In the circuit shown in Fig. 19-69, the 33 ohms resistor dissipates 0.44 W. What is the battery voltage?

In the circuit shown in Fig. 19-69, the 33 ohms resistor dissipates 0.44 W. What is the battery voltage?

_______ V



Figure 19-69 link....

http://img.photobucket.com/albums/v706/E...

In the circuit shown in Fig. 19-69, the 33 ohms resistor dissipates 0.44 W. What is the battery voltage?
If the 33-ohm resistor dissipates 0.44W, the current is given by W = I^2*R, so I = √[W/R] = √[0.44/33] = 0.115A



The overall resistance of the circuit is 33 + (68*75)/(68+75) = 33 + 35.6 = 68.66 ohm The voltage is then 68.66*0.115 = 7.9V
Reply:Calculate the total resistance of the circuit.



Rt = R1 + [R2 * R3/ R2 + R3]

Rt = 33 + [68*75/ 68 + 75]

Rt = 33 + 5100/143

Rt = 33 + 35?664 335 6...Ω

Rt = 68?664 335 66...Ω



33Ω = 0?44w

1 Ω = 0?44/ 33

1Ω = 0?013 333 333...w

35?664 335 6......Ω = (35?664 335 6......)( 0?013 333 333...)

35?664 335 6......Ω = 0?475 524 474...w



P = V2/R

V2 = PR

V = √(PR)



Total voltage:

Vt = √(P1R1) + √(P2R2)

Vt = √(0?44)(33) + √(0?475 524 474...)(35?664 335 6...)

Vt = 3?810 511... + 4?118 162 632...

Vt = 7?928 674 409...

Vt ≈ 7?93v