Let the resistance of an electrical component remains constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it?
According to Ohm's law, the current through an electrical component is given by I = V/R. When the potential difference across the component decreases to half of its former value while the resistance remains constant, the current will also decrease to half of its original value.
This can be understood mathematically. If the initial potential difference is V and the resistance is R, then the initial current is I = V/R. When the potential difference becomes V/2 (half), the new current I' = (V/2)/R = V/(2R) = I/2. Thus, the current becomes half of its former value when the potential difference is reduced to half while resistance remains unchanged.
Explanation
The textbook context provides the fundamental relationship I = V/R from Eq. (11.5). The question tests the student's understanding of the direct proportionality between current and potential difference when resistance is constant. The context also mentions that 'If the resistance is doubled the current gets halved' which demonstrates the inverse relationship between current and resistance. Similarly, when potential difference is halved (with constant resistance), current is also halved due to the direct proportionality.
Solution Steps
Step 1: Recall Ohm's law: I = V/R
Step 2: Initial current I₁ = V/R (where V is the original potential difference)
Step 3: New potential difference V₂ = V/2 (half of former value)
Step 4: New current I₂ = V₂/R = (V/2)/R = V/(2R) = I₁/2
Step 5: Therefore, the current decreases to half of its former value.