Question

a parallel circuit contains the following resistor values $r_1 = 82 \text{k}omega$, $r_2 = 75 \text{k}omega$, $r_3 = 56 \text{k}omega$, $r_4 = 62 \text{k}omega$, and the total current $(i_t)$ is $0.0143 \text{a}$. the voltage applied $(e_t)$ across the circuit is $square \text{v}$.

Explanation:

Step1: Find total resistance ($R_T$) in parallel

For parallel resistors, $\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4}$.
$R_1 = 82\ \text{k}\Omega = 82000\ \Omega$, $R_2 = 75000\ \Omega$, $R_3 = 56000\ \Omega$, $R_4 = 62000\ \Omega$.
$\frac{1}{R_T} = \frac{1}{82000} + \frac{1}{75000} + \frac{1}{56000} + \frac{1}{62000}$.
Calculate each term:
$\frac{1}{82000} \approx 1.2195\times10^{-5}$, $\frac{1}{75000} \approx 1.3333\times10^{-5}$, $\frac{1}{56000} \approx 1.7857\times10^{-5}$, $\frac{1}{62000} \approx 1.6129\times10^{-5}$.
Sum: $1.2195 + 1.3333 + 1.7857 + 1.6129 \approx 5.9514\times10^{-5}$.
Thus, $R_T = \frac{1}{5.9514\times10^{-5}} \approx 16802.7\ \Omega$.

Step2: Use Ohm's Law ($E_T = I_T \times R_T$)

$I_T = 0.0143\ \text{A}$, $R_T \approx 16802.7\ \Omega$.
$E_T = 0.0143 \times 16802.7 \approx 240.3$.

Answer:

$\approx 240$ (or more precisely, $\approx 240.3$)