Question 734: Current Electricity

Question

Consider a cylindrical element as shown in the figure. The current that flows through the element is I and resistivity of material of the cylinder is $\rho$. Choose the correct option out the following

Accepted Answer

Resistance of $1^{	ext {st }}$ half part is $\mathrm{R}_1=\frac{ho l_1}{\mathrm{~A}_1}=\frac{ho(1 / 2)}{\pi(4 r)^2}$
Resistance of $2^{	ext {nd }}$ part is, $\mathrm{R}_2=\frac{ho l_2}{\mathrm{~A}_2}=\frac{ho(1 / 2)}{(2 r)^2 \pi}$
$$Thus, \frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{\frac{ho(1 / 2)}{\pi 16 r^2}}{\frac{ho(1 / 2)}{4 \pi r^2}}=\frac{1}{4} $$
(A) Power loss : $P_1=I^2 R_1$ and $P_2=I^2 R_2$,
So $\frac{\mathrm{P}_1}{\mathrm{P}_2}=\frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{1}{4} \Rightarrow \mathrm{P}_1 .4=\mathrm{P}_2$
(B) Voltage drop : $\mathrm{V}_1=\mathrm{IR}_1$ and $\mathrm{V}_2=\mathrm{IR}_2$
So $\frac{\mathrm{V}_1}{\mathrm{~V}_2}=\frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{1}{4}$
(C) Current density : $\mathrm{J}_1=\frac{\mathrm{I}}{\mathrm{A}_1}$ and $\mathrm{J}_2=\frac{\mathrm{I}}{\mathrm{A}_2}$
So $\frac{\mathrm{J}_1}{\mathrm{~J}_2}=\frac{\mathrm{A}_2}{\mathrm{~A}_1}=\frac{\pi r^2}{\pi(2 r)^2}=\frac{1}{4}$
(D) Electric field: $\frac{E_1}{E_2}=\frac{J_1}{J_2}=\frac{1}{4} \quad(J=\sigma E)$

Answer

Power loss in first half is four times the power loss in second half.

Answer

Voltage drop in first half is 4 times of voltage drop in second half.

Answer

Current density in both halves are equal.

Answer

Electric field in both halves is equal

Current Electricity

Consider a cylindrical element as shown in the figure. The current that flows through the element is I and resistivity of material of the cylinder is \rho. Choose the correct option out the following

Choose the correct answer: