Question

1. an electrical circuit has a variable voltage source and as the voltage (volts, v) in the circuit increases, so does the current (milliamps, ma) flowing in the circuit. the table below gives voltages and the corresponding currents, based on direct measurements of the circuit.

volts	5	6	7	8	9
milliamps	7.4	9.7	12.0	14.3	16.6

(a) plot a graph of this data with voltage on the x - axis.
(b) write the equation of the resultant line.
(c) what current would flow if a voltage of 3.0 v was supplied to this circuit?
(d) approximately what voltage would be necessary to produce a current of 25 ma?
(e) if the voltage doubles in this circuit, what happens to the current?

Explanation:

Step1: Determine slope (m)

Let $(x_1,y_1)=(5,7.4)$ and $(x_2,y_2)=(6,9.7)$. Using the slope - formula $m=\frac{y_2 - y_1}{x_2 - x_1}$, we have $m=\frac{9.7 - 7.4}{6 - 5}=\frac{2.3}{1}=2.3$.

Step2: Determine y - intercept (b)

Using the point - slope form $y - y_1=m(x - x_1)$ with $(x_1,y_1)=(5,7.4)$ and $m = 2.3$, we get $y-7.4=2.3(x - 5)$. Expanding gives $y-7.4=2.3x-11.5$. Then $y=2.3x - 4.1$.

Step3: Find current for 3.0 V

Substitute $x = 3$ into $y=2.3x - 4.1$. So $y=2.3\times3-4.1=6.9 - 4.1 = 2.8$ mA.

Step4: Find voltage for 25 mA

Set $y = 25$ in $y=2.3x - 4.1$. Then $25=2.3x-4.1$. Rearranging gives $2.3x=25 + 4.1=29.1$. So $x=\frac{29.1}{2.3}\approx12.65$ V.

Step5: Analyze effect of doubling voltage

If the original equation is $y = 2.3x-4.1$, let the new voltage be $x' = 2x$. Then the new current $y'=2.3x'-4.1=2.3(2x)-4.1=4.6x-4.1$. Comparing with the original $y = 2.3x-4.1$, the current also approximately doubles (since the linear relationship dominates for reasonable values of $x$).

Answer:

(a) (Graph plotting is not possible in this text - based format, but you would plot points (5,7.4), (6,9.7), (7,12.0), (8,14.3), (9,16.6) with voltage on x - axis and current on y - axis).
(b) $y = 2.3x-4.1$
(c) 2.8 mA
(d) Approximately 12.65 V
(e) The current approximately doubles.