A) \[{{R}_{1}}=\,\,{{R}_{2}}\,\,=\,\,{{R}_{3}}\]
B) \[{{R}_{2}}={{R}_{3}}\,\,and\,\,{{R}_{1}}=4{{R}_{2}}\]
C) \[{{R}_{2}}={{R}_{3}}\,\,and\,\,{{R}_{1}}=\left( 1/4 \right){{R}_{2}}\]
D) \[{{R}_{1}}={{R}_{2}}+{{R}_{3}}\]
Correct Answer: C
Solution :
When resistors are connected in parallel potential difference across them is the same. In the given circuit the resistor?s \[{{R}_{2}}\,\,and\,\,{{R}_{3}}\] are connected in parallel, hence potential difference (V) across them is the same. In order that they undergo the same energy loss, \[H=\frac{{{V}^{2}}}{R}t,\,\,{{R}_{2}}\,\,\] must be equal to\[{{R}_{3}}\]. That is, \[{{R}_{2}}={{R}_{3}}\] Now resistor \[{{R}_{1}}\] is in series with \[{{R}_{2}}\], hence energy through them is \[H={{i}^{2}}{{R}_{1}}t\,\,=\,\,i_{1}^{2}{{R}_{2}}t\] where \[{{i}_{1}}\] is current across \[{{R}_{2}}\]. Since \[{{R}_{2}}={{R}_{3}}\], therefore current through them is \[\frac{i}{2}={{i}_{1}}\] \[\therefore \,\,\,\,\,\,\,\,{{i}^{2}}{{R}_{1}}t=\frac{{{i}^{2}}}{4}{{R}_{2}}t\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,R{{ & }_{1}}=\frac{{{R}_{2}}}{4}\]
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