Question

1. For the circuit below: (a) Show the complete calculation of the equivalent resistance for the four resistors. (b) Show the complete calculation of the current from the battery. (c) Show the complete calculation of the power dissipated in the 36 Ω resistor. 252 V, 25 Ω, 35 Ω; 36 Ω and 18 Ω in parallel.

Answer 1

Solution. According to the condition of the problem R₁=25Ω, R₂=35Ω, R₃=36Ω, R₄=18Ω, U=252V

a) For resistors connected in parallel, the formula of the equivalent resistance 

$$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\frac{1}{R_4}$$

$$R=\frac{R_1R_2R_3R_4}{R_2R_3R_4+R_1R_3R_4+R_1R_2R_4+R_1R_2R_3}$$ $$R=\frac{25\times 35\times36\times18}{ 35\times36\times18+25\times 36\times18+25\times 35\times18+25\times 35\times36}\approx6.583\Omega$$

b) Using Ohm's law we find the total current

$$I=\frac{U}{R}=\frac{252V}{6.583\Omega}\approx38.28A$$

c) The power dissipated in the 36 Ω resistor can be calculated by the formula

$$P=U_3I_3=\frac{U_3^2}{R_3}=\frac{U^2}{R_3}=\frac{252^2}{36}=1764W$$

(the voltage across each resistor for parallel connection is equal to the total voltage).

Answer. a) R=6.583Ω b) 38.28A c) 1764W