A 10 V battery of negilgible internal resistance is connected across a 200V battery and a resistance of `38 Omega` as shown in figure.
. Find the value of current in the circuit.

Explanation

Since, the positive terminal of the batteries are connected together, so the equivalent emf of the batteries is given by `epsilon= 200 - 10 = 190v`.

Hence, the current in the circuit is given by `I=(epsilon)/(R)`.
`I=(190)/(38)=5A`.