A) 8V
B) 10 V
C) 22V
D) 52V
Correct Answer: A
Solution :
Key Idea: The potential difference across the series combination of L and C is the difference in potential differences across individuals. Current through the circuit, \[i=\frac{{{V}_{rms}}}{\sqrt{{{R}^{2}}+{{({{X}_{L}}-{{X}_{c}})}^{2}}}}\] Given,\[R=3\,\Omega ,\,{{X}_{L}}=15\,\Omega ,\,{{X}_{C}}=11\,\Omega ,\] \[{{V}_{rms}}=10\,V\] \[\therefore \] \[i=\frac{10}{\sqrt{{{(3)}^{2}}+{{(15-11)}^{2}}}}\] Or \[i=\frac{10}{\sqrt{9+16}}\] Or \[i=\frac{10}{5}=2A\] Since L, C and R are connected in series combination, then potential difference across R is \[{{V}_{R}}=iR=2\times 3=6V\] Across \[L,{{V}_{L}}=i{{X}_{L}}=2\times 15=30V\] Across \[C,{{V}_{C}}=i{{X}_{C}}=2\times 11=22V\] So, potential difference across series combination of L and C \[={{V}_{L}}-{{V}_{C}}\] \[=30-22\] \[=8\,V\] Note: In\[LCR\]circuit whenever voltage across various elements is asked, find rms values unless stated in the question for the peak or instantaneous value.You need to login to perform this action.
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