Question

A 3.50 Ω resistor is connected in parallel with a 0.830 m length of uniform resistance wire, of cross-sectional area 2.09 ×10^-9 m^2. When a power supply is connected, the p.d across the parallel combination is 5.00 V and the total current through the resistor and the resistance wire is 4.11A.


1) determine the current flowing through the resistance wire.

2) determine the resistivity of the resistance wire.

Answer 1

Notations

• Please refer to the sketch attached.

Assumption

• All the connecting wires are of zero resistance.

Calculations

1). Considering the system,

$\qquad \qquad i+i_1=4.11\cdots\cdots(1)$

Considering the resistor alone,

$\qquad \begin{aligned} \end{aligned}$ $\qquad \begin{aligned} \footnotesize V&=\footnotesize iR\\ \footnotesize 5V&=\footnotesize i\times3.5\Omega\\ \footnotesize i&=\footnotesize \frac{10}{7}A\\ \footnotesize \therefore i_1&=\footnotesize 4.11-\frac{10}{7}\,A \cdots\cdots(from \,1)\\ &= \footnotesize \bold {2.681\,A} \end{aligned}$

2). Considering the wire alone,

• Now the voltage applied & the current flowing through the wire are known hence the resistance could be found.

$\qquad \qquad \begin{aligned} \footnotesize V&= \footnotesize iR\\ \footnotesize R&= \footnotesize \frac{5V}{2.681A}\\ &= \footnotesize \bold{1.865\Omega} \end{aligned}$

$\qquad \begin{aligned} \footnotesize resistance (R) &= \footnotesize \frac{\rho l}{Area(A)}\\ \footnotesize \therefore \rho &= \footnotesize \frac{R \times A}{l}\\ &= \footnotesize \frac{1.865\Omega \times 2.09\times10^{-9}m^2}{0.830m}\\ \footnotesize resistivity (\rho) &= \footnotesize \bold{4.696\times10^{-9}\Omega m} \end{aligned}$

Good luck!