using the laws of circuit theory, solve for rt , it , vr1 , vr2 , vr3 , and vr4 for the circuit shown in…

Explanation:

Step1: Calculate total resistance

For a series - circuit, the total resistance $R_T$ is the sum of individual resistances. $R_T=R1 + R2+R3 + R4$. Given $R1 = 2.7k\Omega$, $R2 = 10k\Omega$, $R3 = 3.3k\Omega$, $R4 = 8.2k\Omega$. So, $R_T=(2.7 + 10+3.3 + 8.2)k\Omega=24.2k\Omega$.

Step2: Calculate total current

Using Ohm's law $I_T=\frac{V_T}{R_T}$. Given $V_T = 6V$ and $R_T=24.2\times10^{3}\Omega$. Then $I_T=\frac{6V}{24.2\times10^{3}\Omega}\approx248\mu A$.

Step3: Calculate voltage across each resistor

Using Ohm's law $V = IR$. For $R1$: $V_{R1}=I_T\times R1$. Substituting $I_T\approx248\times10^{- 6}A$ and $R1 = 2.7\times10^{3}\Omega$, we get $V_{R1}=248\times10^{-6}A\times2.7\times10^{3}\Omega\approx0.67V$. For $R2$: $V_{R2}=I_T\times R2$. Substituting $I_T\approx248\times10^{-6}A$ and $R2 = 10\times10^{3}\Omega$, we get $V_{R2}=248\times10^{-6}A\times10\times10^{3}\Omega\approx2.48V$. For $R3$: $V_{R3}=I_T\times R3$. Substituting $I_T\approx248\times10^{-6}A$ and $R3 = 3.3\times10^{3}\Omega$, we get $V_{R3}=248\times10^{-6}A\times3.3\times10^{3}\Omega\approx0.82V$. For $R4$: $V_{R4}=I_T\times R4$. Substituting $I_T\approx248\times10^{-6}A$ and $R4 = 8.2\times10^{3}\Omega$, we get $V_{R4}=248\times10^{-6}A\times8.2\times10^{3}\Omega\approx2.03V$.

Step4: Verify with Kirchhoff's voltage law

$V_T=V_{R1}+V_{R2}+V_{R3}+V_{R4}$. $0.67V + 2.48V+0.82V + 2.03V=6V$, which verifies our results.

Answer:

$R_T = 24.2k\Omega$, $I_T\approx248\mu A$, $V_{R1}\approx0.67V$, $V_{R2}\approx2.48V$, $V_{R3}\approx0.82V$, $V_{R4}\approx2.03V$