NEET UG PHYSICS MCQ: CURRENT ELECTRICITY

Solution approach

1. Correct Answer + Core Reason: Decreases. Adding a resistor in series increases the total resistance, decreasing the current and hence the power dissipated by R1.
   
   Step-by-Step Solution:
   
   Step 1: Initially, the power dissipated by $R_1$ is $P = I^2 R_1$, where $I$ is the current through $R_1$.
   
   Step 2: When $R_2$ is added in series with $R_1$, the total resistance becomes $R_{total} = R_1 + R_2$.
   
   Step 3: The new current in the circuit is $I' = \frac{V}{R_1 + R_2}$, where $V$ is the voltage of the generator.
   
   Step 4: The new power dissipated by $R_1$ is $P' = (I')^2 R_1 = \left(\frac{V}{R_1 + R_2}\right)^2 R_1 = \frac{V^2 R_1}{(R_1 + R_2)^2}$.
   
   Step 5: Since $(R_1 + R_2)^2 > R_1^2$, we have $P' < P$.
   
   Step 6: Therefore, the power dissipated by $R_1$ decreases.
   
   Why other options are wrong:
   - Option B (Increases): Incorrect because adding resistance in series reduces the current and hence the power dissipated.
   - Option C (Remains the same): Incorrect because the current changes when a resistor is added in series.
   - **Option D (Any of the above depending upon the relative values of ${R_1}$ and ${R_2}$): Incorrect because the power always decreases when a resistor is added in series.
   
   Key Concepts:**
   - Power dissipation in a resistor: $P = I^2 R = \frac{V^2}{R}$
   - Series resistance: $R_{total} = R_1 + R_2$
   - Ohm's Law: $V = IR$
   
   Memory Aids:
   - Power is proportional to the square of the current.
   
   Quick Tips:
   - Remember that adding resistance in series decreases the current in the circuit.