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Step-by-Step Solution
Step 1: State the Problem Clearly
Two wires, made of the same material, are connected in parallel. Their lengths are in the ratio $4:3$, and their radii are in the ratio $2:3$. We need to find the ratio of the electric current passing through these two wires.
Step 2: Recall the Formula for Resistance
The resistance $R$ of a wire of length $l$, cross-sectional area $A$, and resistivity $\rho$ is given by:
$ R = \rho \dfrac{l}{A} $
Step 3: Express the Cross-Sectional Area in Terms of Radius
The cross-sectional area of a wire (cylindrical) is $ A = \pi r^2 $. Given the radii ratio $r_1 : r_2 = 2 : 3$, we can write:
$ A_1 : A_2 = \pi (2^2) : \pi (3^2) = 4 : 9 $
Step 4: Determine the Ratio of Resistances
Because both wires have the same material (same $\rho$) and are connected in parallel (thus same voltage across each), the relevant comparison is their resistance ratio. For wire 1 and wire 2:
$ R_1 = \rho \dfrac{l_1}{A_1} \quad \text{and} \quad R_2 = \rho \dfrac{l_2}{A_2}. $
The ratio $ \dfrac{R_1}{R_2} $ is:
$ \dfrac{R_1}{R_2}
= \dfrac{\rho\, \dfrac{l_1}{A_1}}{\rho\, \dfrac{l_2}{A_2}}
= \dfrac{l_1}{l_2} \times \dfrac{A_2}{A_1}. $
We know $l_1 : l_2 = 4 : 3$ and $A_1 : A_2 = 4 : 9$, so $A_2 : A_1 = 9 : 4$. Hence:
$ \dfrac{R_1}{R_2} = \dfrac{4}{3} \times \dfrac{9}{4} = 3. $
Step 5: Relate Resistance to Current
In a parallel connection, the same voltage $V$ is applied across each wire. The current in each wire is $ I = \dfrac{V}{R}.$ Thus the ratio of currents is:
$ \dfrac{I_1}{I_2} = \dfrac{V/R_1}{V/R_2} = \dfrac{R_2}{R_1}. $
Substituting $R_1 : R_2 = 3 : 1$, we get:
$ \dfrac{I_1}{I_2} = \dfrac{1}{3}. $
Step 6: State the Final Ratio
The ratio of the current passing through the two wires (wire 1 : wire 2) is $ 1 : 3 $, which is often written as $ \dfrac{1}{3}. $
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