356282
A transformer of efficiency \(90 \%\) draws an input power of \(4\;kW\). An electrical appliances connected across the secondary draws a current of \(6\;A\). The impedance of the device is:
1 \(60\,\Omega \)
2 \(50\,\Omega \)
3 \(80\,\Omega \)
4 \(100\,\Omega \)
Explanation:
Given that: Input power \( = 4\,kW\): Efficiency is \(90 \%\) \(\Rightarrow\) Output power \( = 0.9 \times 4\;kW = 3.6\;kW\) \(3.6\;kW = {I^2}R = 36 \times R\) \(R = \frac{{3.6\;kW}}{{36}} = \frac{{3.6 \times {{10}^3}}}{{36}} = {10^2}\Omega = 100\,\Omega \) So, correct option is (4).
PHXII07:ALTERNATING CURRENT
356283
A transformer having efficiency of 90% is working on 200 \(V\) and 3 \(KW\) power supply. The current in the secondary coil is 6 \(A\) and voltage in the primary coil is 200 \(V\). The voltaege in the secondary coil and current in the primary coil respectively are
1 \(600\,V,15\,A\)
2 \(450\,V,13.5\,A\)
3 \(450\,V,15\,A\)
4 \(300\,V,15\,A\)
Explanation:
Intial power \( = 3000{\kern 1pt} W\) As efficiency is 90%, then final power \( = 3000 \times \frac{{90}}{{100}} = 2700{\kern 1pt} W\) \( \Rightarrow \left. \begin{array}{l} \quad {V_1}{I_1} = 3000W\\ \quad {V_2}{I_2} = 2700W \end{array} \right]\) \({V_2} = \frac{{2700}}{6} = \frac{{900}}{2}\) \([\because {I_2} = 6A]\) \( \Rightarrow \quad {V_2} = 450V\) and \({I_1} = \frac{{3000}}{{200}}\) \(\left[ {\because {V_1} = 200V} \right]\) \( \Rightarrow \quad {I_1} = 15A\)
PHXII07:ALTERNATING CURRENT
356284
A current of 5\(A\) is flowing at 220 \(V\) in the primary coil of a transformer. If the voltage produced in the secondary coil is 2200 \(V\) and 50% of power is lost, then the current in the secondary will be
1 \(0.25\,A\)
2 \(0.5\,A\)
3 \(2.5\,A\)
4 \(5\,A\)
Explanation:
Given the power output is 50% of the input power, \({i_s}{V_s} = (1/2){i_p}{V_p}\) Also given \({i_p} = 5A,{V_p} = 220V\) and \({V_s} = 2200V\) \({i_S} = \frac{1}{2}\frac{{{i_P}{V_P}}}{{{V_S}}} \Rightarrow {i_S} = \frac{1}{2} \times \frac{{5A \times 220V}}{{2200\,V}}\) \({i_S} = 0.25\,A\)
KCET - 2008
PHXII07:ALTERNATING CURRENT
356285
A 220 \(V\) input is supplied to a transformer. The output circuit draws a current of 2.0 \(A\) at 440 \(V\). If the efficiency of the transformer is 80%, the current drawn by the primary windings of the transformer is
1 \(2.5\,A\)
2 \(3.6\,A\)
3 \(2.8\,A\)
4 \(5.0\,A\)
Explanation:
Efficiency is defined as the ratio of output power and input power i.e., \(\eta \% = \frac{{{P_{out}}}}{{{P_{input}}}} \times 100 = \frac{{{V_s}{i_s}}}{{{V_p}{i_p}}} \times 100\) \(80 = \frac{{2 \times 440}}{{220 \times {i_p}}} \times 100 \Rightarrow \quad {i_p} = 5A\)
PHXII07:ALTERNATING CURRENT
356286
A step-up transformer works on \(220\;V\) and gives \(2\;A\) to an external resistor. The turn ratio between the primary and secondary coils is 2:25. Assuming 100% efficiency, find the secondary voltage. Primary current and power delivered.
