In the circuit shown in the figure , EMF = 60.0 V, R1 = 36.0 %u03A9, R2 = 25.0 %
ID: 2117173 • Letter: I
Question
In the circuit shown in the figure , EMF = 60.0 V, R1 = 36.0 %u03A9, R2 = 25.0 %u03A9, and L = 0.310 H. Switch S is closed at t=0.Just after the switch is closed, what is the potential difference v_{ab} across the resistor R1?
Which point, a or b, is at a higher potential?
What is the potential difference vcd across the inductor L?
Which point, c or d, is at a higher potential?
The switch is left closed a long time and then opened. Just after the switch is opened, what is the potential difference vab across the resistor R1?
Which point, a or b, is at a higher potential?
What is the potential difference vcd across the inductor L?
Explanation / Answer
1)
Just after the switch is closed , the series connection of the c=inductor and R2 acts as open circuit , so only R1 comes into the connection, to emf
So, Vab = emf = 60 V <--------------answer
2)
a is at higher potential than b
3)
potential difference across inductor = 60 V
4)
c is at higher potential
5)
after a long time, the series circuit of inductor and R2 comes into picture,
So, after a LONG TIME, R1 and R2 are in parallel,
So, potential across R1 = EMF = 60 V
6)
a is at higher potential than b
7)
potential differecnce across inductor = 0 <---as it gets shorted(current is constant)