The Question is from neet-physics current-electricity electric-power-and-joules-law, and it was created by the author Mr. ANIL KUMAR.G

Question

Two wires of the same material and length but different cross-sectional areas are connected in series across a battery. If the ratio of their radii is 2:1, what is the ratio of the power dissipated in them?

Mr. ANIL KUMAR.G · Accepted Answer

## CORE INFORMATION
✓ **Answer**: Option A: 1:4
## CONCEPT FRAMEWORK
 **Primary Concept**: Power dissipation in a resistor and its relation to resistance and current.
 **Related Topics**: Ohm's law, resistance of a wire, series circuits.
 **Prerequisites**: Understanding of electrical resistance, power, and series circuits.
## SOLUTION PATHWAY
**Given Information**:
  * Two wires of the same material and length.
  * Ratio of radii: $r_1 : r_2 = 2 : 1$
  * Connected in series across a battery.
**Method & Steps**:
 1. Find the resistance of each wire.
  Reasoning: The resistance of a wire is given by $R = \frac{ho l}{A}$, where $ho$ is the resistivity, $l$ is the length, and $A$ is the cross-sectional area. Since the wires are of the same material and length, the resistivity and length are constant.  The area is proportional to the square of the radius ($A = \pi r^2$).
  Formula/Rule: $R = \frac{ho l}{\pi r^2}$
 2. Determine the ratio of resistances.
  Working: Let $R_1$ and $R_2$ be the resistances of the wires with radii $r_1$ and $r_2$ respectively. Then:
       $\frac{R_1}{R_2} = \frac{\frac{ho l}{\pi r_1^2}}{\frac{ho l}{\pi r_2^2}} = \frac{r_2^2}{r_1^2} = \left(\frac{1}{2}ight)^2 = \frac{1}{4}$
  Key Point:  In a series circuit, the current is the same through both resistors.
 3. Calculate the ratio of power dissipated.
  Reasoning: Power dissipated in a resistor is given by $P = I^2R$. Since the current $I$ is the same for both wires, the ratio of power dissipated is equal to the ratio of resistances.
  Result: $\frac{P_1}{P_2} = \frac{I^2 R_1}{I^2 R_2} = \frac{R_1}{R_2} = \frac{1}{4}$
  Verification: The power dissipated is inversely proportional to the square of the radius, so a larger radius implies less power dissipation.
 ## OPTION ANALYSIS
A ✓: 1:4 - Correct. The ratio of power dissipated is 1:4, as derived above.
B :x: 4:1 - Incorrect. This would imply that the thicker wire dissipates more power, which is incorrect in a series circuit.
C :x: 1:2 - Incorrect. This ratio doesn't account for the area's relationship with resistance.
D :x: 2:1 - Incorrect. This is the inverse of the correct ratio.
 ## LEARNING AIDS
 **Common Mistakes**:
  1. [Error 1]: Confusing the relationship between resistance and radius (using a linear relationship instead of an inverse square relationship):  Remember that resistance is inversely proportional to the square of the radius ($R \propto \frac{1}{r^2}$).
  2. [Error 2]: Incorrectly applying power formulas for parallel circuits: The problem specifies a series circuit, so the current is the same in both wires.
 **Quick Tips**:
  1. [Helpful shortcut]: Remember the formula for resistance of a wire and how it relates to the area.  Then, directly find the ratio of resistances.
  2. [Memory aid]:  Think about a thicker wire having lower resistance and therefore lower power dissipation (for the same current).

Mr. ANIL KUMAR.G · Answer

The correct Option is "Option A: 1:4"

Mr. ANIL KUMAR.G · Answer

Option B: 4:1

Mr. ANIL KUMAR.G · Answer

Option C: 1:2

Mr. ANIL KUMAR.G · Answer

Option D: 2:1

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