356282
A transformer of efficiency \(90 \%\) draws an input power of \(4\;kW\). An electrical appliances connected across the secondary draws a current of \(6\;A\). The impedance of the device is:
1 \(60\,\Omega \)
2 \(50\,\Omega \)
3 \(80\,\Omega \)
4 \(100\,\Omega \)
Explanation:
Given that: Input power \( = 4\,kW\): Efficiency is \(90 \%\) \(\Rightarrow\) Output power \( = 0.9 \times 4\;kW = 3.6\;kW\) \(3.6\;kW = {I^2}R = 36 \times R\) \(R = \frac{{3.6\;kW}}{{36}} = \frac{{3.6 \times {{10}^3}}}{{36}} = {10^2}\Omega = 100\,\Omega \) So, correct option is (4).
PHXII07:ALTERNATING CURRENT
356283
A transformer having efficiency of 90% is working on 200 \(V\) and 3 \(KW\) power supply. The current in the secondary coil is 6 \(A\) and voltage in the primary coil is 200 \(V\). The voltaege in the secondary coil and current in the primary coil respectively are
1 \(600\,V,15\,A\)
2 \(450\,V,13.5\,A\)
3 \(450\,V,15\,A\)
4 \(300\,V,15\,A\)
Explanation:
Intial power \( = 3000{\kern 1pt} W\) As efficiency is 90%, then final power \( = 3000 \times \frac{{90}}{{100}} = 2700{\kern 1pt} W\) \( \Rightarrow \left. \begin{array}{l} \quad {V_1}{I_1} = 3000W\\ \quad {V_2}{I_2} = 2700W \end{array} \right]\) \({V_2} = \frac{{2700}}{6} = \frac{{900}}{2}\) \([\because {I_2} = 6A]\) \( \Rightarrow \quad {V_2} = 450V\) and \({I_1} = \frac{{3000}}{{200}}\) \(\left[ {\because {V_1} = 200V} \right]\) \( \Rightarrow \quad {I_1} = 15A\)
PHXII07:ALTERNATING CURRENT
356284
A current of 5\(A\) is flowing at 220 \(V\) in the primary coil of a transformer. If the voltage produced in the secondary coil is 2200 \(V\) and 50% of power is lost, then the current in the secondary will be
1 \(0.25\,A\)
2 \(0.5\,A\)
3 \(2.5\,A\)
4 \(5\,A\)
Explanation:
Given the power output is 50% of the input power, \({i_s}{V_s} = (1/2){i_p}{V_p}\) Also given \({i_p} = 5A,{V_p} = 220V\) and \({V_s} = 2200V\) \({i_S} = \frac{1}{2}\frac{{{i_P}{V_P}}}{{{V_S}}} \Rightarrow {i_S} = \frac{1}{2} \times \frac{{5A \times 220V}}{{2200\,V}}\) \({i_S} = 0.25\,A\)
KCET - 2008
PHXII07:ALTERNATING CURRENT
356285
A 220 \(V\) input is supplied to a transformer. The output circuit draws a current of 2.0 \(A\) at 440 \(V\). If the efficiency of the transformer is 80%, the current drawn by the primary windings of the transformer is
1 \(2.5\,A\)
2 \(3.6\,A\)
3 \(2.8\,A\)
4 \(5.0\,A\)
Explanation:
Efficiency is defined as the ratio of output power and input power i.e., \(\eta \% = \frac{{{P_{out}}}}{{{P_{input}}}} \times 100 = \frac{{{V_s}{i_s}}}{{{V_p}{i_p}}} \times 100\) \(80 = \frac{{2 \times 440}}{{220 \times {i_p}}} \times 100 \Rightarrow \quad {i_p} = 5A\)
PHXII07:ALTERNATING CURRENT
356286
A step-up transformer works on \(220\;V\) and gives \(2\;A\) to an external resistor. The turn ratio between the primary and secondary coils is 2:25. Assuming 100% efficiency, find the secondary voltage. Primary current and power delivered.
NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXII07:ALTERNATING CURRENT
356282
A transformer of efficiency \(90 \%\) draws an input power of \(4\;kW\). An electrical appliances connected across the secondary draws a current of \(6\;A\). The impedance of the device is:
1 \(60\,\Omega \)
2 \(50\,\Omega \)
3 \(80\,\Omega \)
4 \(100\,\Omega \)
Explanation:
Given that: Input power \( = 4\,kW\): Efficiency is \(90 \%\) \(\Rightarrow\) Output power \( = 0.9 \times 4\;kW = 3.6\;kW\) \(3.6\;kW = {I^2}R = 36 \times R\) \(R = \frac{{3.6\;kW}}{{36}} = \frac{{3.6 \times {{10}^3}}}{{36}} = {10^2}\Omega = 100\,\Omega \) So, correct option is (4).
PHXII07:ALTERNATING CURRENT
356283
A transformer having efficiency of 90% is working on 200 \(V\) and 3 \(KW\) power supply. The current in the secondary coil is 6 \(A\) and voltage in the primary coil is 200 \(V\). The voltaege in the secondary coil and current in the primary coil respectively are
1 \(600\,V,15\,A\)
2 \(450\,V,13.5\,A\)
3 \(450\,V,15\,A\)
4 \(300\,V,15\,A\)
Explanation:
Intial power \( = 3000{\kern 1pt} W\) As efficiency is 90%, then final power \( = 3000 \times \frac{{90}}{{100}} = 2700{\kern 1pt} W\) \( \Rightarrow \left. \begin{array}{l} \quad {V_1}{I_1} = 3000W\\ \quad {V_2}{I_2} = 2700W \end{array} \right]\) \({V_2} = \frac{{2700}}{6} = \frac{{900}}{2}\) \([\because {I_2} = 6A]\) \( \Rightarrow \quad {V_2} = 450V\) and \({I_1} = \frac{{3000}}{{200}}\) \(\left[ {\because {V_1} = 200V} \right]\) \( \Rightarrow \quad {I_1} = 15A\)
PHXII07:ALTERNATING CURRENT
356284
A current of 5\(A\) is flowing at 220 \(V\) in the primary coil of a transformer. If the voltage produced in the secondary coil is 2200 \(V\) and 50% of power is lost, then the current in the secondary will be
1 \(0.25\,A\)
2 \(0.5\,A\)
3 \(2.5\,A\)
4 \(5\,A\)
Explanation:
Given the power output is 50% of the input power, \({i_s}{V_s} = (1/2){i_p}{V_p}\) Also given \({i_p} = 5A,{V_p} = 220V\) and \({V_s} = 2200V\) \({i_S} = \frac{1}{2}\frac{{{i_P}{V_P}}}{{{V_S}}} \Rightarrow {i_S} = \frac{1}{2} \times \frac{{5A \times 220V}}{{2200\,V}}\) \({i_S} = 0.25\,A\)
KCET - 2008
PHXII07:ALTERNATING CURRENT
356285
A 220 \(V\) input is supplied to a transformer. The output circuit draws a current of 2.0 \(A\) at 440 \(V\). If the efficiency of the transformer is 80%, the current drawn by the primary windings of the transformer is
1 \(2.5\,A\)
2 \(3.6\,A\)
3 \(2.8\,A\)
4 \(5.0\,A\)
Explanation:
Efficiency is defined as the ratio of output power and input power i.e., \(\eta \% = \frac{{{P_{out}}}}{{{P_{input}}}} \times 100 = \frac{{{V_s}{i_s}}}{{{V_p}{i_p}}} \times 100\) \(80 = \frac{{2 \times 440}}{{220 \times {i_p}}} \times 100 \Rightarrow \quad {i_p} = 5A\)
PHXII07:ALTERNATING CURRENT
356286
A step-up transformer works on \(220\;V\) and gives \(2\;A\) to an external resistor. The turn ratio between the primary and secondary coils is 2:25. Assuming 100% efficiency, find the secondary voltage. Primary current and power delivered.
