Let's analyze the circuit shown in the image and solve for the current $I$.

Step-by-step explanation:

1. Given components:

   • The circuit is a quadrilateral loop $ABCD$.
   • There is a 9V battery between points $A$ and $B$, supplying a voltage of 9V.
   • Two resistors are present:
     • One resistor of $6\Omega$ between points $A$ and $D$.
     • Another resistor of $3\Omega$ between points $C$ and $D$.

2. Resistor Connections:

   • The two resistors seem to be connected in series, and they are the only components providing resistance in the loop.

3. Total Resistance (R_total):

   • Since the resistors are in series, the total resistance is simply the sum of the individual resistances: $R_{total} = R_1 + R_2 = 6\Omega + 3\Omega = 9\Omega$

4. Current Calculation (Using Ohm’s Law):

   • The formula for current is given by Ohm’s Law: $I = \frac{V}{R}$
   • Here, the total voltage $V = 9V$ and the total resistance $R_{total} = 9\Omega$.
   • Substituting these values into Ohm’s law: $I = \frac{9V}{9\Omega} = 1A$

Thus, the current $I$ in the circuit is $1A$.

Would you like more details on the steps or any clarification? Here are some related questions:

1. What happens to the current if we add another resistor in parallel?
2. How do we calculate the equivalent resistance for resistors in parallel?
3. Can we analyze the voltage drops across each resistor?
4. How would the current change if the battery's voltage increases?
5. What would happen if one of the resistors were removed?

Tip: Always double-check if the resistors are connected in series or parallel, as this will affect the total resistance in the circuit.