Question:

Published on: 22 June, 2022

**A coil of resistance 10 Ω and inductance 0.02 Hs connected in series with another coil of resistance 6 Ω and inductance 15 mH across a 230 V, 50 Hz supply. calculate**

**Impedance of the coil****Voltage drop across each coil****The total power consumed by the circuit**

Answer:

Fig. 27

The circuit diagram of question 6(b) is given in Fig. 27.

Total impedance for coil 1 is \(Z_1=\left(10+j\times2\pi\times50\times0.02\right)=(10+j6.282)\ \mathrm{\Omega}\)

\(Z_1=\sqrt{{10}^2+{6.282}^2}\ \tan^{-1}{\frac{6.282}{10}}=11.81{(32.14)}^0\)

Total impedance for coil 2 is \(Z_2=\left(6+j\times2\pi\times50\times15\times{10}^{-3}\right)=(6+j4.71)\ \mathrm{\Omega}\)

\(Z_2=\sqrt{6^2+{4.71}^2}\ \tan^{-1}{\frac{4.71}{6}}=7.63{(38.13)}^0\)

Total impedance of the circuit is Z=Z_1+Z_2=\left(16+j11\right)\mathrm{\Omega}=19.4\ {(34.5)}^0\)

The current through the circuit is \(I=\frac{230{(0)}^0}{19.4\ {(34.5)}^0}=11.86{(-34.5)}^0\)

The current follow through the circuit is 11.86 Amp at {\(34.5}^0\) lagging the supply voltage.

The voltage drop across the coil 1 is \(11.86\times11.81\ {(-34.5+32.14)}^0=140\ {(-2.36)}^0\)

The voltage drop across the coil 2 is \(11.86\times7.63\ {(38.13-34.5)}^0=90.49\ {(3.63)}^0\)

The total apparent power consume by the circuit is \({11.86}^2\times19.4=2728.796\ Watts\)

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**A flux of 0.0006Wb is required in the air – gap of an iron ring of cross – section 5.0 cm2and mean length 2.7 m with an air – gap of 4.5 mm. Determine the ampere turns required. Six H values and corresponding B values are noted from the magnetization curve of iron and given below.**

## H(At/m) |
## 200 |
## 400 |
## 500 |
## 600 |
## 800 |
## 1000 |

## B(Wb/m2) |
## 0.4 |
## 0.8 |
## 1.0 |
## 1.09 |
## 1.17 |
## 1.19 |

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22 June, 2022

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