Step 1: Recall Ohm’s Law. \[ I = \frac{V}{R} \] Let original voltage be \( V \), and resistance be \( R \), then the original current is: \[ I = \frac{V}{R} \] Step 2: Modify the values. New voltage \( V' = \frac{V}{2} \), and new resistance \( R' = 2R \)
Step 3: Substitute into the current formula. \[ I' = \frac{V'}{R'} = \frac{V/2}{2R} = \frac{V}{4R} \] Step 4: Compare with original current. \[ I' = \frac{I}{4} \] So the new current becomes one-fourth of the original.
The Wheatstone bridge is an arrangement of four resistances, say \(R_1, R_2, R_3\), and \(R_4\). The null point condition is given by: