Answer to Question #185603 in Electric Circuits for James

Question

A coil is joined in series with a pure resistor of resistance 80002 across a 100V, 50 Hz supply. The reading on a voltmeter across the coil is 45V and across the resistor is 80V. Determine the inductance and resistance of the coil. (ANS: 98.43812, 439.112 (1.398H)).

Answer 1

Explanations & calculations

• Consider the phasor diagram attached

• The current flowing in the circuit can be calculated using the given information for the pure resistor.

$\qquad\qquad \begin{aligned} \small V&=\small iR\\ \small I_{total}&=\small \frac{80V}{80002\Omega}=\bold{9.99\times10^{-4}A} \end{aligned}$

• For this value of current, the given results are not received but for a value of $\small R=800\Omega$ , the given results are obtained.

$\qquad\qquad \begin{aligned} \small I&=\small \frac{80V}{800\Omega}=0.1A \end{aligned}$

• To find the angle by which the coil voltage leads the circuit current, apply the parallelogram theorem.

$\qquad\qquad \begin{aligned} \small R&=\small \sqrt{P^2+Q^2+2PQ\cos\theta}\\ \small 100\Omega&=\small \sqrt{80^2+45^2+2(80)(45)\cos\theta}\\ \small \theta&=\small 77.36437^0 \end{aligned}$

• Then considering the embedded sub-phasor diagram of the coil in the main diagram,

Voltage dropped for the coil resistance,

$\qquad\qquad \begin{aligned} \small V_{R,C}&=\small 45V\times\cos\theta=45\cos(77.36437)\\ &=\small 9.84375V \end{aligned}$

Since the same current flows through this resistance,

$\qquad\qquad \begin{aligned} \small R_{coil}&=\small \frac{9.84375V}{0.1A}=\bold{98.4375\,\Omega} \end{aligned}$

Voltage dropped for the inductance of the coil,

$\qquad\qquad \begin{aligned} \small V_{X,C}&=\small 45V\times\sin\theta=45\times\sin(77.36437)\\ &=\small 43.910V \end{aligned}$

For a coil of inductance L,

$\qquad\qquad \begin{aligned} \small V&=\small 2\pi fL.i \end{aligned}$

Using this, the inductance can be calculated (also the same current flows here)

$\qquad\qquad \begin{aligned} \small 43.910V&=\small 2\pi\times50Hz\times L\times0.1A\\ \small L&=\small \bold{1.3977H} \end{aligned}$

Impedance of the coil if needed

$\qquad\qquad \begin{aligned} \small Z_{coil}&=\small 2\times\pi\times50Hz\times1.3977H=439.1004\Omega \end{aligned}$