Question

using the laws of circuit theory, solve for rt, it, ir1, ir2, and ir3 for the circuit shown in figure 13. put your answer in proper engineering notation and use the correct units. 12v vt 8.2kω r1 6.8kω r2 4.7kω r3

Explanation:

Step1: Calculate total resistance ($R_T$)

For resistors in parallel, the formula is $\frac{1}{R_T}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$. Substituting $R_1 = 8.2\times10^{3}\Omega$, $R_2=6.8\times 10^{3}\Omega$, $R_3 = 4.7\times10^{3}\Omega$:
$\frac{1}{R_T}=\frac{1}{8.2\times10^{3}}+\frac{1}{6.8\times10^{3}}+\frac{1}{4.7\times10^{3}}$
$=\frac{6.8\times4.7 + 8.2\times4.7+8.2\times6.8}{8.2\times6.8\times4.7\times10^{9}}$
$=\frac{31.96+38.54 + 55.76}{259.912\times10^{9}}$
$=\frac{126.26}{259.912\times10^{9}}$
$R_T=\frac{259.912\times10^{9}}{126.26}\approx2.06\times10^{3}\Omega = 2.06k\Omega$

Step2: Calculate total current ($I_T$)

Using Ohm's law $I_T=\frac{V_T}{R_T}$, with $V_T = 12V$ and $R_T=2.06\times 10^{3}\Omega$.
$I_T=\frac{12}{2.06\times10^{3}}\approx5.83\times10^{-3}A=5.83mA$

Step3: Calculate current through $R_1$ ($I_{R1}$)

Using Ohm's law $I_{R1}=\frac{V_T}{R_1}$, with $V_T = 12V$ and $R_1 = 8.2\times10^{3}\Omega$.
$I_{R1}=\frac{12}{8.2\times10^{3}}\approx1.46\times10^{-3}A = 1.46mA$

Step4: Calculate current through $R_2$ ($I_{R2}$)

Using Ohm's law $I_{R2}=\frac{V_T}{R_2}$, with $V_T = 12V$ and $R_2=6.8\times10^{3}\Omega$.
$I_{R2}=\frac{12}{6.8\times10^{3}}\approx1.76\times10^{-3}A=1.76mA$

Step5: Calculate current through $R_3$ ($I_{R3}$)

Using Ohm's law $I_{R3}=\frac{V_T}{R_3}$, with $V_T = 12V$ and $R_3 = 4.7\times10^{3}\Omega$.
$I_{R3}=\frac{12}{4.7\times10^{3}}\approx2.55\times10^{-3}A = 2.55mA$

Answer:

$R_T = 2.06k\Omega$, $I_T=5.83mA$, $I_{R1}=1.46mA$, $I_{R2}=1.76mA$, $I_{R3}=2.55mA$