Question

Find the value of unknown resistance X in the circuit shown in the figure if no current flows through the section AO. Also calculate the current drawn by the circuit from the battery of emf. 6V and negligible internal resistance.

Answer 1

Part 1: Finding the value of the unknown resistance X

The problem states that no current flows through the section AO. Let's label the central junction point as 'O'.

1. Identify the Principle: The condition that no current flows through the resistor between points A and O (I_AO = 0) means that the electric potential at point A is equal to the electric potential at point O (V_A = V_O). This is the defining condition for a balanced Wheatstone bridge.

2. Identify the Wheatstone Bridge: The resistors connected between points A, B, C, and O form a Wheatstone bridge.
   The four arms of the bridge are the resistors R_AB = 2Ω, R_AC = 4Ω, R_BO = 3Ω, and the unlabeled resistor between O and C, which we will assume is the unknown resistance X (i.e., R_OC = X).
   The resistor between A and O acts as the galvanometer arm. Its resistance value is not given, but since the bridge is balanced, its value is irrelevant for this calculation.

3. Apply the Balance Condition: For a balanced Wheatstone bridge, the ratio of resistances in the adjacent arms is equal.

   R_AB / R_BO = R_AC / R_OC

4. Substitute the known values and solve for X:

   2Ω / 3Ω = 4Ω / X

   To solve for X:

   2 X = 4 3
   2X = 12
   X = 12 / 2
   X = 6Ω

Therefore, the value of the unknown resistance is 6Ω.

---

Part 2: Calculating the current drawn from the battery

Now that we know the bridge is balanced, we can simplify the circuit to find the total equivalent resistance and then the total current.

1. Simplify the Circuit: Since no current flows through the arm AO, we can remove it from the circuit for the purpose of calculating the equivalent resistance. The circuit simplifies as follows:
   The resistors R_AB (2Ω) and R_AC (4Ω) are in series, forming one branch.
   The resistors R_BO (3Ω) and R_OC (6Ω) are in series, forming a second branch.
   These two branches are connected in parallel between points B and C.
   This entire parallel combination is in series with the 2.4Ω resistor.

2. Calculate the Resistance of the Series Branches:
   Resistance of the upper branch (R_upper): R_AB + R_AC = 2Ω + 4Ω = 6Ω
   Resistance of the lower branch (R_lower): R_BO + R_OC = 3Ω + 6Ω = 9Ω

3. Calculate the Equivalent Resistance of the Parallel Combination (R_p):

   1 / R_p = 1 / R_upper + 1 / R_lower
   1 / R_p = 1/6 + 1/9

   To add these fractions, find a common denominator (18):

   1 / R_p = 3/18 + 2/18 = 5/18

   R_p = 18 / 5 = 3.6Ω

4. Calculate the Total Equivalent Resistance of the Circuit (R_eq):
   The parallel combination (R_p) is in series with the 2.4Ω resistor.

   R_eq = R_p + 2.4Ω
   R_eq = 3.6Ω + 2.4Ω = 6.0Ω

5. Calculate the Total Current Drawn from the Battery:
   Using Ohm's Law (I = V / R), with the battery's EMF (V) of 6V and the total equivalent resistance (R_eq) of 6.0Ω. The internal resistance is negligible.

   I_total = V / R_eq
   I_total = 6V / 6.0Ω
   I_total = 1A

Final Answer

• The value of the unknown resistance X is 6Ω.
• The current drawn by the circuit from the battery is 1A.