Question

3.) calculate the value of rt, it, vr1, vr2, vr3. use units and proper engineering notation for full credit. 9v vt it 470ω r1 1.2kω r2 270ω r3

Explanation:

Step1: Convert resistance units

First, convert $R_2 = 1.2k\Omega$ to $\Omega$. Since $1k\Omega=1000\Omega$, then $R_2 = 1.2\times10^{3}\Omega$.

Step2: Calculate total resistance

The resistors are in series. The formula for total resistance $R_T$ in a series - circuit is $R_T=R_1 + R_2+R_3$. Given $R_1 = 470\Omega$, $R_2 = 1.2\times10^{3}\Omega$, and $R_3 = 270\Omega$. Then $R_T=470\Omega+1.2\times10^{3}\Omega + 270\Omega=(470 + 1200+270)\Omega=1940\Omega = 1.94\times10^{3}\Omega$.

Step3: Calculate total current

Using Ohm's law $I_T=\frac{V_T}{R_T}$. Given $V_T = 9V$ and $R_T=1.94\times10^{3}\Omega$. Then $I_T=\frac{9V}{1.94\times10^{3}\Omega}\approx4.64\times10^{- 3}A = 4.64mA$.

Step4: Calculate voltage across each resistor

Using Ohm's law $V = IR$.
For $R_1$: $V_{R1}=I_T\times R_1$. Substitute $I_T = 4.64\times10^{-3}A$ and $R_1 = 470\Omega$, then $V_{R1}=4.64\times10^{-3}A\times470\Omega\approx2.18V$.
For $R_2$: $V_{R2}=I_T\times R_2$. Substitute $I_T = 4.64\times10^{-3}A$ and $R_2 = 1.2\times10^{3}\Omega$, then $V_{R2}=4.64\times10^{-3}A\times1.2\times10^{3}\Omega\approx5.57V$.
For $R_3$: $V_{R3}=I_T\times R_3$. Substitute $I_T = 4.64\times10^{-3}A$ and $R_3 = 270\Omega$, then $V_{R3}=4.64\times10^{-3}A\times270\Omega\approx1.25V$.

Answer:

$R_T = 1.94\times10^{3}\Omega$, $I_T=4.64\times10^{-3}A$, $V_{R1}\approx2.18V$, $V_{R2}\approx5.57V$, $V_{R3}\approx1.25V$