Free Study Material

NCERT Solutions

NCERT Books

Revision Notes

Practice Problems

This is the second of our many quizzes on Transformers. Click on this to access more such competitive quizzes on Current Electricity and EMI.

Transformers Quiz 2

physics

The ratio of power loss to output power in transformer one is equal to $\dfrac{1}{3}$ and for transformer two is $\dfrac{1}{6}$. The ratio of efficiency of these two transformer will be given by- 

 If power $P$ is applied to a transformer then, some amount will be dissipated in the form of loss, so input power will be equal to sum of output power and loss or we can say that some power given to the transformer will not be utilized further. 

 $\text{Output power of transformer = O} \\  \text{Loss in transformer = L,}\\  \text{Efficiency of transformer is given by,}\\  E=\dfrac{O}{O+L}\\  \Rightarrow E=\dfrac{1}{1+\dfrac{L}{O}}\\  \text{For transformer one ratio of power loss to output power is}\,\dfrac{1}{3}\\  {{E}_{1}}=\dfrac{1}{1+\dfrac{L}{O}}\\  \Rightarrow {{E}_{1}} =\dfrac{1}{1+\dfrac{1}{3}}\\\Rightarrow {{E}_{1}}=\dfrac{3}{4}\\  \text{For transformer two ratio of power loss to output power is}\,\dfrac{1}{6}\\  {{E}_{2}}=\dfrac{1}{1+\dfrac{L}{O}}\\\Rightarrow {{E}_{2}}=\dfrac{1}{1+\dfrac{1}{6}}\\  \Rightarrow {{E}_{2}}=\dfrac{6}{7}\\\Rightarrow \dfrac{{{E}_{1}}}{{{E}_{2}}}=\dfrac{{}^{3}/{}_{4}}{{}^{6}/{}_{7}}\\  \therefore\dfrac{{{E}_{1}}}{{{E}_{2}}}=\dfrac{7}{8}$

 A step down transformer having a number of turns in primary coil equal to 1000 and in secondary coil equal to 50. Two resistance of $5\Omega $ each are connected in the secondary coil and supply of 50 V is connected in the primary coil. The value of magnitude of current in primary coil will be-<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/ddf69bd9-6a29-4734-8a92-3707a21ddb5048891046162827014.png"/> 

 The relation between resistance of primary coil and secondary coil with number of turns can easily derived by the common ratios for transformers, i.e.$\dfrac{{{N}_{P}}}{{{N}_{S}}}=\dfrac{{{V}_{P}}}{{{V}_{S}}}=\dfrac{{{I}_{S}}}{{{I}_{P}}}$.

 $\text{Equivalent resistance at secondary coil is,}\\  R={{R}_{1}}+{{R}_{2}}\\  \Rightarrow {{R}_{S}}=5+5=10\Omega\\  \text{Now we know that,}\\  \dfrac{{{R}_{P}}}{{{R}_{S}}}={{\left( \dfrac{{{N}_{P}}}{{{N}_{S}}} \right)}^{2}}\\\Rightarrow {{R}_{P}}={{\left( \dfrac{{{N}_{P}}}{{{N}_{S}}} \right)}^{2}}{{R}_{S}}\\  \Rightarrow {{R}_{P}} ={{\left( \dfrac{1000}{50} \right)}^{2}}\times 10\\  \Rightarrow {{R}_{P}} =4000\Omega\\\text{Now, current in primary coil is}\\  {{I}_{P}}=\dfrac{V}{{{R}_{P}}}\\\Rightarrow{{I}_{P}} =\dfrac{50}{4000}\\ \therefore{{I}_{P}} =\dfrac{1}{80}A$

 If an old transformer has an efficiency of 90 %. This efficiency can be increased by some modern methods and due to which power loss is reduced to half of its initial value. The new efficiency of transformer will be- 

If the transformer's power loss is reduced, more input power can be used to produce output power, increasing performance.

