A) 10 V
B) \[10\sqrt{2}\,V\]
C) \[\left( \frac{10}{\sqrt{2}} \right)V\]
D) 20 V
Correct Answer: B
Solution :
\[R={{X}_{L}}={{X}_{c}}\](\[\because \] voltage across them is same) Here Total voltage in the circuit \[V=I{{[{{R}^{2}}+{{({{X}_{L}}-{{X}_{c}})}^{2}}]}^{\frac{2}{3}}}\] \[=IR=10V\] when capacitor is short circuited, \[I=\frac{10}{{{({{R}^{2}}+X_{L}^{2})}^{\frac{1}{2}}}}=\frac{10}{\sqrt{2}R}\] \[\therefore \] potential drop across inductance \[=I{{X}_{L}}=IR=10\sqrt{2}\,\text{volt}\text{.}\] Hence, the correction option is [b].
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