A) \[\frac{{{\alpha }_{1}}+{{\alpha }_{2}}}{2},\frac{{{\alpha }_{1}}+{{\alpha }_{2}}}{2}\]
B) \[\frac{{{\alpha }_{1}}+{{\alpha }_{2}}}{2},{{\alpha }_{1}}+{{\alpha }_{2}}\]
C) \[{{\alpha }_{1}}+{{\alpha }_{2}},\frac{{{\alpha }_{1}}+{{\alpha }_{2}}}{2}\]
D) \[{{\alpha }_{1}}+{{\alpha }_{2}},\frac{{{\alpha }_{1}}{{\alpha }_{2}}}{{{\alpha }_{1}}+{{\alpha }_{2}}}\]
Correct Answer: A
Solution :
| [a] In series, \[{{R}_{1}}+{{R}_{2}}=Rs\] |
| \[R(1+{{\alpha }_{1}}T)+R(1+{{\alpha }_{2}}T)=2R(1+{{\alpha }_{s}}T)\] |
| \[2R+RT({{\alpha }_{1}}+{{\alpha }_{2}})=2R+2R\text{ }{{\alpha }_{s}}T\] |
| \[{{\alpha }_{s}}=\frac{{{\alpha }_{1}}+{{\alpha }_{2}}}{2}\] |
| In parallel\[\frac{1}{Rp}=\frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}}\] |
| \[\frac{1}{\frac{R}{2}(1+{{\alpha }_{p}}T)}=\frac{1}{R(1+{{\alpha }_{1}}T)}+\frac{1}{R(1+{{\alpha }_{2}}T)}\] |
| \[2(1-{{\alpha }_{p}}T)=1-{{\alpha }_{1}}T+1-{{\alpha }_{2}}T\] |
| \[{{\alpha }_{p}}=\frac{{{\alpha }_{1}}+{{\alpha }_{2}}}{2}\] |
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