Name: Purely Inductive Circuit
Uploaded: 2019-07-19
Duration: 5 min 40 s
Description: How does an inductor behave in a circuit? This video will explain every parameter involved in the derivation of a pure inductor working with AC supply.

Question

An ac generator has emf

E = E_{m} \sin (ω_{d} t - π / 4),

where

E = 25.0 V

and

ω_{d} = 270 rad / s .

The current produced in a connected circuit is

i (t) = I \sin (ω_{d} t - 3 π / 4),

where

I = 620 mA

. At what time after

t = 0

does
(a) the generator emf first reach a maximum and (b) the current first reach a maximum?
(c) The circuit contains a single element other than the generator. Is it a capacitor, an inductor, or a resistor? Justify your answer.
(d) What is the value of the capacitance, inductance, or resistance, as the case may be?

Accepted Answer

Our circuit consists of an ac generator that produces an alternating current, as well as a load that could be purely resistive, capacitive, or inductive. The nature of the load can be determined by the phase angle between the current and the emf.

The generator emf and the current are given by

$ε = ε_{m} \sin (ω_{d} t - π / 4), i (t) = I \sin (ω_{d} t - 3 π / 4)$

The expressions show that the emf is maximum when $\sin (ω_{d} t - \frac{π}{4}) = 1$ or

$ω_{d} t - \frac{π}{4} = (\frac{π}{2}) \pm 2 n π$ [n = integer].

Similarly, the current is maximum when $\sin (ω_{d} t - 3 π / 4) = 1,$ or

$ω_{d} t - \frac{3 π}{4} = (\frac{π}{2}) \pm 2 n π$ [n = integer]

(a) The first time the emf reaches its maximum after $t = 0$ is when $ω_{d} t - \frac{π}{4} = \frac{π}{2}$ (that is, $n = 0$ ).

Therefore,

$t = \frac{3 π}{4 ω_{d}} = \frac{3 π}{4 \times 270 rad s^{- 1}} = 8.73 \times 10^{- 3} s$

(b) Current reaches maximum, when
$\Rightarrow ω_{d} t - \frac{3 π}{4} = (\frac{π}{2}) \pm 2 n π$
First time, it will be maximum, when $n = 0$
$\Rightarrow ω_{d} t = \frac{π}{2} + \frac{3 π}{4}$
$\Rightarrow ω_{d} t = \frac{5 π}{4}$
$\Rightarrow t = \frac{5 π}{4 ω_{d}} = \frac{5 π}{4 \times 270 rad s^{- 1}} = 14.54 \times 10^{- 3} s$

(c) The current lags the emf by $+ \frac{π}{2} rad$ , so the circuit element must be an inductor.

(d) The current amplitude $I$ is related to the voltage amplitude $V_{L}$ by $V_{L} = I X_{L},$ where $X_{L}$ is the inductive reactance, given by $X_{L} = ω_{d} L$ . Furthermore, since there is only one element in the circuit, the amplitude of the potential difference across the element must be the same as the amplitude of the generator emf: $V_{L} = ε_{m} .$ Thus, $ε_{m} = I ω_{b} L$ and

$L = \frac{ε_{m}}{I ω_{d}} = \frac{25.0 V}{(620 \times 10^{- 3} A) (270 rad s^{- 1})} = 0.149 H$

The current in the circuit can be rewritten as

$i (t) = I \sin (ω_{d} t - \frac{3 π}{4}) = I \sin (ω_{d} t - \frac{π}{4} - ϕ)$

where $ϕ = + \frac{π}{2}$ . In a purely inductive circuit, the current lags the voltage by $90 °$ .

Important Questions on Alternating Current

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