Given: C = 2F. Voltage v(t) is piecewise-linear (from graph):
Reading piecewise v(t) from the graph (straight lines between points):
0<t<2: v = 5t V (slope = +5 V/s)
2<t<4: v = 10 V (flat)
4<t<5: v = −5t+30 V (slope = −5)
5<t<6: v = 5 V (flat)
6<t<7: v = −5t+35 V (slope = −5)
i(t) = C·dv/dt = 2 × slope:
0<t<2: i = 2×5 = +10 A
2<t<4: i = 2×0 = 0 A
4<t<5: i = 2×(−5) = −10 A
5<t<6: i = 2×0 = 0 A
6<t<7: i = 2×(−5) = −10 A
Sketch of i(t): current is +10A when v is rising, 0 when v is constant, −10A when v is falling. i = C·dv/dt = 2 × slope.
Tutorial 10 · Problem 4 (6.2) — L=500mH triangular current pulse → v, p, w
Given: L = 0.5H. Triangular current pulse (peak=100mA at t=25ms).
(a) i(t) expressions in 4 intervals:
Using straight-line equation between (0,0)→(25ms,100mA) and (25ms,100mA)→(50ms,0):
t<0: i = 0 mA
0<t<25ms: i = 4t mA (slope = 100mA/25ms = 4 A/s)
25ms<t<50ms: i = −4t + 200 mA (slope = −4 A/s)
t>50ms: i = 0 mA
(b) Voltage v = L·di/dt:
t<0: v = 0.5 × 0 = 0 V
0<t<25ms: v = 0.5 × 4×10⁻³ = +2 mV (wait: di/dt = 4 A/s → v = 0.5×4 = 2V... checking units: i=4t mA → di/dt=4×10⁻³/10⁻³ = 4 A/s → v = 2V)
25ms<t<50ms: v = 0.5 × (−4) = −2 V
t>50ms: v = 0 V
Power p = v·i (passive sign convention):
t<0: p = 0
0<t<25ms: p = 2 × 4t×10⁻³ = 8t mW
25ms<t<50ms: p = (−2) × (−4t+200)×10⁻³ = (8t−400) mW
t>50ms: p = 0
Energy w = ∫p dt:
0<t<25ms: w = ∫8t dt = 4t² mJ = ½Li² = ½×0.5×(4t×10⁻³)² ✓
At t=25ms: w_max = 4×(25)² = 2500 μJ = 2.5 mJ
v: {0, +2V, −2V, 0} | p: {0, 8t mW, (8t−400)mW, 0} Energy stored: 0→25ms | Energy extracted: 25ms→50ms
Problem 4 sketches: i(t) triangular pulse, v(t) rectangular pulse (L·di/dt), p(t) = v·i. Energy stored when p>0 (0→25ms), extracted when p<0 (25→50ms).
Tutorial 10 · Problem 5 — 47μF capacitor with sinusoidal voltage → sketch i(t)
Given: C = 47μF, V(t) = 20 sin(200πt) V. Find i(t) and sketch.
Key points:
t=0: i = 0.591 A (maximum, v=0 rising)
t=T/4=2.5ms: i = 0 (v=maximum, flat)
t=T/2=5ms: i = −0.591 A (minimum, v=0 falling)
t=3T/4=7.5ms: i = 0 (v=minimum, flat)
t=T=10ms: i = 0.591 A (cycle repeats)
i(t) = 0.591 cos(200πt) A | Amplitude = 0.591A | Period = 10ms Note: current LEADS voltage by 90° — capacitor property!
Problem 5: i(t) = 0.591 cos(200πt) A. Period = 10ms. Current is a cosine while voltage is a sine — current leads voltage by 90° in a capacitor.
Tutorial 10 · Problem 6 — C=0.5μF: find i, p, w and sketch all four waveforms