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