98 J of heat is produced each second in 2 `Omega` resistor.Find the potential difference.

500 J of energy is produced each second in a 20 ohm bulb. Find the potential differecne across the bulb.

The circuit in Fig. EP 3.24 shows two cells connected in opposition to each other. Cell E_1 is of emf 6V and internal resistance 2 Omega , the 8 Omega . Find the potential difference between the points A and B.

Two cells of emf 2V and 4V and internal resistance 1Omega and 2Omega respectively are connected in parallel so as to send the current in the same direction through an external resistance of 10Omega. Find the potential difference across 10Omega resistor.

In the circuit shown, the current through the 4Omega resistor is 1 A, when the points P and M are connected to a d.c. source. The potential difference between the points M and N is

A cell of e.m.f. 2 V and negligible internal resistance is connected to a potentiometer wire of resistance 10 Omega and length 4 m. The potential difference per unit length (potential gradient ) of the wire is

A 2Omega resistor is connected in series with R Omega resistor. This combination is connected across a cell. When the potential difference across 2 Omega resistor is balanced on potentiometer wire, null point is obtained at length of 300cm. When the same procedure is repeated for R Omega resistor, null point is obtained at length 350cm, value of R is