In Fig.30.11 in the textbook, suppose that E = 61.0 V , R = 200 ? and L = 0.140
ID: 1508517 • Letter: I
Question
In Fig.30.11 in the textbook, suppose that E = 61.0 V , R = 200 ? and L = 0.140 H . With switch S2open, switch S1 is left closed until a constant current is established. Then S2 is closed and S1opened, taking the battery out of the circuit.
Part A
What is the initial current in the resistor, just after S2 is closed and S1 is opened?
I = .305A (V = I*R ; I = V/R)
Part B
What is the current in the resistor at t = 4.20×10?4 s ?
I = .167
Part C
What is the potential difference between points b and c at t = 4.20×10?4 s ?
need some explanation here!!!
Part D
Which point is at a higher potential? ( B or C)
Part E
How long does it take the current to decrease to half its initial value?
Explanation / Answer
(a) Once current is established in the circuit , then steady current will be established
When we disconnect the battery then current will start decreasing exponentially
I(t) = Ioe-(t/T)
where Io is the maximum current , t is the time and T is the time constant
where T = (L/R) , L is inductance and R is the resistance
Since we are calculating the current at the istant when we disconnected the battery that means
at t= 0
hence from the above formula
I(0) = Ioe-(0/T) = Io
And we know that Io = V/R = 61/200 = 0.305 A
(b) at t= 4.2*10-4 s
T = L/R = o.14/200 = 7*10-4
t/T = 4.2*10-4 / 7*10-4 = 0.6
I(0) = Ioe-(t/T) = 0.305*e-(0.6) = 0.167 A
(c) Voltage across inductor is given by
V =- Vo e-(t/T) = 61*e-(0.6) = -33.48 Volt
(d) Here current is flowing in the direction of a to b
therefore a will be at higher potential than b.
But since the voltage across the inductor is negative therefore the potential at C will be higher than B.
(e) We will use the same formula
I(t) = Ioe-(t/T)
I(t) = Io/2 is given , therefore
o.5Io = Ioe-(t/T)
0.5 = e-(t/T)
taking natural log both side
ln(0.5) = -(t/T)
-0.6931 = -(t/T)
t = 0.6931*T = 0.6931*7*10-4 = 4.852*10-4 s