In the circuit shown in the figure ( Figure 1 ) , E = 60.0V , R 1 = 39.0? , R 2
ID: 1291001 • Letter: I
Question
In the circuit shown in the figure (Figure 1) , E = 60.0V ,R1 = 39.0? , R2 = 27.0? , and L = 0.295H . Switch S is closed at t=0.
Just after the switch is closed, what is the potential difference vab across the resistor R1?
What is the potential difference v_cd across the inductor L?
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?
What is the potential difference vcd across the inductor L?
In the circuit shown in the figure (Figure 1) , E = 60.0V ,R1 = 39.0? , R2 = 27.0? , and L = 0.295H . Switch S is closed at t=0. Just after the switch is closed, what is the potential difference vab across the resistor R1? What is the potential difference v_cd across the inductor L? 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? What is the potential difference vcd across the inductor L?Explanation / Answer
a)
Immediately after switch is closed inductor acts as open circuit ,so potential difference across R1 is
Vab=E=60 Volts
b)
Potential difference across inductor is
VCd=E=60 V
c)
afte long time inductor acts as short circuit ,so current flowing through inductor is
IL=E/R2=60/27=2.22 A
After switch is reopened ,inductor doesn't allow sudden change ,so
Vab=-IL*R1=-2.22*39 =-86.67 Volts
d)
VCd=-86.67-60=-146.67 Volts