Question

Question #f7e9a

Answer 1

Please See Below .

Explanation:

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

Similarly
For 'n' such connet
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..)`

Answer 2

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