The current flow in the circuit of two battery with internal resistance and two external resistance as shown in figure will be

The correct option is 0.67 A.

Step-by-step Explanation:

To find the current flowing in the given circuit, we can analyze the components shown in the diagram:
1. A battery on the left with electromotive force (emf) ${\epsilon }_1=15\text{V}$ and internal resistance $r_1=1.2\Omega$.
2. A battery on the right with emf ${\epsilon }_2=5\text{V}$ and internal resistance $r_2=0.8\Omega$.
3. Two external resistors: one at the top with resistance $R_1=9\Omega$ and another at the bottom with resistance $R_2=4\Omega$.

1. Determine the Net Electromotive Force (emf):
Looking at the orientation of the two batteries in the circuit:
- The positive terminal (longer line) of both batteries points upwards.
- When tracing the circuit in a single loop, these batteries oppose each other. The 15 V battery attempts to drive current clockwise, while the 5 V battery attempts to drive current counterclockwise.
Therefore, the net emf ($E_{\text{net}}$) in the loop is the difference between the two emfs:

$$E_{\text{net}}={\epsilon }_1-{\epsilon }_2$$

$$E_{\text{net}}=15\text{V}-5\text{V}=10\text{V}$$

2. Calculate the Total Resistance of the Circuit:
Since all components (external resistors and internal resistances of the batteries) are connected in series, we find the total resistance ($R_{\text{total}}$) by taking their sum:

$$R_{\text{total}}=R_1+r_2+R_2+r_1$$

$$R_{\text{total}}=9\Omega +0.8\Omega +4\Omega +1.2\Omega$$

$$R_{\text{total}}=15\Omega$$

3. Calculate the Current:
Using Ohm's law, the current ($i$) flowing through the circuit is given by:

$$i=\frac{E_{\text{net}}}{R_{\text{total}}}$$

$$i=\frac{10\text{V}}{15\Omega }=\frac{2}{3}\text{A}\approx 0.67\text{A}$$

Thus, the current flowing in the circuit is approximately 0.67 A, flowing in the clockwise direction.