Question and Answer

Added 2 years ago

Active

Viewed 1698

Basic Electrical Engineering Electrical Engineering

Ans

Equivalent resistance of the circuit $R_{t h} = \frac{10 \times 15}{10 + 15} + \frac{12 \times 16}{12 + 16} = 6 + 6.86 = 12.86 Ω$

Total current follow through the circuit $I = \frac{10}{12.86} = 0.78 A m p$ .

The current passing through the branch resistances $12 Ω$ and $16 Ω$ is $I_{1} = \frac{10}{28} = 0.357 A m p$ .

The current passing through the branch resistances $10 Ω$ and $15 Ω$ is $I_{2} = \frac{10}{25} = 0.4 A m p$ .

$V_{x} = 10 - 12 \times 0.36 = 10 - 4.286 = 5.714 A m p$ .

And $V_{y} = 10 - 0.4 \times 10 = 10 - 4 = 6 V$

Hence $V_{o . c} = V_{y} - V_{x} = 6 - 5.714 = 0.286 V$

Fig. 22(b)

Fig. (c)

Fig. 22(d)

Here the galvanometer resistance $R = 5 Ω$

The current through the galvanometer is $I_{L} = \frac{V_{o . c}}{R_{t h} + R} = \frac{0.286}{12.86 + 5} = 0.016 A m p$ .

Related Questions