Four resistances `40Omega, 60Omega, 90Omega` and `100Omega` make the arms of a quadrilateral ABCD. Across AC is a battery of emf 40 V and internal resistance negligible. The potential difference across BD in V is ________.

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

A capacitor of capacitance 1muF is connected in parallel with a resistance 10Omega and the combination is connected across the terminals of a battery of EMF 50 V and internal resistance 1Omega . The potential difference across the capacitor at steady state is (in Volt)

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 battery of emf 2 V and initial resistance 1 Omega is connected across terminals A and B of the circuit shown in figure .

Two resistance of 400 Omega and 800 Omega are connected in series with 6 volt battery of negligible internal resistance. A voltmeter of resistance 10,000 Omega is used to measure the potential difference across 400 Omega . The error in measurement of potential difference in volts approximatley is

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

In the circuit shown below, E is a battery of emf 6 V and internal resistance 1Omega . Find the reading of the ammeter A. if it has negligible resistance :

Find potential difference across the resistance 300 Omega in A and B.

Four resistence of 10 Omega, 60Omega,100Omega and 200Omega , respectively taken in order are used to form a Wheatstone's bridge . A 15 V battery is connected to the ends of a 200Omega resistance, the current through it will be

The four arms of a Wheatstone bridge (Fig. 3.26) have the following resistances: AB = 100 Omega, BC = 10 Omega, CD = 5Omega, and DA = 60Omega . A galvanometer of 15W resistance is connected across BD. Calculate the current through the galvanometer when a potential difference of 10 V is maintained across AC.