Question

the circuit shown in figure 15 is a compound circuit which contains a series and parallel circuit. that is, some of its resistors are connected in series and some are in parallel. using the laws of circuit theory, solve for rt and it. 12v vt 150ω r1 200ω r2 1.8kω r4 470ω r3 270ω r5 rt

Explanation:

Step1: Calculate parallel - resistance of R2 and R3

First, find the equivalent resistance $R_{23}$ of $R2 = 200\Omega$ and $R3=470\Omega$ in parallel. The formula for two - resistors in parallel is $R_{eq}=\frac{R_1\times R_2}{R_1 + R_2}$. So, $R_{23}=\frac{200\times470}{200 + 470}=\frac{94000}{670}\approx140.3\Omega$.

Step2: Calculate total parallel - resistance of the parallel part

Now, find the equivalent resistance of $R_{23}$ and $R4 = 1800\Omega$ in parallel. Let $R_{p}$ be the equivalent resistance of the parallel part. Using the parallel - resistance formula again, $R_{p}=\frac{R_{23}\times R4}{R_{23}+R4}=\frac{140.3\times1800}{140.3 + 1800}=\frac{252540}{1940.3}\approx130.2\Omega$.

Step3: Calculate total resistance RT

The resistors $R1 = 150\Omega$, $R5 = 270\Omega$ and $R_{p}$ are in series. The total resistance $R_T=R1 + R5+R_{p}=150+270 + 130.2=550.2\Omega$.

Step4: Calculate total current IT

Using Ohm's law $I=\frac{V}{R}$, where $V = 12V$ and $R = R_T$. So, $I_T=\frac{V_T}{R_T}=\frac{12}{550.2}\approx0.0218A = 21.8mA$.

Answer:

$R_T\approx550.2\Omega$, $I_T\approx21.8mA$