The peak value of an alternating e.m.f. is 141.4 V. What would be the reading of an a.c. voltmeter when connected across this e.m.f. ?

The r.m.s value of alternating e.m.f. is

The peak value of alternating e.m.f. in a generator is given by e_(0) =

In the given figure, the instantaneous value of alternating e.m.f. is e = 14.14 sin omega t . The reading of voltmeter in volt will be

A capacitor of 10 mu F is connected to an a.c. source of e.m.f. E = 220 sin 100 pi t . Write the equaiton of instananeous current through the circuit. What will be the reading of a.c. ammeter connected in the circuit ?

The diagram in Fig. shows a cell of e.m.f. epsi = 2 volt and internal resistance r = 1 ohm connected to an external resistance R = 4 ohm. The ammeter A measures the current in the circuit and the voltmeter V measures the terminal voltage across the cell. What will be the readings of the ammeter and voltmeter when the key K is closed.

The diagram in Fig. shows a cell of e.m.f. epsi = 2 volt and internal resistance r = 1 ohm connected to an external resistance R = 4 ohm. The ammeter A measures the current in the circuit and the voltmeter V measures the terminal voltage across the cell. What will be the readings of the ammeter and voltmeter when the key K is open

What is peak value and r.m.s. value of alternating e.m.f.in the electric circuit?

An alternating e.m.f. is given by 100 sin 100 pi t o a pure resistance of 50 Omega . calculate:- peak value of e.m.f