How will you connect three resistances of 4 ohms each to get `12 Omega`, `6 Omega` and `4/3 Omega` respectively?

How will you connect three resistors of resistances 3 Omega each to obtain a resistance of 4.5 Omega ?

How will you connect three resistors (A=2 Omega), (B=3Omega) and (C =6 Omega) to obtain a resistance of 4 Omega ?

Three cells are connected in parallel with their like poles connected together with wires of negligible resistance. The e.m.f. of the cells are 2V, 3V and 4V respectively and their internal resistances are 1 Omega , 2Omega and 3Omega , respectively. Find the current through each branch.

Two batteries with e.m.f 12 V and 13 V are connected in parallel across a load resistor of 10 Omega . The internal resistances of the two batteries are 1 Omega and 2 Omega respectively. The voltage across the load lies between :

How will you connect three resistors (A=4 Omega) , (B=6Omega) and (C=2Omega) to obtain a resistance of 5.5Omega ?

Choose correct alternative. The emf of a cell of negligible internal resistance is 2 V. It is connected to the series combination of 2 Omega , 3 Omega and 5 Omega resistances. The potential difference across 3 Omega resistance will be (in volt)

Find the effective resistance of the series . combination of three resistors of resistances 10 Omega , 20 Omega and 30 Omega .

The e.m.f. e and the internal resistance r of the battery shwn in figure are 4.3 V and 1Omega respectively. The external resistance R is 50Omega . The resistance of the ammeter and voltmeter are 2.0Omega and 200Omega respectively. If the switch is thrown to the other side, the reading of the ammeter will be

Three resistors having resistances of 12 Omega , 9 Omega , 4 Omega and 3 Omega are connected in series. What is the effective resistance in the circuit?