Directions : In the following question more than one of the answers given may be correct. Select the correct answer and mark it according to the code:
Two different coils have self-inductance\[{{L}_{1}}=8\] \[mH,\text{ }{{L}_{2}}=2\text{ }mH\]. The current in one coil is increased at a constant rate. The current in the second coil is also increased at the same rate. At a certain instant of time the power given to the coil is same. At that time the current, the induced voltage and the energy stored in the first coil are\[{{i}_{1}},{{v}_{1}}\]and\[{{w}_{1}}\]respectively. Corresponding values for the second coil at the same instant are\[{{i}_{1}},{{v}_{2}}\]and \[{{w}_{2}}\].Then (1) \[\frac{{{i}_{1}}}{{{i}_{2}}}=\frac{1}{4}\] (2) \[\frac{{{w}_{2}}}{{{w}_{1}}}=4\] (3) \[\frac{{{v}_{2}}}{{{v}_{1}}}=\frac{1}{4}\] (4) \[\frac{{{i}_{1}}}{{{i}_{2}}}=48\]A) 1, 2 and 3 are correct.
B) 1 and 2 are correct.
C) 2 and 4 are correct.
D) 1 and 3 are correct.
Correct Answer: A
Solution :
From Farada/s law, the induced voltage\[V\propto L\]rate of change of current in constant \[\left[ V=-\frac{Ldi}{dt} \right]\] \[\therefore \] \[\frac{{{V}_{2}}}{{{V}_{1}}}=\frac{{{L}_{2}}}{{{L}_{1}}}=\frac{2}{8}=\frac{1}{4}\] \[\Rightarrow \] \[\frac{{{V}_{1}}}{{{V}_{2}}}=4\] Power given to the two coils is same ie, \[{{V}_{1}}{{i}_{1}}={{V}_{2}}{{i}_{2}}\] \[\Rightarrow \] \[\frac{{{i}_{1}}}{{{i}_{2}}}=\frac{{{V}_{2}}}{{{V}_{1}}}=\frac{1}{4}\] Energy stored \[W=\frac{1}{2}L\,{{i}^{2}}\] \[\Rightarrow \] \[\frac{{{W}_{2}}}{{{W}_{1}}}=\left[ \frac{{{L}_{2}}}{{{L}_{1}}} \right]{{\left[ \frac{{{i}_{2}}}{{{i}_{1}}} \right]}^{2}}\] \[=\left[ \frac{1}{4} \right]{{(4)}^{2}}=4\] \[\Rightarrow \] \[\frac{{{W}_{1}}}{{{W}_{2}}}=\frac{1}{4}\]
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