using the laws of circuit theory, solve for rt, it, ir1, ir2, ir3, and ir4 for the circuit shown in figure…

Explanation:

Step1: Calculate total resistance ($R_T$)

For parallel - connected resistors $R_1 = 2.7\times10^{3}\Omega$, $R_2 = 10\times10^{3}\Omega$, $R_3 = 3.3\times10^{3}\Omega$, $R_4 = 8.2\times10^{3}\Omega$, the formula for total resistance of parallel resistors is $\frac{1}{R_T}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\frac{1}{R_4}$. $\frac{1}{R_T}=\frac{1}{2.7\times 10^{3}}+\frac{1}{10\times 10^{3}}+\frac{1}{3.3\times 10^{3}}+\frac{1}{8.2\times 10^{3}}$ $=\frac{10\times3.3\times8.2 + 2.7\times3.3\times8.2+2.7\times10\times8.2 + 2.7\times10\times3.3}{2.7\times10\times3.3\times8.2\times10^{3}}$ $=\frac{2706 + 734.58+2214+891}{729.18\times10^{3}}$ $=\frac{6545.58}{729.18\times10^{3}}$ $R_T=\frac{729.18\times10^{3}}{6545.58}\approx111.4\Omega = 1.114\times 10^{2}\Omega$

Step2: Calculate total current ($I_T$)

Using Ohm's law $I_T=\frac{V_T}{R_T}$, where $V_T = 6V$. $I_T=\frac{6}{1.114\times 10^{2}}\approx53.95mA=5.395\times 10^{-2}A$

Step3: Calculate current through each resistor

Since the voltage across each resistor in a parallel - circuit is the same ($V = V_T=6V$), using Ohm's law $I=\frac{V}{R}$ for each resistor:

For $R_1$:

$I_{R1}=\frac{V_T}{R_1}=\frac{6}{2.7\times 10^{3}}\approx2.22mA = 2.22\times 10^{-3}A$

For $R_2$:

$I_{R2}=\frac{V_T}{R_2}=\frac{6}{10\times 10^{3}} = 0.6mA=6\times 10^{-4}A$

For $R_3$:

$I_{R3}=\frac{V_T}{R_3}=\frac{6}{3.3\times 10^{3}}\approx1.82mA = 1.82\times 10^{-3}A$

For $R_4$:

$I_{R4}=\frac{V_T}{R_4}=\frac{6}{8.2\times 10^{3}}\approx0.73mA = 7.3\times 10^{-4}A$

Answer:

$R_T = 1.114\times 10^{2}\Omega$, $I_T=5.395\times 10^{-2}A$, $I_{R1}=2.22\times 10^{-3}A$, $I_{R2}=6\times 10^{-4}A$, $I_{R3}=1.82\times 10^{-3}A$, $I_{R4}=7.3\times 10^{-4}A$