Question 4: The voltage across a 0.01 μF capacitor is 240sin(1.4×104t – 300) volts. Write the mathematical expression for current through it.
Capacitance, C = 0.01 μF = 0.01×10-6 F,
Instantaneous voltage, V = 240sin(1.4×104t – 300) volts
Instantaneous Current, I = ?
(i) We know the general sinusoidal voltage is given by the equation V = Vmsinωt, where V is the instantaneous value of the voltage, Vm is the maximum value of the voltage in a cycle and ωt is the angle at a particular instant.
The given voltage is V = 240sin(1.25×104t-30) volts. Therefore, comparing with the general voltage equation,
Vm = 240 V … (i),
ω = 1.25×104 … (ii)
(ii) Similarly, in a capacitor, the current is leading the voltage by 900 and the general equation for the sinusoidal current in a capacitive circuit is
I = CVmω sin(ωt + 90) … (1)
Put values in equation (1) from the given data and equations (i) and (ii),
I = 0.01×10-6 × 240 × 1.25×104 sin(1.25×104-30+90)
= 0.01×240×1.25×10-2 sin(1.25×104t+60)
I = 0.03sin(1.25×104t+60) Amp (Answer)
Pingback:Numerical Problem 5, Alternating Current Circuit, Physics 12 … msa – msa
Pingback:numerical-problem-3-alternating-current-physics-12 – msa
Pingback:solved-numerical-problems-alternating-current-circuit-physics-12-msa – msa