A series battery of 6 lead accumulators, each of $2V$ EMF and internal resistance $0.50 ohm$ is charged by a $100V$ dc supply. The series resistance should be used in the charging circuit in order to limit the current to $8.0 A$.
A) 4 ohm
B) 6 ohm
C) 8 ohm
D) 10 ohm

Hint: In order to solve the above given problem we will apply the OHM's law which is related as:
$V=IR$ ( V is the voltage, I is the current, R is the resistance), voltage applied in the circuit is directly proportional to current flowing, where proportionality constant is R resistance of the circuit.
By calculating the total voltage in the circuit using Kirchhoff's voltage law.
With the help of above mentioned two laws we will calculate the resistance of the circuit.

Complete step by step solution:
Now let’s apply the KVL in the circuit, which states that:
“Sum of all the voltage drops in the circuit is equal to zero “.
Sign conventions used in the circuit to apply Kirchhoff's law:
If we move from positive to negative sign then voltage is dropped then we will use negative sign while applying KVL ,If we move from negative to positive sign then voltage is grained then we will use positive sign in the equation.
Voltage applied in the circuit is 100V.
Voltage of each accumulator is 2V, 6 accumulators are connected then the total voltage of the accumulator is: $2 \times 6 = 12$V
Total internal resistor of the accumulators: $6 \times 0.5 = 3$
On applying Kirchhoff's voltage law we have:
$100-12=I(R +3)$ ( R is the series resistance which is added with internal resistance)
$ \Rightarrow 100 - 12 = 8(3 + R)$ (We have substituted the value of voltage and current as per the question)
$ \Rightarrow 88 = 8(3 + R)$ (On rearranging the terms)
$ \Rightarrow \dfrac{{88}}{8} = 3 + R$
$ \Rightarrow 11 - 3 = R $
$ \Rightarrow R = 8 $ (Value of resistance of the series circuit is 8 Ohm)

Thus, option (C) is correct.

Note: When the resistors are connected in series the magnitude of the resistors are connected with simple addition and when the resistors are connected in parallel reciprocal of the magnitude of the resistors is added to find the equivalent resistance of the circuit given.