An a.c. source of internal resistance `9000 Omega` is to supply current to a load resistor of 10 ohm. How should the source be matched to the load and what is the ratio of the currents passing through the load and the source ?

An a.c. source of internal resistance 9000 Omega is to supply current to a load resistor of 10 Omega . How should the source the matched to the load and what is then the ratio of the currents passing through the load and the source ?

4 cells each of emf 2 V and internal resistance of 1Omega are connected in parallel to a load resistor of 2Omega . Then the current through the load resistor is

An AC source is connected to a diode and a resistor in series. Is the current through the resistor AC or DC?

An AC source is connected to a diode and a resistor in series. Is the current through the resistor AC or DC?

An AC source is connected to a diode and a resistor in series. Is the current through the resistor AC or DC?

An ac source is connected across a resistance of 10 Omega The power dissipated in the resistor is 100W The rms valuse of the current and voltge are .

20Omega resistor is connected with the A.C. source of V = 282sin (120 pi t), the current passing through the resistor will be ……..

A source voltage of 8 V drives the diode in fig. through a current- limiting resistor of 100 ohm . Then the magnitude of the slope load line on the V - I characteristics of the diode is

Resistor of resistance R and capacitor of capacitance C are connected in series to an AC source of angular frequency omega . Rms current in the circuit is I. When frequency of the source is changed to one - third of initial value keeping the voltage same then current is found to be halved. Find the ratio of reactance of capacitor to that with resistance at the original frequency omega .