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

Let us use notation: $I = 4.11 A$ - total current through both resistors, $U$ - voltage across two resistors.

$I_1, R_1 = 3.5 \Omega; I_2, R_2$ - currents and voltages across resistor and resistance wire respectively.

Since both resistors are connected in parallel, voltage across them is the same, and total current is the sum of currents through the first and second resistor: $I = I_1 + I_2$.

According to Ohm's law, $I_1 = \frac{U}{R_1}$, and hence from the formula above, current through the resistance wire is $I_2 = I - I_1 = I - \frac{U}{R_1} \approx 2.68 A$.

Resistance of the wire is $R = \frac{\rho l}{S}$, where $\rho$ is resistivity, $l$ - length of the wire, $S$ - cross-sectional area of the wire.

Using Ohm's law and the last formula:

$I_2 = \frac{U}{R_2} = \frac{U S}{\rho l}$, from where $\rho = \frac{U S}{I_2 l} \approx 4.7 \cdot 10^{-9} \Omega \cdot m$.