A Wheatstone bridge consists of AB = 4 Ω, BC = 3 Ω, CD = 6 Ω and DA = 5 Ω, A 2.4 V battery is connected between points B and D. A galvanometer of 8 Ω resistances is connected between A and C. Using Thevenin’s theorems find the current through the galvanometer.
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Fig.28(a)
Total resistance of parallel pair \(\frac{9 \times 9}{ 9 + 9}=4.5\ \mathrm{\Omega}\)
Source current is \(\frac{2.4}{4.5}=0.53\ Amp\).
Current the branch BAD is \(\frac{2.4}{4.5 \times 2} = \frac{1.2}{4.5}=0.27\ Amp\).
Current through the branch BCD is \(\frac{2.4}{4.5 \times 2}=\frac{1.2}{4.5}=0.27\ Amp\).
Voltage drop across BA is \(\frac{4 \times 1.2}{4.5}=1.07\ V\)
Voltage drop across BC is \(\frac{3 \times 1.2}{4.5}=0.8\ V\)
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Fig. 28(b)
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Fig. 28(c)
Hence voltage across AC is \(V_{o.c}=1.07-0.8=0.27\ V\)
To find the Rth the circuit diagram may be redrawn as
Hence Rth =\(\left(8+2.2+2\right)=12.2\ \mathrm{\Omega}\)
The current through the galvanometer is \(\frac{0.27}{12.2}=0.02\ Amp\).
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 |
Derive the expression of quality factor of series R-L-C circuit at resonance.
Find VAB from the circuit if all the resistances are of same value of 1 Ω.
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Fig. 20(a)
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Fig. 18(a)