 9) + 5}}I = \frac{3}{4}I\) Ratio of heat generated at \(2\,\Omega \) and \(5\,\Omega \) resistors (\(Q\) and \({Q_1}\) respectively) at same time \(t\): \(\frac{Q}{{{Q_1}}} = \frac{{{I^2} \cdot 2 \cdot t}}{{I_1^2 \cdot 5 \cdot t}} = {\left( {\frac{4}{3}} \right)^2} \times \frac{2}{5} = \frac{{32}}{{45}}\) Given \({Q_1} = 4.05\,J,\) \(Q = \frac{{32}}{{45}} \times 4.05 = 2.88\,\,J\)
KCET - 2012
PHXII03:CURRENT ELECTRICITY
357175
Two lamps, each with a resistance of \(50\,\Omega \), are connected in series. The lamps will fuse if a power of more than 200 \(W\) is dissipated in it. What is the maximum voltage that can be applied to the circuit ?
1 \(140V\)
2 \(100V\)
3 \(200V\)
4 None of these
Explanation:
Let \({V_0}\) be the maximum voltage The current through a resistor is \(i = \frac{{{V_0}}}{{100}}\) The power through a resistor is \(P = {i^2}R\) \(200 = {\left( {\frac{{{V_0}}}{{100}}} \right)^2}50 \Rightarrow {V_0} = 200V\)
