ENGR 1990 Engineering Mathematics Application of Derivatives in Electrical Engineering The diagram shows a typical element (resistor, capacitor, inductor, etc.) in an electrical circuit. Here, v (t ) represents the voltage across the element, and i (t ) represents the current flowing through the element. Both are generally functions of time, t . For any such element, the following equations apply. dw v (t ) = dq i (t ) = dq dt p (t ) = v (t ) + circuit element − i (t ) ⎧ v(t ) is the voltage (volts) ⎪ ⎨ w is the energy (joules) ⎪q is the charge (coulombs) ⎩ ⎧i (t ) is the current (amps) ⎨ ⎩ t is the time (sec) dw ⎛ dw ⎞ ⎛ dq ⎞ =⎜ ⎟ ⎜ ⎟ = v(t ) ⋅ i (t ) dt ⎝ dq ⎠ ⎝ dt ⎠ { p(t ) is the power (joules/sec) or (watts) This last equation is an application of the chain rule. Example 1: Given: The charge and voltage for a given circuit element are given by the following 1 sin(250 π t ) (coulombs) and v(t ) = 100sin(250π t ) (volts) . equations: q(t ) = 50 Find: a) i (t ) the current passing through the element, b) p (t ) the power dissipated by the element, and c) pmax the maximum power dissipated by the element. Solution: dq d 1 1 × cos(250 π t ) × 250 π = 5π cos(250 π t ) (amps) = ( 50 sin(250 π t ) ) = 50 dt dt b) p (t ) = v(t ) ⋅ i (t ) = (100sin(250π t ) )( 5π cos(250 π t ) ) a) i (t ) = = 500π sin(250π t )cos(250 π t ) (watts) c) To find maximum power, we use the trigonometric identity: 2sin(θ )cos(θ ) = sin(2θ ) p(t ) = 500π sin(250π t )cos(250π t ) = 12 ( 500π sin(500π t ) ) = 250π sin(500π t ) (watts) ⇒ pmax = 250 π ≈ 785 (watts) 1/4 Current–Voltage Relationships for Resistors, Capacitors and Inductors The voltage across and the current through a resistor are related simply by its resistance. R v (t ) v(t ) = Ri (t ) or i (t ) = v(t ) R Given a voltage v (t ) applied to a capacitor, the corresponding current i (t ) can be calculated as i (t ) = C i (t ) dv dt i (t ) v (t ) C Given the current i (t ) passing through an inductor, the corresponding voltage v (t ) can be calculated as v (t ) = L di dt i (t ) L v (t ) Example 2: Given: A voltage v (t ) = 110cos(120π t ) (volts) is applied to a capacitor with C = 100 (μ f ) . Find: a) i (t ) , the current through the capacitor, and b) the maximum power pmax . i (t ) v (t ) C Solution: a) The current may be found by differentiating the voltage. dv d = (100 × 10−6 ) (110cos(120 π t ) ) dt dt = (100 × 10−6 ) (110 )(120 π )( − sin(120 π t ) ) i (t ) = C = −4.147sin(120 π t ) (amps) b) p (t ) = v(t ) ⋅ i(t ) = (110cos(120 π t ) )( −4.147sin(120 π t ) ) = −456sin(120 π t )cos(120 π t ) = −456 × 12 sin(240 π t ) = −228sin(240 π t ) (watts) ⇒ pmax = 228 (watts) 2/4 Alternate Solution for part (a): (without using derivatives) We could have solved this problem using complex numbers. Z C = − j ωC = − j (120 π ) (100 × 10−6 ) = − j (106 ) (120 π )(100 ) = − j 26.526 (ohms) = 26.526∠ ( −90o ) I= 110∠ ( 0o ) 26.526∠ ( −90 o ) = 4.147∠ ( 90o ) i (t ) = 4.147cos(120 π t + 90o ) (amps) ⎛ ⎞ = 4.147 ⎜ cos(120 π t )cos(90o ) − sin(120 π t )sin(90o ) ⎟ ⎟ ⎜ 0 1 ⎝ ⎠ = −4.147sin(120 π t ) (amps) Example 3: Given: A voltage v (t ) = 110 e−10 t cos(120π t ) (volts) is applied to a capacitor with C = 100 (μf ) . i (t ) v (t ) Find: i (t ) , the current through the capacitor C Solution: To find the current we differentiate the voltage using the product and chain rules. dv d i (t ) = C = (100 × 10−6 ) (110 e −10t cos(120 π t ) ) dt dt d ⎡⎛ d ⎞ ⎛ ⎞⎤ = (100 × 10−6 ) (110 ) ⎢⎜ ( e −10t ) × ( cos(120 π t ) ) ⎟ + ⎜ ( e −10t ) × ( cos(120 π t ) ) ⎟ ⎥ dt ⎠ ⎝ ⎠⎦ ⎣⎝ dt = 0.011 ⎡⎣( −10e −10t cos(120 π t ) ) + ( −120 π e −10t sin(120 π t ) ) ⎤⎦ = −0.011e −10t [10cos(120 π t ) + 120 π sin(120 π t ) ] (amps) The term in square brackets can be reduced to a single, phase-shifted sine or cosine function. For example, 120 π sin(120 π t ) + 10cos(120 π t ) = M cos(120 π t + φ ) where M = (120π )2 + 102 ≈ 377.1 and φ = tan −1 (−120π / 10) = −88.48o = −1.544 (rad) . So, i (t ) = −4.15 e−10 t cos(120 π t − 1.544) (amps) 3/4 Example 4: Given: A current i (t ) = 5 t e−10 t (amps) is applied to an inductor with L = 250 (mh) . i (t ) v (t ) L Find: v (t ) , the voltage across the inductor. Solution: To find the voltage we differentiate the current using the product and chain rules. v (t ) = L di d d ⎛d ⎞ ⎛ ⎞ = 0.25 ( 5 t e−10 t ) = 14 ⎜ ( 5t ) ⎟ ( e−10 t ) + 14 ⎜ ( 5t ) ( e−10 t ) ⎟ dt dt dt ⎝ dt ⎠ ⎝ ⎠ ( ) = 14 ( 5e−10 t ) + 14 5t ( −10e−10 t ) = 14 ( 5e−10 t − 50 t e−10 t ) = 5 e−10 t 4 (1 − 10 t ) (volts) 4/4