Potential, Internal Resistance, Dissipated Power When charges q_a, q_b, and q_c

Question

Potential, Internal Resistance, Dissipated Power

When charges q_a, q_b, and q_c are placed respectively at the corners a, b, and c of a right triangle, the potential hi the midpoint of the hypotenuse is 20 V When the charge q, is removed the potential at the midpoint becomes 15 V When, instead, the charge q_b removed (q_a and q_c both in place), the potential al the midpoint becomes 12 V, What is the potential at the midpoint charges q_a and q_c are removed? 8 V 5 V 7 V 12 V 13 V A battery rated at 9.00 V yields a 4.00 A current when a 2.00 Ohm resistor is connected in series with the battery. We can conclude that the internal resistance of the battery is 0.25 V 0.25 Ohm. 0.50 V 0.50 Ohm not known. A heater with 2 Ohm resistance is plugged into a 120 V circuit. How much power is dissipated? 7200 W 3600 W 1800 W 900 W 450 W

Explanation / Answer

Potential at midpoint of hypotenuse = V = Va + Vb + Vc
20 = Va + Vb + Vc
15 = Vb + Vc, Vc = 15 - Vb
12 = Va + Vc, Va = 12 - Vc = 12 - 15 + Vb = Vb - 3
so, Vb = 20 - Va - Vc = 20 - Vb + 3 - 15 + Vb
Vb = 8 V

Internal resistance = r
9 = 4*(r + 2)
r = 0.25 ohm

P = V^2/R = 7200 W

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