7
 A galvanometer of 15W resistance is connected across BD Calculate
the current through the galvanometer when a potential difference of
10 V is maintained across AC Solution  Considering the mesh BADB, we have
100I1 + 15Ig – 60I2 = 0
or   20I1 + 3Ig – 12I2= 0
[3 65(a)]
Considering the mesh BCDB, we have
10 (I1 – Ig) – 15Ig – 5 (I2 + Ig) = 0
10I1 – 30Ig –5I2 = 0
2I1 – 6Ig – I2 = 0
[3