A) LR
B) \[{{(1+A)}^{n}}=I+\lambda A,\]
C) \[\lambda \]
D) \[2n-1\]
Correct Answer: B
Solution :
The instantaneous current in L-R circuit during growth is given by\[\gamma =\frac{3}{2}=\] Since, all the exponential terms are dimensionless, therefore, it follows that factor \[\frac{{{C}_{p}}}{{{C}_{v}}}\] should be dimensionless i.e. \[{{C}_{v}}=\frac{{{n}_{1}}{{C}_{{{v}_{1}}}}+{{n}_{2}}{{C}_{{{v}_{2}}}}}{{{n}_{1}}+{{n}_{2}}}\]\[{{C}_{p}}=\frac{{{n}_{1}}{{C}_{{{p}_{1}}}}+{{n}_{2}}{{C}_{{{p}_{2}}}}}{{{n}_{1}}+{{n}_{2}}}\]\[\therefore \]\[\therefore \]\[\gamma =\frac{{{C}_{p}}}{{{C}_{v}}}=\frac{{{n}_{1}}{{C}_{{{p}_{1}}}}+{{n}_{2}}{{C}_{{{p}_{2}}}}}{{{n}_{1}}{{C}_{{{v}_{1}}}}+{{n}_{2}}{{C}_{{{v}_{2}}}}}\] Hence, the time constant of E-R circuit is \[\frac{3}{2}=\frac{2\left( \frac{5}{2}R \right)+n\left( \frac{7}{2}R \right)}{2\left( \frac{3}{2}R \right)+n\left( \frac{5}{2}R \right)}\]
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