Explanation:
(a) A circuit has 50 Ω, a 100Ω and a 150 Ω resistor are connected in series. We know that in series combination, current through each resistor is same. So, current through a 50 Ω, a 100Ω and a 150 Ω resistor is same.
(b) Ohm's law of given by :
V = I R
V is potential difference. As I is same, so, the resistor having highest resistance will have highest potential difference. So,
.
Hence, this is the required solution.
Answer:
the final potential energy of this system is 3U0/10
Explanation:
We are given
charge at left end and another test charge at point p
Potential energy is given by =
where k is electrostatics constant =
Q1 = first charge , Q2= test charge
R= distance between charges
potential at point p
U0 = k*Q1*Q2 /3 ⇒ kq1q2 = 3U0 ..............1
now the test charge moves to point R
using Pytahgoreou theorem
R(distance) = = 10
New Potential energy
U1 = kq1*q2 / 10
substituting kq1q2 = 3U0 from 1
U1 = 3U0/10
So this is the final potential energy of this system.