Question

calculate the equivalent resistance between two points A and b in the circuit shown in figure​

Answer 1

Step-by-step explanation:

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Answer 2

$\large\bf\underline{Question:-}$

Calculate the equivalent resistance between the points A and B in the circuits shown in the figures.

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$\large\bf\underline{Given:-}$

• Two figures are given :-
• In 1st figure Two resistors are in series and one in parallel.
• In 2nd figure Two resistors are in parallel and connected to series.

$\large\bf\underline {To \: find:-}$

• Equivalent resistance.

$\huge\bf\underline{Solution:-}$

In 1st figure :-

Firstly we find the equivalent resistance of resistors connected in series .

$\bf \: R_s = R_1 + R_2$

R1 = 6 , R2 = 6

Rs = 6 + 6

$\rm \: Rs = 12 \Omega$

Now we find equivalent resistance of resistors connected in parallel

$\bf \: \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2}$

Now,

R1 = 12 ohm

R2 = 6 ohm

$\rm \: \frac{1}{R_p} = \frac{1}{12} + \frac{1}{6} \\ \\ \rm \: \frac{1}{R_p} = \frac{1 + 2}{12} \\ \\ \rm \: \frac{1}{R_p} = \cancel{\frac{3}{12} } \\ \\ \rm \: \frac{1}{R_p} = \frac{1}{4} \\ \\ \rm \: {R_p} = 4 \Omega$

So, equivalent resistance between two points A and B is 4 ohm.

In 2nd figure :-

firstly we find the resistance in parallel combination

Resistors = 6 ohm ,6 ohm

$\bf \: \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2}$

$\rm \: \frac{1}{R_p} = \frac{1}{6} + \frac{1}{6} \\ \\ \rm \: \frac{1}{R_p} = \frac{2}{6} \\\\ \rm \: \frac{1}{R_p} = \frac{1}{3}\\ \\ \rm \: {R_p} = 3 \Omega$

For another parallel combination :-

$\rm \: \frac{1}{R_p} = \frac{1}{6} + \frac{1}{6} \\ \\ \rm \: \frac{1}{R_p} = \frac{2}{6} \\\\ \rm \: \frac{1}{R_p} = \frac{1}{3}\\ \\ \rm \: {R_p} = 3 \Omega$

Now, equivalent resistance between points A and B

In which resistors are connected in series:-

$\bf \: R_s = R_1 + R_2$

$\rm \: \: R_s = 3+ 3 \\ \\ \rm \: R_s = 6 \Omega$

Hence

equivalent resistance between points A and B in figure 1st is 4 ohm .

equivalent resistance between points A and B in figure 2nd is 6ohm .