question_answer

A 50 Hz AC source of 20 V is connected across R and C as shown in the figure. The voltage across R is 12 V. The voltage across C is

A) 8 V

B) 16 V

C) 10 V

D) not possible to determine unless values of R and C are given

Correct Answer: B

Solution :

Voltage across capacitor lags behind current by\[90{}^\circ \]. The given circuit is a \[\operatorname{C} -R\] series circuit, \[{{V}_{R}}\] is in phase with i while \[{{V}_{C}}\] behind i by\[90{}^\circ \]. Hence, resultant potential is \[E={{E}_{o}}\,sin\,\omega t\] \[V=\sqrt{V_{R}^{2}+V_{C}^{2}}\] Given, \[{{V}_{R}}=12\,\,V,\,\,V=20\,\,V\] \[\therefore \,\,\,\, {{\left( 20 \right)}^{2}}={{\left( 12 \right)}^{2}}+V_{C}^{2}\] \[\Rightarrow \,\,\,\,\,\,V_{C}^{2}={{\left( 20 \right)}^{2}}-{{\left( 12 \right)}^{2}}\] \[=400-144=256\] \[={{V}_{C}}\,\,=\,\,\sqrt{256} \,=\,\,16\,V\]