Question

Using Kirchhoff's rules, calculate the current (Ig) that flows through the
galvanometer of resistance 15 in the circuit diagram shown in the
figure.
41022
100225
wa -
10 V​

Answer 1

Answer:

Value of current through galvanometer is given as

$i_g = \frac{11}{64} A$

Explanation:

Let the current in left loop is i1 in 10 ohm

now we will have

$10 i_1 + 15i_g - 60(i - i_1) = 0$

$14i_1 + 3i_g - 12i = 0$

for other loop we have

$10(i_1 - i_g) - 5(i - i_1 + i_g) - 15 i_g = 0$

$3i_1 - 6i_g - i = 0$

for lower side loop we have

$60(i - i_1) + 5 (i - i_1 + i_g) = 10$

$13i -13i_1 + i_g = 2$

so we have value of current through galvanometer is given as

$i_g = \frac{11}{64} A$

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Topic : Kirchoff's law

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