Question

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Answer 1

Given Question :

Calculate the net resistance between the two points a and b in the given circuit.

To Find :

Net Resistance of circuit

Formula used :

Resistance in series,

$\sf \: R = R_{1} + R_{2} + R_{3} + ......... R_{n}$

Resistance in parallel,

$\sf \: \frac{1}{R} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + .......... \frac{1}{R_{n}}$

Solution :

Refer to the attachment for better understanding.

Let in the first circuit,

The resistance 12 ohm and 6 ohm are connected in parallel, so

$\rightarrow \: \sf \: \frac{1}{R} = \frac{1}{12} + \frac{1}{6} \\ \\ \rightarrow \sf \: \frac{1}{R} = \frac{6 + 12}{12 \times 6} \\ \\ \rightarrow \sf \: \frac{1}{ R} = \frac{1}{4} \\ \\ \rightarrow \sf \: \boxed{R = 4 ohm.}$

Then, this 4 ohm and 5 ohm resistance gets connected in series, so

$\sf \: R \: = 4 + 5 \: ohm \\ \\ \sf \: R = 9 \: ohm$

Now, on solving the parallel series of resistance 4 ohm and 8 ohm,

$\rightarrow \sf \: \frac{1}{R} = \frac{1}{4} + \frac{1}{8} \\ \\ \rightarrow \sf \: \frac{1}{R} = \frac{8 + 4}{8 \times 4} \\ \\ \rightarrow \sf \: \frac{1}{R} = \frac{3}{8} \\ \\ \rightarrow \sf \boxed{R = 8/3}$

Now, 8/3 ohm resistance gets connected in series connection with 9 ohm, then the resistance will be

R = 9 + 8/3 ohm

R = 11.6 ohm.

Therefore, the total resistance of the given circuit is 11.6 ohm.

Answer 2

To find:-

Calculate the net resistance between point a and b

• Resistance in series

$\bigstar{\boxed{\sf { R_{series}= R_1+R_2+R_3....+R_n}}}$

• Resistance in parallel

$\bigstar{\boxed{\sf{\dfrac{1}{R_p}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}.....+\dfrac{1}{R_n}}}}$

Now Solve the first part

• This is connected in parallel

$\longmapsto\sf \dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}\\ \\ \longmapsto\sf \dfrac{1}{R}= \dfrac{1}{12}+\dfrac{1}{6}\\ \\\longmapsto\sf \dfrac{1}{R}=\dfrac{1+2}{12}\\ \\ \longmapsto\sf \dfrac{1}{R}=\cancel{\dfrac{3}{12}}\\ \\ \longmapsto\sf \dfrac{1}{R}=\dfrac{1}{4}\\ \\\boxed{\sf{R=4ohm}}-------(i)$

• And 5ohm is already in series -----(ii)

• Now the second part

$\longmapsto\sf \dfrac{1}{R}=\dfrac{1}{4}+\dfrac{1}{8}\\ \\\longmapsto\sf \dfrac{1}{R}=\dfrac{2+1}{8}\\ \\ \longmapsto\sf \dfrac{1}{R}=\dfrac{3}{8}\\ \\ \boxed{\sf{R=\dfrac{8}{3}ohm}}------(iii)$

Now the total resistance

$\longmapsto\sf R=R_i+R_{ii}+R_{iii}\\ \\ \longmapsto\sf R= 4+5+\dfrac{8}{3}\\ \\\longmapsto\sf R=\dfrac{12+15+8}{3}\\ \\\longmapsto\sf R=\cancel{\dfrac{35}{3}}\\ \\\longmapsto\sf R= 11.66...ohm$

$\bigstar{\boxed{\sf{Net \ resistance = 11.66ohm}}}$