# Question #f7e9a

Apr 8, 2017

#### Explanation:

For determining the resistance of 'n' Resistors in parallel.We know that for two resistance in parrallel we have formula
$\frac{1}{R} _ \left(N e t\right) = \frac{1}{R} _ 1 + \frac{1}{R} _ 2$
${R}_{N e t} = \frac{{R}_{1} \cdot {R}_{2}}{{R}_{1} + {R}_{2}}$

Similarly
For 'n' such connet
we can Say :-
$\frac{1}{R} _ \left(N e t\right) = \frac{1}{R} _ 1 + \frac{1}{R} _ 2. . . . \frac{.1}{R} _ n$
${R}_{N e t} = \frac{{R}_{1} \cdot {R}_{2} \cdot {R}_{3.} \ldots {R}_{n}}{{R}_{2} \cdot {R}_{3.} . {R}_{n} + {R}_{1} \cdot {R}_{3} \cdot {R}_{4} \cdot {R}_{n} . .}$

Apr 8, 2017

The diagram in your mind must be very clear ..
First we need to find out the Net resistance ..
which is
$\frac{1}{R} _ \left(N e t\right) = \frac{1}{4} + \frac{1}{6} + \frac{1}{10}$
$\implies 1.93 \Omega$
Net current in the circuit is given by ohms law..
$V = I \cdot R$
net Current =6.21 A
Power = ${I}^{2} R$
${\left(6.21\right)}^{2} \cdot 1.93 = P$
P=74.42 watt