An $A.C.$ source is connected to a series $LCR$ circuit. If the voltage across $R$ is $40 \,V$, the voltage across $L$ is $80 \,V$, and the voltage across $C$ is $40 \,V$, then the e.m.f. '$e$' of the $A.C.$ source is:
- A$40 \,V$
- B$40 \sqrt{2} \,V$
- C$80 \,V$
- D$160 \,V$
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