Question

4. A long, narrow resistor is placed in a series circuit along with a short, wide resistor made of the same material. Which will have a greater electric potential drop across it? Explain your reasoning.

Answer 1

Given data:

One long, narrow resistor and short wide resistor.

The expression for the resistance is given by,

`R=\rho(L)/(A)`

Here, ρ is the resistivity that depends on material, L is the length, and A is the area of cross-section.

The expression for the voltage drop according to Ohm's law is,

`\begin{gathered} V=IR \\ V=I\rho(L)/(A) \end{gathered}`

The ratio of voltage drop will be,

`\begin{gathered} (V_1)/(V_2)=(I_1\rho(L_1)/(A_1))/(I_2\rho(L_2)/(A_2)) \\ (V_(1))/(V_(2))=(I_1L_1A_2)/(I_2L_2A_1) \end{gathered}`

Here, suffix 1 shows long narrow resistor and suffix 2 shows wide short resistor.

The current will also be same as the resistors are in series I₁=I₂.

`(V_(1))/(V_(2))=(L_1A_2)/(L_2A_1)`

From above equation we can observe that V₁ will have more voltage drop than V₂ as L₁/A₁ is more for long narrow resistor and A₂/L₂ will be less for short and wide resistor.