356282
A transformer of efficiency \(90 \%\) draws an input power of \(4\;kW\). An electrical appliances connected across the secondary draws a current of \(6\;A\). The impedance of the device is:
1 \(60\,\Omega \)
2 \(50\,\Omega \)
3 \(80\,\Omega \)
4 \(100\,\Omega \)
Explanation:
Given that: Input power \( = 4\,kW\): Efficiency is \(90 \%\) \(\Rightarrow\) Output power \( = 0.9 \times 4\;kW = 3.6\;kW\) \(3.6\;kW = {I^2}R = 36 \times R\) \(R = \frac{{3.6\;kW}}{{36}} = \frac{{3.6 \times {{10}^3}}}{{36}} = {10^2}\Omega = 100\,\Omega \) So, correct option is (4).
PHXII07:ALTERNATING CURRENT
356283
A transformer having efficiency of 90% is working on 200 \(V\) and 3 \(KW\) power supply. The current in the secondary coil is 6 \(A\) and voltage in the primary coil is 200 \(V\). The voltaege in the secondary coil and current in the primary coil respectively are
1 \(600\,V,15\,A\)
2 \(450\,V,13.5\,A\)
3 \(450\,V,15\,A\)
4 \(300\,V,15\,A\)
Explanation:
Intial power \( = 3000{\kern 1pt} W\) As efficiency is 90%, then final power \( = 3000 \times \frac{{90}}{{100}} = 2700{\kern 1pt} W\) \( \Rightarrow \left. \begin{array}{l} \quad {V_1}{I_1} = 3000W\\ \quad {V_2}{I_2} = 2700W \end{array} \right]\) \({V_2} = \frac{{2700}}{6} = \frac{{900}}{2}\) \([\because {I_2} = 6A]\) \( \Rightarrow \quad {V_2} = 450V\) and \({I_1} = \frac{{3000}}{{200}}\) \(\left[ {\because {V_1} = 200V} \right]\) \( \Rightarrow \quad {I_1} = 15A\)
PHXII07:ALTERNATING CURRENT
356284
A current of 5\(A\) is flowing at 220 \(V\) in the primary coil of a transformer. If the voltage produced in the secondary coil is 2200 \(V\) and 50% of power is lost, then the current in the secondary will be
1 \(0.25\,A\)
2 \(0.5\,A\)
3 \(2.5\,A\)
4 \(5\,A\)
Explanation:
Given the power output is 50% of the input power, \({i_s}{V_s} = (1/2){i_p}{V_p}\) Also given \({i_p} = 5A,{V_p} = 220V\) and \({V_s} = 2200V\) \({i_S} = \frac{1}{2}\frac{{{i_P}{V_P}}}{{{V_S}}} \Rightarrow {i_S} = \frac{1}{2} \times \frac{{5A \times 220V}}{{2200\,V}}\) \({i_S} = 0.25\,A\)
KCET - 2008
PHXII07:ALTERNATING CURRENT
356285
A 220 \(V\) input is supplied to a transformer. The output circuit draws a current of 2.0 \(A\) at 440 \(V\). If the efficiency of the transformer is 80%, the current drawn by the primary windings of the transformer is
1 \(2.5\,A\)
2 \(3.6\,A\)
3 \(2.8\,A\)
4 \(5.0\,A\)
Explanation:
Efficiency is defined as the ratio of output power and input power i.e., \(\eta \% = \frac{{{P_{out}}}}{{{P_{input}}}} \times 100 = \frac{{{V_s}{i_s}}}{{{V_p}{i_p}}} \times 100\) \(80 = \frac{{2 \times 440}}{{220 \times {i_p}}} \times 100 \Rightarrow \quad {i_p} = 5A\)
PHXII07:ALTERNATING CURRENT
356286
A step-up transformer works on \(220\;V\) and gives \(2\;A\) to an external resistor. The turn ratio between the primary and secondary coils is 2:25. Assuming 100% efficiency, find the secondary voltage. Primary current and power delivered.
