using the laws of circuit theory, solve for rt, it, ir1, ir2, and ir3 for the circuit shown in figure 13…

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:

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