Question

a parallel circuit contains the following resistor values ( r_1 = 82space komega ), ( r_2 = 75space komega ), ( r_3 = 56space komega ), ( r_4 = 62space komega ), and the total current (( i_t )) is ( 0.0143space a ). the voltage applied (( e_t )) across the circuit is ( 240\times v ).

Explanation:

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

For parallel resistors, \(\frac{1}{R_T}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\frac{1}{R_4}\).
\(R_1 = 82\,\text{k}\Omega=82000\,\Omega\), \(R_2 = 75\,\text{k}\Omega=75000\,\Omega\), \(R_3 = 56\,\text{k}\Omega=56000\,\Omega\), \(R_4 = 62\,\text{k}\Omega=62000\,\Omega\).
\(\frac{1}{R_T}=\frac{1}{82000}+\frac{1}{75000}+\frac{1}{56000}+\frac{1}{62000}\).
Calculate each term:
\(\frac{1}{82000}\approx1.2195\times10^{-5}\), \(\frac{1}{75000}\approx1.3333\times10^{-5}\), \(\frac{1}{56000}\approx1.7857\times10^{-5}\), \(\frac{1}{62000}\approx1.6129\times10^{-5}\).
Sum: \(1.2195 + 1.3333 + 1.7857 + 1.6129 = 5.9514\times10^{-5}\).
Thus, \(R_T=\frac{1}{5.9514\times10^{-5}}\approx16802.7\,\Omega\approx16.8\,\text{k}\Omega\).

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

\(I_T = 0.0143\,\text{A}\), \(R_T\approx16802.7\,\Omega\).
\(E_T = 0.0143\times16802.7\approx240.28\,\text{V}\approx240\,\text{V}\).

Answer:

\(240\) (The calculated value confirms the given voltage of \(240\,\text{V}\) is correct.)