Circuit A in a house has a voltage of 208 V and is limited by a 40.0-A circuit breaker. Circuit B is at 120.0 V and has a 20.0-A
1 answer:
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Complete Question:
Two 3.0µC charges lie on the x-axis, one at the origin and the other at 2.0m. A third point is located at 6.0m. What is the potential at this third point relative to infinity? (The value of k is 9.0*10^9 N.m^2/C^2)
Answer:
The potential due to these charges is 11250 V
Explanation:
Potential V is given as;
where;
K is coulomb's constant = 9x10⁹ N.m²/C²
r is the distance of the charge
q is the magnitude of the charge
The first charge located at the origin, is 6.0 m from the third charge; the potential at this point is:
The second charge located at 2.0 m, is 4.0 m from the third charge; the potential at this point is:
Total potential due to this charges = 4500 V + 6750 V = 11250 V