356282
A transformer of efficiency \(90 \%\) draws an input power of \(4\;kW\). An electrical appliances connected across the secondary draws a current of \(6\;A\). The impedance of the device is:
1 \(60\,\Omega \)
2 \(50\,\Omega \)
3 \(80\,\Omega \)
4 \(100\,\Omega \)
Explanation:
Given that: Input power \( = 4\,kW\): Efficiency is \(90 \%\) \(\Rightarrow\) Output power \( = 0.9 \times 4\;kW = 3.6\;kW\) \(3.6\;kW = {I^2}R = 36 \times R\) \(R = \frac{{3.6\;kW}}{{36}} = \frac{{3.6 \times {{10}^3}}}{{36}} = {10^2}\Omega = 100\,\Omega \) So, correct option is (4).
PHXII07:ALTERNATING CURRENT
356283
A transformer having efficiency of 90% is working on 200 \(V\) and 3 \(KW\) power supply. The current in the secondary coil is 6 \(A\) and voltage in the primary coil is 200 \(V\). The voltaege in the secondary coil and current in the primary coil respectively are
1 \(600\,V,15\,A\)
2 \(450\,V,13.5\,A\)
3 \(450\,V,15\,A\)
4 \(300\,V,15\,A\)
Explanation:
Intial power \( = 3000{\kern 1pt} W\) As efficiency is 90%, then final power \( = 3000 \times \frac{{90}}{{100}} = 2700{\kern 1pt} W\) \( \Rightarrow \left. \begin{array}{l} \quad {V_1}{I_1} = 3000W\\ \quad {V_2}{I_2} = 2700W \end{array} \right]\) \({V_2} = \frac{{2700}}{6} = \frac{{900}}{2}\) \([\because {I_2} = 6A]\) \( \Rightarrow \quad {V_2} = 450V\) and \({I_1} = \frac{{3000}}{{200}}\) \(\left[ {\because {V_1} = 200V} \right]\) \( \Rightarrow \quad {I_1} = 15A\)
PHXII07:ALTERNATING CURRENT
356284
A current of 5\(A\) is flowing at 220 \(V\) in the primary coil of a transformer. If the voltage produced in the secondary coil is 2200 \(V\) and 50% of power is lost, then the current in the secondary will be
1 \(0.25\,A\)
2 \(0.5\,A\)
3 \(2.5\,A\)
4 \(5\,A\)
Explanation:
Given the power output is 50% of the input power, \({i_s}{V_s} = (1/2){i_p}{V_p}\) Also given \({i_p} = 5A,{V_p} = 220V\) and \({V_s} = 2200V\) \({i_S} = \frac{1}{2}\frac{{{i_P}{V_P}}}{{{V_S}}} \Rightarrow {i_S} = \frac{1}{2} \times \frac{{5A \times 220V}}{{2200\,V}}\) \({i_S} = 0.25\,A\)
KCET - 2008
PHXII07:ALTERNATING CURRENT
356285
A 220 \(V\) input is supplied to a transformer. The output circuit draws a current of 2.0 \(A\) at 440 \(V\). If the efficiency of the transformer is 80%, the current drawn by the primary windings of the transformer is
1 \(2.5\,A\)
2 \(3.6\,A\)
3 \(2.8\,A\)
4 \(5.0\,A\)
Explanation:
Efficiency is defined as the ratio of output power and input power i.e., \(\eta \% = \frac{{{P_{out}}}}{{{P_{input}}}} \times 100 = \frac{{{V_s}{i_s}}}{{{V_p}{i_p}}} \times 100\) \(80 = \frac{{2 \times 440}}{{220 \times {i_p}}} \times 100 \Rightarrow \quad {i_p} = 5A\)
PHXII07:ALTERNATING CURRENT
356286
A step-up transformer works on \(220\;V\) and gives \(2\;A\) to an external resistor. The turn ratio between the primary and secondary coils is 2:25. Assuming 100% efficiency, find the secondary voltage. Primary current and power delivered.