Question

Two wires P ad Q, each of the same length and same material, are connected in parallel to a battery. The diameter of P is half that of Q. What fraction of the total current passes through P?

A. 0.2
B. 0.25
C. 0.33
D. 0.5

Answer 1

Two wires P and Q, each of the same length and same material, are connected in parallel to a battery. The diameter of P is half that of Q. What fraction of the total current passes through P?

A. 0.2

B. 0.25

C. 0.33

D. 0.5

The resistance $R$ of a conductor of uniform cross section can be computed as

$$R = \frac{\rho l}{A} = \frac{\rho l}{\pi d^{2} / 4}$$

where $l$ is the length of the conductor, measured in metres [m], $A = \frac{\pi d^{2}}{4}$ is the cross-section area of the conductor measured in square metres $[\mathfrak{m}^2]$, and $\rho$ (rho) is the electrical resistivity (also called specific electrical resistance) of the material, measured in ohm-metres $(\Omega \cdot \mathfrak{m})$.

Therefore:

$$R_{P} = \frac{\rho l}{\pi d_{P}^{2} / 4}$$ $$R_{Q} = \frac{\rho l}{\pi d_{Q}^{2} / 4}$$

Or:

$$\frac{R_{P}}{R_{Q}} = \frac{d_{Q}^{2}}{d_{P}^{2}} = 4$$

Ohm's law states that the current through a conductor between two points is directly proportional to the potential difference across the two points:

$$I = \frac{V}{R}$$

where I is the current through the conductor in units of amperes, V is the potential difference measured across the conductor in units of volts, and R is the resistance of the conductor in units of ohms.

Therefore, current passes through P equals:

$$I _ {P} = \frac {V}{R _ {P}}$$

Current passes through Q equals:

$$I _ {Q} = \frac {V}{R _ {P}}$$

Total current:

$$I = I _ {Q} + I _ {P} = V \left(\frac {1}{R _ {Q}} + \frac {1}{R _ {P}}\right)$$

Fraction of the total current passes through P:

$$\frac {I _ {P}}{I} = \frac {\frac {V}{R _ {P}}}{V \left(\frac {1}{R _ {Q}} + \frac {1}{R _ {P}}\right)} = \frac {1}{\frac {R _ {P}}{R _ {Q}} + 1} = \frac {1}{5} = 20 \%$$

Answer: 20 %