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

We know that in series connection, the current is the same in all resistors. Therefore, we can find the resistance of the first resistor where the voltage drops to 18.85 V:

$$R_1=\frac{V_1}I=\frac{18.85}{0.75}=25.1\space\Omega.$$

The second resistance can be found from the dissipated power:

$$P_2=I^2R_2,\\ R_2=\frac{P_2}{I^2}=3.73\space\Omega.$$

The equivalent resistance of resistors connected in series equals total voltage over total current. This will help us to find the remaining resistance:

$$R_1+R_2+R_3=\frac{V}I,\\ R_3=\frac{V}I-R_1-R_2=\\ =\frac{100}{0.75}-25.1-3.73=104.5\space\Omega.\\$$