Question

find the equivalent resistance between a andb​

Answer 1

➤Given:-

• A circuit diagram is given to us .

➤To Find:-

• The net resistance in the series.

➤Formula Used:-

When n resistances are connected in series ,

$\large{\underline{\boxed{\red{\sf{\dag R_{net}=R_1+R_2+R_3+.......+R_n }}}}}$

When n resistances are in parallel ,

$\large{\underline{\boxed{\red{\sf{\dag \dfrac{1}{R_{net}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}+.......+\dfrac{1}{\R_n}}}}}}$

➤Calculation:-

[ For step by step simplified circuit diagram refer to attachment:]

Here in initial circuit we can see that 15Ω resistances and 10Ω resistances are in series . Let R is net resistance. So ,

➥ R = 15Ω + 10Ω .

➥ R = 25Ω .

Again we can see that the resistances of 15Ω and 20Ω are in series . Let R be the net resistance . So ,

➥ R = 15Ω + 20Ω .

➥ R = 35Ω .

Now , we can see that 35Ω resistances , 30Ω resistance and 25Ω resistance are in parallel . Let net resistance be R . So ,

➥ 1/R = 1/30Ω + 1/35Ω + 1/25Ω .

➥ 1/R = 35 + 30 + 42 / 1050Ω .

➥ 1/R = 107 /1050Ω.

➥ 1/R = 1 / 9.81 Ω .

➥ 1/R = 1 / 9.9 Ω [ 9.81 ≈ 9.9 ]

➥ R = 9.9Ω .

Hence the net resistance is 9.9Ω ≈ 10Ω .