In the network shown the potential difference between `A` and `B` is `(R = r_(1) = r_(2) = r_(3) = 1 Omega, E_(1) = 3V, E_(2) = 2V, E_(3) = 1V`)

In the network shown the potential difference between A and B is R(=r_(1)=r_(2)=r_(3)=1Omega,E_(1)=3V,E_(2)=2V,E_(3)=1V)

In Fig. key K is closed at t = 0 . After a long time, the potential difference between A and B is zero, the value of R will be [r_(1) = r_(2) = 1 Omega, E_(1) = 3 V and E_(2) = 7 V , C = 2 mu F , L = 4 mH , where r_(1) and r_(2) are the internal resistances of cells E_(1) and E_(2) , respectively].

In Fig. key K is closed at t = 0 . After a long time, the potential difference between A and B is zero, the value of R will be [r_(1) = r_(2) = 1 Omega, E_(1) = 3 V and E_(2) = 7 V , C = 2 mu F , L = 4 mH , where r_(1) and r_(2) are the internal resistances of cells E_(1) and E_(2) , respectively].

In the given network of ideal cells and resistors, R_(1) = 1.0 Omega, R_(2) = 2.0 Omega, E_(1) = 2V and E_(2) = E_(3) = 4V . The potential difference between the point a and b is

Find a potentail difference varphi_(1) - varphi_(2) between points 1 and 2 of the circuit shown in Fig if R_(1) 1= 10 Omega, R_(2) = 20 Omega, E_(1) = 5.0 V , and E_(2) = 2.0 V . The internal resisances of the current sources are negligible.

Four resistors R_(1),R_(2),R_(3) and R_(4) are connected between two terminals A and B iin a network as shown in the diagram. A key K can connect the two points (see diagram) C and D. A constant potential difference (V_(A)-V_(B))=V is maintained between thhe points A and B Q. If R_(1)=R_(4)=1Omega and R_(2)=R_(3)=2Omega , the effective resistance between the terminals A and B will change, on closing the key K, by

In the circuit shown below E_(1) = 4.0 V, R_(1) = 2 Omega, E_(2) = 6.0 V, R_(2) = 4 Omega and R_(3) = 2 Omega . The current I_(1) is

In the circuit shown below E_(1) = 4.0 V, R_(1) = 2 Omega, E_(2) = 6.0 V, R_(2) = 4 Omega and R_(3) = 2 Omega . The current I_(1) is