Question #f7e9a

2 Answers
Apr 8, 2017

Answer:

Please See Below .

Explanation:

For determining the resistance of 'n' Resistors in parallel.We know that for two resistance in parrallel we have formulaenter image source here
#1/R_(Net)=1/R_1+1/R_2#
#R_(Net)=(R_1*R_2)/(R_1+R_2)#

Similarly
For 'n' such connetenter image source here
we can Say :-
#1/R_(Net)=1/R_1+1/R_2. . . . .1/R_n#
#R_(Net)=(R_1*R_2*R_3....R_n)/(R_2*R_3..R_n+R_1*R_3*R_4*R_n..)#

Apr 8, 2017

The diagram in your mind must be very clear ..
First we need to find out the Net resistance ..
which is
#1/R_(Net)=1/4+1/6+1/10#
#=>1.93 Omega#
Net current in the circuit is given by ohms law..
#V=I*R#
net Current =6.21 A
Power = #I^2R#
#(6.21)^2*1.93=P#
P=74.42 watt