Calculate the voltage difference V_a - V_b as a function of time if the coil has

Question

Calculate the voltage difference V_a - V_b as a function of time if the coil has 3 Henries and a I(t) = 5 t-3t^2 [in SI units] In the circuit shown, the charge on the capacitor changes according to the formula Q = 5 t [t is in seconds, Q is in Coulons] epsilon(t) is a time-dependent voltage source (a) Calculate the current in the circuit and the potential difference V_b - V_c. (b) Here R = 3 ohms. Calculate the potential difference V_a - V_b. (c) Calculate the voltage of the voltage source epsilon(t) at t = 0 sec.

Explanation / Answer

Q3. A. Given:

Inductance = L = 3 H

I(t) = 5t - 3t2

potential differene = L x dI(t)/dt

= 3 x d/dt[5t - 3t2]

= 3 x [ 5 -6t]

= 15 -18t

B. Here charge is changing Q=5t

so, current=i dQ/dt =5 A

a.potential differene across inductor= Vb-Vc= L x dI(t)/dt

=L x d/dt[5]

= L x 0 =0

b. resistance R= 3ohm

potential difference across the resistor = current x resistance= 5 x 3=15 V

so, Va-Vb = 15 V

c. At any point of time,

the total voltage of the voltage source= potential difference across resistor + potential difference across capacitor + potential difference across inductor

= 15 + charge/capacitance +0

= 15 + 5t/C

at time t=0, the total voltage of the voltage source= 15 + 5 x0/C=15+0=15 V

ALL THE BEST IN THE COURSE WORK

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