A series combination of `L` and `R` is connected to a battery of emf `E` having negligible internal resistance. The final value of current depends upon

A series combination of L and R is connected to a battery of emf E having negligible internal resistance. The final value of current depeds upon

Three resistances 2Omega, 4Omega, 5Omega are combined in series and this combination is connected to a battery of 12 V emf and negligible internal resistance. The potential drop across these resistances are

Three resistors 2Omega, 4Omega and 5Omega are combined in parallel. This combination is connected to a battery of emf 20V and negligible internal Resistance. The total current drawn from the battery is

(a) Three resistors 2 Omega, 4 Omega and 5 Omega are combined in series. What is the total resistance of the combination ? (b) If the combination is connected to a battery of emf 20 V and negligible internal resistance, determine the current through each resistor, and the total current drawn from the battery.

Three resistors of resistances 3Omega, 4Omega and 5Omega are combined in parallel. This combination is connected to a battery of emf 12V and negligible internal resistance, current through each resistor in ampere is

What is the maximum rate at which energy is stored in inductor (L) when it is connected in series with resistance R and one battery of emf E and negligible internal resistance? Refer to growth and decay of current in a coil, explained in competition file.

A voltmeter of resistance R_(V) and an ammeter of resistance R_(A) are connected in series across a battery of emf E and of negligible internal resistance . When a resistance R is connected in parallel to volmeter , reading of ammeter increases to three times while that of volmeter reduces to one - third. Calculate R_(A)andR_(V) in terms of R.

Shows a network of eight resistors battery of resistance R(= 2 Omega) connected in a 3V battery of negligible internal resistance. The current I in the circuit is

Two resistors of resistances 2Omega and 6Omega are connected in parallel. This combination is then connected to a battery of emf 2 V and internal resistance 0.5 Omega . What is the current flowing thrpough the battery ?