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 ?

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 ?

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?

A galvanometer of resistance 90 Omega is shunted by a resistance of 10 Omega . What fraction of main current passes through the shunt ?

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 .

An Ac source of volatage V=100 sin 100pit is connected to a resistor of ressistance 20 Omega The rsm value of current through resistor is

A p-n junction diode has a forward bias resistance of 10 Omega and a reverse bias resistance of 10 k Omega . It is connected in series with a resistance of 990 Omega and a.d.c. source of 5V in such a way that the negative terminal of the source is connected to then region of the p-n junction. What is the current flowing in the circuit?

There are a large number of cells available, each marked (6 V, 0.5 Omega ) to be used to supply current to a device of resistance 0.75 Omega requiring 24 A current . How should the cells be arranged , so that maximum power is transmitted to the load using minimum number of cells ?