8B-Soln07-v1.0.5 - box delivers power to the circuit and...

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d) U dissipated ,5 R ( t ' = " ) = U stored , C ( t ' = 0) = Q 2 2 C = 25 98 CV batt 2 e) The presence of a dielectric increases C by a factor of the dielectric constant, k , so (a) will remain unchanged, (b) will increase, (c) will remain the same, and (d) will increase. 9. What we know: P batt = 36 W V batt = 12 V P fan = 24 W We can figure out I left by using P batt = V batt I left to get I left = 3 A, going CW around the left side of the circuit. Since we know the current going through the left resistor, we can figure out the power dissipated by that resistor: P Rleft = I left 2 R = 18 W. We can also figure out the voltage drop over the entire left-hand side: V left = V batt - I left R left = 6 V. The left side of the circuit, the right side of the circuit, and the fan are all in parallel with eachother, so V right = 6V. Energy is conserved, so the total power delivered to the circuit must equal the power used and dissipated. Therefore, P batt = P Rleft + P Rright + P fan + P box . We then get that P box + I right 2 R = -6 W. I right 2 R is inherently positive, so P box is negative. Therefore, the
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Unformatted text preview: box delivers power to the circuit and must be a battery, whose voltage we will call V box . We get the following three equations, then, for the three unknowns I fan , V box , I right : (note that we define V box to be the voltage gain from the right to the left side of the box, define I right as moving CCW in the right-hand portion of the circuit, and define I fan as going downward.) I left + I right = I fan 6 V = V box- I right R-I right V box + I right 2 R = -6 W We can rewrite the third equation as I right * ( V box- I right R ) = 6 W. Whats in the parentheses, though, is just the right-hand side of the second equation! Therefore, the equation becomes I right * 6 V = 6 W, which means I right = +1 A. This lets us solve everything else, so we get: a) I fan = 4 Amps, moving downward b) The box contains a battery of V box = 8 V, oriented so that its positive terminal is on the left-hand side of the box....
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