An ideal battery sends a current of 5A in a resistor.When another resistor of value `10 Omega`is connected in parallel ,the current through the battery is increased to 6A.Find the resistance of the first resistor.

An ideal battery passes a current of 5A through a resistor. When it is connected to another resistance of 10Omega in parallel, the current is 6A. Find the resistance of the first resistor.

Two identical cells connected in series send 1.0A current through a 5 Omega resistor. When they are connected in parallel, they send 0.8 A current through the same resistor. What is the internal resistance of the cell?

Two identical cells connected in series send 10 A current through a 5 Omega resistor. When they connected in parallel, they send 8 A current through the same resistance, the internal resistance of each cell is

A battery of e.m.f. 15 V and internal resistance 3 Omega is connected to two resistors 3 Omega and 6 Omega connected in parallel. Find : the current through the battery,

A battery of e.m.f 10V and internal resistance 3Omega is connected to a resistor.If the current in the circuit is 0.5A,then the resistance of the resistor is

When a resistor of 20 Omega is connected in series with a battery, the current is 0.5 A. when a resistor of 10 Omega is connected the current becomes 0.8 A. calculate the emf and the internal resistance of the battery.

Six 18.0Omega resistors are connected in parallel across a 12.0V ideal battery. What is the current through the battery?

A battery of emf 12 V and internal resistance 3 Omega is connected to a resistor. If the current in the circuit is 0.5 A, Calculate resistance of resistor.