Expression for internal resistance in terms of I1, I2, R1, R2:

Question

In a meter bridge experiment, a resistance of \( 10 \, \Omega \) is balanced by a resistance \( X \) with a balance point at \( 40 \, \mathrm{cm} \). What is the value of \( X \)?

Answer 1

Internal Resistance Calculation \( r \)

Given:

Current \( I_1 = \frac{E}{R_1 + r} \)
Current \( I_2 = \frac{E}{R_2 + r} \)

Step 1: Rearrange Equations

Rearrange the given current equations to express \( E \):

\[ E = I_1 (R_1 + r) \quad \text{and} \quad E = I_2 (R_2 + r) \]

Step 2: Equate Expressions for \( E \)

Since both expressions equal \( E \), they can be set equal to each other:

\[ I_1 (R_1 + r) = I_2 (R_2 + r) \]

Step 3: Expand and Isolate \( r \) Terms

Expand both sides of the equation:

\[ I_1 R_1 + I_1 r = I_2 R_2 + I_2 r \]

Group terms containing \( r \) on one side and other terms on the other:

\[ I_1 r - I_2 r = I_2 R_2 - I_1 R_1 \]

Factor out \( r \):

\[ r (I_1 - I_2) = I_2 R_2 - I_1 R_1 \]

Step 4: Solve for \( r \)

Divide by \( (I_1 - I_2) \) to find \( r \):

\[ r = \frac{I_2 R_2 - I_1 R_1}{I_1 - I_2} \]

Final Result:

The formula for internal resistance \( r \) is: