In the Wheatstone's network given, P=10 , Q=20 , R=15 , S=30 the current have passing through the battery (of negligible internal resistance) is

Step-by-step solution

The balanced condition for Wheatstones bridge is P Q = R S as is obvious from the given values. No, current flows through galvanometer is zero. Now, P and R are in series, so Resistance R₁=P+R=10+15=25 Similarly, Q and S are in series, so Resistance R₂=R+S=20+30=50 Net resistance of the network as R₁ and R₂ are in parallel 1 R = 1 R₁ + 1 R₂ array ll & R= 25 50 25+50 = 50 3 Hence, & I= V R = 6 50 =0.36 ~A array