Question

Three resistors are connected in series with one another. When a supply of 100V is applied to the circuit, current of 0.75A flows in it. The potential difference across one resistor is found to be 18.85V and the power dissipated by the second is 2.8W. Determine the value of the remaining resistance.

Answer 1

$1)R_1=\frac{U_1}{I}=\frac{18.85}{0.75}=25.133Ohm;\\2)R_2=\frac{P}{I^2}=\frac{2.8}{0.75^2}=4.978Ohm;\\3)R_1+R_2+R_3=\frac{U}{I}; \\R_3=\frac{U}{I}-R_1-R_2=\frac{100}{0.75}-25.133-4.978=\\=133.33-25.133-4.978=103.22Ohm$

Checking the solution:

$I=\frac{U}{R}=\frac{100}{25.133+4.978+103.22}=0.75A$

Answer: The value of the remaining resistance is 103.22 (Ohm)