Question

4. Three 100 Sresistors are connected, as shown in Figure 4. The maximum power that can safely be dissipated in any one resistor is 26.5 W. (a) What is the maximum voltage that can be applied to the terminals a and b? (b) For the voltage determined in part (a), what is the power dissipated in each resistor? (c) What is the total power dissipation? 5. A battery has an emf of 15.0 V. The terminal voltage of the battery is 11.8 V when it is delivering 14.0 W of power to an external load resistor R. (a) What is the value of R? (b) What is the internal resistance of the battery? 6. The quantity of charge q (in coulombs) that has passed through a surface area 2.00cm2 varies with time according to the equation q=4t3+5+6, where t is in seconds (a) What is the instantaneous current across the surface at t=2:00s? (b) what is the value of the current density? 15.0 V 7.002 7.00 2 W 9.00 12 5.00 W 10.0 52 2.00 Figure 2 Figure 3 Figure 1 1000 M 1002 1002 Figure 4

Answer 1

4)

(a) Each resistor can safely bear a power input = 26.5 W.

Hence the maximum current allowed through any of the given resistor

( I ) = √[ power / Resistance ]

=> I = √[ 26.5 / 100 ] = 0.5148 A

Hence considering the assumed circuit the resistance in series can have maximum current = 0.5196 A, while each of the two connected in parallel will have current = (1/2) ( 0.5196 ) = 0.2574 A

Again total circuit resistance = 100 + (100/2) = 150 Ω

Hence the maximum permissible voltage = 0.5148 x 150 = 77.22 V

(b)

Power delivered to the series resistor = (current)^2 x Resistance

=> (0.5148)^2 x 100 = 26.5 W

Power delivered to each of the resistances in parallel = (0.2574)^2 x 100 Ώ = 6.63 W

(c) Total Power delivered = 26.5 + (1/2) (26.5) = 39.75 W