| Two coils have self-inductances \[{{L}_{1}}=8\,\,mH\] and \[{{L}_{2}}=2\,\,mH.\] In both of them currents are increased at the same constant rate. At a certain instant the power given to the two coils is the same. If at that instant, \[{{i}_{1}},\] \[{{V}_{1}},\] \[{{U}_{1}}\] and \[{{i}_{2}},\] \[{{V}_{2}},\] \[{{U}_{2}}\] be the currents, induced voltages and energies stored in the two coils respectively, then- | |
| [A] \[{{i}_{1}}/{{i}_{2}}=1/4\] | [B] \[{{i}_{1}}/{{i}_{2}}=4\] |
| [C] \[{{U}_{2}}/{{U}_{1}}=4\] | [D] \[{{V}_{2}}/{{V}_{1}}=1/4\] |
A) a, b, c
B) a, b, d
C) a, c, d
D) b, c, d
Correct Answer: C
Solution :
[C]| \[e=\frac{LdI}{dt}\] |
| \[\Rightarrow e\propto L\]{as\[\frac{di}{dt}\] is same for both coil} |
| \[\frac{{{e}_{1}}}{{{e}_{2}}}=\frac{{{L}_{1}}}{{{L}_{2}}}\] |
| Power in both coil is same |
| \[\therefore Power={{e}_{1}}{{I}_{1}}={{e}_{2}}{{I}_{2}}\] |
| \[\Rightarrow \frac{{{I}_{1}}}{{{I}_{2}}}=\frac{{{L}_{2}}}{{{L}_{1}}}\] |
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