A) 200 V, 1 A
B) 800 V, 2 A
C) 100 V,2A
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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