A) \[200\,\,V,\,\,1\,\,A\]
B) \[800\,\,V,\,\,2\,\,A\]
C) \[100\,\,V,\,\,2\,\,A\]
D) \[220\,\,V,\,\,2.2\,\,A\]
Correct Answer: D
Solution :
In the series \[L-C-R\] circuit \[{{V}_{R}}\] (voltage across resistance) is in the phase with current\[(i)\], while \[{{V}_{C}}\] (voltage across inductance) leads \[i\] by\[{{90}^{o}}\], while \[{{V}_{C}}\] lags behind \[i\] by\[{{90}^{o}}\]. Resultant emf \[E\] is \[{{E}^{2}}=V_{R}^{2}+{{({{V}_{L}}-{{V}_{C}})}^{2}}\] Since, \[{{V}_{L}}={{V}_{C}}=300\,\,V\] \[\therefore \] \[E={{V}_{R}}=220\,\,V\] Reading of ammeter,\[i=\frac{E}{R}=\frac{220}{100}=2.2A\]
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