91影视

For circuit 1,

A Nichrome wire (鈥渨ire 1鈥�) of length $L_1$and cross-sectional area $A_1$is connected in series with a battery and an ammeter.

For circuit 2,

Now the Nichrome wire is removed and replaced with a different wire (鈥渨ire 2鈥�), which is 2.5 times as long and has 5.5 times the cross-sectional area of the original wire, which means the length now is $L_2=2.5L_1$and the cross-sectional area is $A_2=5.5A_1$.

The current flowing in a circuit can be written mathematically as,

$\mathit{I}\mathbf{=}\frac{\mathbf{V}\mathbf{A}}{\mathbf{蚁}\mathbf{L}}$ (1)

Here V is the applied voltage in the circuit; A is the cross-sectional area of the connecting wire that is used in the circuit; L is the length of the wire, role="math" localid="1663003300566" $\mathit{蚁}\mathbf{}$is the resistivity of the material out of which the wire was made.

For circuit 1, use equation 1 to calculate the flowing current as,

$I_1=\frac{V{A}_1}{蚁L_1}$ (2)

For circuit 2, use equation 1 to calculate the flowing current as,

$I_2=\frac{V{A}_2}{蚁L_2}$ (3)

Divide equation 3 by equation 2 as,

$\begin{array}{rcl}\frac{I_2}{I_1}& =& \frac{\frac{V{A}_2}{蚁L_2}}{\frac{V{A}_1}{蚁L_1}}\\ \frac{I_2}{I_1}& =& \frac{\frac{A_2}{L_2}}{\frac{A_1}{L_1}}\end{array}$

Substitute the values in the above expression, and we get,

$\begin{array}{rcl}\frac{I_2}{I_1}& =& \frac{\frac{5.5A_1}{2.5L_1}}{\frac{A_1}{L_1}}\\ & =& \frac{5.5}{2.5}\\ \frac{I_2}{I_1}& =& 2.2\\ & & \end{array}$

Thus, the ratio of conventional currents in the two circuits is$\frac{I_2}{I_1}=2.2$ .