A) 8V
B) 10 V
C) 22V
D) 52V
Correct Answer: A
Solution :
Given, \[R=3\,\Omega ,\,{{X}_{L}}=15\,\Omega ,\,{{X}_{C}}=11\,\Omega \], \[{{V}_{R}}=10\,V\] \[\therefore \] Current through the circuit \[i=\frac{{{V}_{rms}}}{\sqrt{{{R}^{2}}+{{({{X}_{L}}-{{X}_{C}})}^{2}}}}\] \[=\frac{{{V}_{rms}}}{\sqrt{{{(3)}^{2}}+{{(15-11)}^{2}}}}\] \[=\frac{10}{\sqrt{9+16}}\] \[=\frac{10}{5}=2\,A\] Since, L, C and R are connected in series combination the potential difference across R is \[{{V}_{R}}=i\times R=2\times 3\] = 6 V Across L, \[{{V}_{L}}=i{{X}_{L}}=2\times 15=30\,N\] Across C, \[{{V}_{C}}=i\,{{X}_{C}}\] \[=2\times 11=22\,V\] So, potential difference across series combination of L and C is \[={{V}_{L}}-{{V}_{C}}\] \[=30-22=8\,V\]You need to login to perform this action.
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