(c) The voltage, v, across an inductor L, in an electrical circuit drops Exponentially over time, t. The relation is v = Ee-t/T where T = /R: %3D A partially complete table of values of v for t = 0.00, 0.01, 0.02 ...0.05 s, in the case where L = 3H, R = 1202 and E = 6.5OV, is given in table Q2(b). Table Q2(b) Time (s) 0.00 0.01 0.02 0.03 0.04 0.05 Voltage(V) 4.36 1.96 0.88 (i) Complete table Q2(b) and plot the curve of v against t in the given range. (ii) Use the graph to find the voltage at 0.025 s. (iii) Use the graph to estimate the time taken to halve the initial voltage, E.

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(c) The voltage, v, across an inductor L, in an electrical circuit drops
Exponentially over time, t. The relation is v = Ee¬t/T where T =
A partially complete table of values of v for t= 0.00, 0.01, 0.02 .0.05 s,
in the case where L = 3H, R = 1202 and E = 6.50V, is given in table
Q2(b).
Table Q2(b)
Time (s)
Voltage(V)
0.00 0.01
0.02 | 0.03
0.04
0.05
4.36
1.96
0.88
(i) Complete table Q2(b) and plot the curve of v against t in the given
range.
(ii) Use the graph to find the voltage at 0.025 s.
(iii) Use the graph to estimate the time taken
to halve the initial voltage, E.
Transcribed Image Text:(c) The voltage, v, across an inductor L, in an electrical circuit drops Exponentially over time, t. The relation is v = Ee¬t/T where T = A partially complete table of values of v for t= 0.00, 0.01, 0.02 .0.05 s, in the case where L = 3H, R = 1202 and E = 6.50V, is given in table Q2(b). Table Q2(b) Time (s) Voltage(V) 0.00 0.01 0.02 | 0.03 0.04 0.05 4.36 1.96 0.88 (i) Complete table Q2(b) and plot the curve of v against t in the given range. (ii) Use the graph to find the voltage at 0.025 s. (iii) Use the graph to estimate the time taken to halve the initial voltage, E.
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