for each of the resistors shown in figures 1 - 6, use ohms law to calculate the unknown quantity. be sure to…

for each of the resistors shown in figures 1 - 6, use ohms law to calculate the unknown quantity. be sure to put your answer in proper engineering notation and use the correct units.\n\ni =?\nr = 470ω\nv = 3.5 v\n\ni = 15.5 ua\nr =?\nv = 3.5 v\n\ni = 1.4 ma\nr = 22kω\nv =

for each of the resistors shown in figures 1 - 6, use ohms law to calculate the unknown quantity. be sure to put your answer in proper engineering notation and use the correct units.\n\ni =?\nr = 470ω\nv = 3.5 v\n\ni = 15.5 ua\nr =?\nv = 3.5 v\n\ni = 1.4 ma\nr = 22kω\nv =

Answer

Explanation:

Step1: Recall Ohm's law

Ohm's law is $V = IR$, where $V$ is voltage, $I$ is current and $R$ is resistance. We can re - arrange it for different unknowns.

Step2: Solve for $I$ in Figure 1

Given $V = 3.5\ V$ and $R=470\ \Omega$. Using $I=\frac{V}{R}$, we have $I=\frac{3.5}{470}\ A\approx7.45\times 10^{-3}\ A = 7.45\ mA$.

Step3: Solve for $R$ in Figure 2

Given $V = 3.5\ V$ and $I = 15.5\ \mu A=15.5\times 10^{- 6}\ A$. Using $R=\frac{V}{I}$, we get $R=\frac{3.5}{15.5\times 10^{-6}}\ \Omega\approx2.26\times10^{5}\ \Omega = 226\ k\Omega$.

Step4: Solve for $V$ in Figure 3

Given $I = 1.4\ mA = 1.4\times 10^{-3}\ A$ and $R = 22\ k\Omega=22\times 10^{3}\ \Omega$. Using $V = IR$, we obtain $V=(1.4\times 10^{-3})\times(22\times 10^{3})\ V = 30.8\ V$.

Answer:

Figure 1: $I = 7.45\ mA$ Figure 2: $R = 226\ k\Omega$ Figure 3: $V = 30.8\ V$