$  \text{Output power of transformer = O}\\  \text{Loss in transformer = L,}\\  \text{Efficiency of transformer is given by,}\\ E=\dfrac{O}{O+L}\times 100=90 \% \\  \Rightarrow \dfrac{O}{\text{O+L}}=0.9\\  \Rightarrow \dfrac{1}{1+\dfrac{L}{O}}=0.9\\  \Rightarrow \dfrac{1}{0.9}=1+\dfrac{L}{\text{O}}\\  \Rightarrow \dfrac{L}{\text{O}}=\dfrac{1}{9}\\  \text{Now, loss is reduced to half so new efficiency will be}\\  E=\dfrac{1}{1+\dfrac{{}^{L}/{}_{2}}{O}}\\  \Rightarrow E=\dfrac{1}{1+\dfrac{L}{\text{2}\times O}}\\  \Rightarrow E=\dfrac{1}{1+\dfrac{1}{2\times 9}}\\  \Rightarrow E=0.947\\  \therefore E=0.947\times 100=94.7$ 

A student of class 12th wants to solve a problem on transformers, in which two transformers are given. First transformer having a number of turns in primary coil equal to 400 and in secondary 20.Second transformer having number of turns in primary coil equal to 600 and in secondary coil 30. The relation between ratio of currents in primary coil to secondary coil in both transformers will be-$[I({{1}^{st}})\,\text{ratio in first transformer and}\\ I({{2}^{nd}})\,\text{ratio in second transformer}]$ 

The relation between number of turns and current can be derived by taking the ratio of power in both coils and ratio of voltage in both coils. 

$  \text{In a transformer the relation between number of turns}\\ \text{and current in primary and secondary coil is given by,}\\  \dfrac{{{N}_{P}}}{{{N}_{S}}}=\dfrac{{{I}_{S}}}{{{I}_{P}}}\\  \text{Ratio of currents in first transformer,}\\ I({{1}^{st}})=\dfrac{{{I}_{P}}}{{{I}_{S}}}\\  \Rightarrow I({{1}^{st}}) =\dfrac{{{N}_{S}}}{{{N}_{P}}}\\  \Rightarrow I({{1}^{st}}) =\dfrac{20}{400}\\  \Rightarrow I({{1}^{st}})=\dfrac{1}{20}\\  \text{Ratio of currents in second transformer,}\\  I({{2}^{nd}})=\dfrac{{{I}_{P}}}{{{I}_{S}}}\\  \Rightarrow I({{2}^{nd}}) =\dfrac{{{N}_{S}}}{{{N}_{P}}}\\  \Rightarrow I({{2}^{nd}}) =\dfrac{30}{600}\\  \Rightarrow I({{2}^{nd}})=\dfrac{1}{20}\\  \therefore I({{2}^{nd}})=I({{1}^{st}})$

The correct efficiency-load current graph according to power factor of transformer is given by- 

 The relation is not in direct proportionality hence straight line curves will not be there, instead of that curved figure can be seen. 

$  \text{Efficiency of transformer by considering power factor is given by,}\\  E=\dfrac{{{V}_{2}}{{I}_{2}}\cos \phi }{{{V}_{2}}{{I}_{2}}\cos \phi +K}\\ \Rightarrow E =\dfrac{1}{1+\dfrac{K}{{{V}_{2}}{{I}_{2}}\cos \phi }}\\  \text{From the equation we can observe that,}\\ \text{if power factor increases}\,\dfrac{K}{{{V}_{2}}{{I}_{2}}\cos \phi }\,\text{this term will decrease}\\  \text{and overall efficiency will increase but the value of power factor will be in range of 0-1 so after some time,}\\ \text{if we keep on increasing the load current the efficiency will eventually start decreasing.}\\\text{So the graph for higher power factor will be higher as for higher efficiency}\\ \text{and for lower power factor low efficiency.}$ 