Question

using the laws of circuit theory, solve for rt , it , vr1 , vr2 , vr3 , and vr4 for the circuit shown in figure 12. be sure to put your answer in proper engineering notation and use the correct units. using your calculations, verify your results using kirchhoff’s voltage law.
figure 12. series circuit
rt = 23.2 k ohms
it = |

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:

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