Name: RC Circuit and Charging Capacitor
Uploaded: 2019-07-19
Duration: 13 min 36 s
Description: In this video, we will discuss the physics of charging the capacitor connected in a circuit and derive the expression and relation between the parameters inv...

Question

A rod of length

L

and cross-section area

A

lies along the

x

-axis between

x = 0

and

x = L

. The material obeys Ohm's law and its resistivity varies along the rod according to

ρ (x) = ρ_{0} e^{- x / L}

. The end of the rod at

x = 0

is at a potential

V_{0}

and it is zero at

x = L

. (i) Find the total resistance of the rod and the current in the wire (ii) Find the electric potential in the rod as a function of

x

.

Accepted Answer

At distance $x$ from $x = 0$ , let us consider shape of area $A and length d x .$

Variation of voltage at $x = x & x = x + d x be V & V + d V$ respectively.

∴ By Ohm's Law,

$V = (\frac{ρ L}{A}) I ∴ d V = \frac{ρ (x) d x}{A} \times I$

Integrating over entire length

$∴ \int_{V_{\circ}}^{0} d V = \int_{0}^{L} L \frac{ρ_{\circ} e^{- \frac{x}{L}} d x \times I}{A}$

We get $- V_{\circ} = {[\frac{ρ_{\circ} e^{- \frac{x}{L}}}{A (\frac{- 1}{L})}]}^{L}_{0}$

$∴ I = \frac{A V_{\circ}}{ρ_{\circ} L} (e^{- 1} - 1)$

∴ For obtaining $V (x)$

Integrating from $0 to x$

$\int_{V_{\circ}}^{V (x)} d x = \int_{0}^{x} \frac{ρ_{\circ} e^{- \frac{x}{L}} d x I}{A}$

$∴ V (x) - V_{\circ} = \frac{ρ_{\circ}}{A} \times \frac{A V_{\circ}}{ρ_{\circ} L} \frac{(e^{- 1} - 1) (e^{- \frac{x}{L}} - 1)}{- \frac{1}{L}} ∴ V (x) - V_{\circ} = - V_{\circ} (e^{- 1} - 1) (e^{- \frac{x}{L}} - 1) V (x) = V_{\circ} - (V_{\circ} e^{- 1} - V_{\circ}) (e^{- \frac{x}{L}} - 1) ∴ V (x) = V_{\circ} - (V_{\circ} e^{- 1} e^{- \frac{x}{L}} - V_{\circ} e^{- 1} - V_{\circ} e^{- \frac{x}{L}} + V_{\circ}) ∴ V (x) = V_{\circ} e^{- 1} + V_{\circ} e^{- \frac{x}{L}} - V_{\circ} e^{- \frac{x}{L} - 1} ∴ V (x) = V_{\circ} (e^{- 1} + e^{- \frac{x}{L}} - e^{- \frac{x}{L} - 1}) Which is the required answer.$

Important Questions on Current Electricity

Learn and Prepare for any exam you want !